Hilbert's Tenth Problem is Unsolvable
Author(s): Martin Davis
Source: The American Mathematical Monthly, Vol. 80, No. 3 (Mar., 1973), pp. 233-269
Published by: Mathematical Association of America
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HILBERT'S TENTH PROBLEM IS UNSOLVABLE
Science
ofMathematical
MARTIN DAVIS, CourantInstitute
would
solved,everymathematician
problemis finally
Whena longoutstanding
whathas been
forhimself
byfollowing
like to sharein thepleasureof discovery
of so muchofcontemporary
bytheabstruiseness
done.Buttoo oftenhe is stymied
The recentnegativesolutionto Hilbert'stenthproblemgivenby
mathematics.
In thisarticle,a complete
(cf. [23], [24]) is a happycounterexample.
Matiyasevic
a readerneedsto followthe
accountofthissolutionis given;theonlyknowledge
aboutdivisibility
basicinformation
theory:specifically
is a littlenumber
argument
(The materialin Chapter1 and the
of positiveintegersand linearcongruences.
threesectionsof Chapter2 of [25] morethansuffices.)
first
whichwilltellof a
algorithm
Hilbert'stenthproblemis to givea computing
or notit
whether
equationwithintegercoefficients
Diophantine
givenpolynomial
provedthatthereis nosuchalgorithm.
Matiyasevic
hasa solutioninintegers.
Hilbert'stenthproblemis thetenthin thefamouslistwhichHilbertgavein his
(cf. [18]). The
Congressof Mathematicians
1900addressbeforetheInternational
wayin whichtheproblemhas beenresolvedis verymuchin thespiritof Hilbert's
"thatevery
amongmathematicians
addressin whichhe spokeof the conviction
ofa precisesettlement,
be susceptible
mustnecessarily
problem
definite
mathematical
an actualanswerto thequestionasked,or by theproofof the
eitherintheform.of
proofs
ofitssolution..." (italicsadded). Concerningsuchimpossibility
impossibility
Hilbertcommented:
hypotheses
"Sometimes
it happensthatwe seekthesolutionunderunsatisfied
unableto reachourgoal. Then the
or in an inappropriate
senseand are therefore
of solvingthe problemunderthe given
task arisesof provingthe impossibility
proofswerealreadygiven
and in thesenserequired.Suchimpossibility
hypotheses
ofan isoscelesrighttriangle
e.g.,thatthehypotenuse
in showing,
bytheancients,
thequestionof theimposhasan irrational
ratiotoitsleg.In modernmathematics
we have acquiredthe
that
role,
so
a
key
played
of
has
solutions
sibility certain
as to provetheparallelaxiom,to
problems
thatsuchold and difficult
knowledge
degreeinradicalshaveno solution
squarethecircle,ortosolveequationsofthefifth
havebeensolvedin a preciseand
in theoriginally
sense,butnevertheless
intended
way."
satisfactory
completely
Ph.D. underAlonzoChurch.He hasheldpositionsat Univ.
MartinDavis receivedhisPrinceton
ofIllinois,IAS, Univ.of Calif.-Davis,Ohio StateUniv.,RensselaerPoly,YeshivaUniv.and New
College,London.He has done researchin various
York Univ.,and he spenta leaveat Westfield
and Unsolvability
of mathematics,
and is theauthorof Computability
aspectsof thefoundations
(editor,RavenPress,1965),Lectureson ModernMathematics
1958),TheUndecidable
(McGraw-Hill,
Analysis(Gordonand Breach,1967).
(Gordonand Breach,1967),and First Coursein Functional
Editor.
233
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234
[March
MARTIN DAVIS
is ofjustthischaracter.
problem
tenth
solutionofHilbert's
negative
Matiyasevic's
a "preciseand
but
rather
sense"
in
intended
It isnota solution Hilbert's"originally
ispossible.Themethods
needed
proofthatno suchsolution
satisfactory"
completely
ofalgorithms
hadnotbeendeveloped
to makeitpossibleto provethenon-existence
functions,
ofrecursive
(orcomputable)
arepartofthetheory
in 1900.Thesemethods
theory).
function
of recursive
developedbylogiciansmuchlater([6] is an exposition
function
is assumed.The
ofrecursive
theory
In thisarticleno previousknowledge
littlethatis neededis developedin thearticleitself.
existsfor
Whatwillbe provedin thebodyof thisarticleis thatno algorithm
or notit has
whether
to determine
withintegercoefficients
testinga polynomial
integersolutions).But
(Hilbertinquiredaboutarbitrary
positiveintegersolutions
forintegersolutions
thenit willfollowat once thattherecan be no algorithm
either.For one could testthe equation
P(Xl,
"'O
'Xn)
forpossessionof positivesolutions<x1, *,x,> bytesting
2
P(1 + p 2+ q1 + r2+ s1,1
+p
2
+q2 + rn+ S2)
=0
ofinteger
solutions<p1,q1,r1,s1,
s ** Pn qn rn,Sn>.Thisis because(by
forpossession
is the sum of four
integer
ofLagrange)everynon-negative
a well-known
theorem
is
exceeded!
Cf.
[17], p. 302.)In the
squares.(Justthisoncethestatedprerequisite
willbe dealt with-except whenthe
bodyof thisarticle,onlypositiveintegers
is explicitly
stated.
contrary
and ingenioussolutionin January
announcedhis beautiful
WhenMatiyasevic
decade
ofHilbert'stenthproblem
known
that
the
unsolvability
for
a
been
1970,ithad
a Diophantine
equationwhosesolutionswere
wouldfollowifone couldconstruct
withanotherof its
grewroughlyexponentially
suchthatone of its components
showedhowthe
Matiyasevic
(In ?9,thisis explainedmoreprecisely.)
components.
suchan equation.In thisarticlethe
couldbe usedto construct
Fibonaccinumbers
theaimhasratherbeento
ofthesubjectwillnotbefollowed;
development
historical
as seemscurrently
an accountofthemainresults
giveas smoothandstraightforward
feasible.A briefappendixgivesthehistory.
equations
Sets. In thisarticletheusualproblemof Diophantine
1. Diophantine
one
Insteadof beinggivenan equationand seekingits solutions,
willbe inverted.
equation.
Diophantine
willbeginwiththesetof"solutions"andseeka corresponding
More precisely:
is calledDiophantine
A setS oforderedn-tuples
ofpositive
integers
..
?
where
m
coefficients
0,
withinteger
ifthereis a polynomial
Ym),
P(xl, ...,9xny1,
suchthata given*n-tuple
<x1,* , xn> belongsto S ifand onlyifthereexistpositive
integersYi1*, ymfor which
DEFINITION.
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1973]
HILBERT'S TENTH PROBLEM IS UNSOLVABLE
235
P(XI*j," ,Xnj,Yl*...sYm) = 0-
Borrowingfromlogic the symbols"3" for"thereexists" and ".*."
for "if and
P canbewritten
thesetS andthepolynomial
between
onlyif",therelation
succinctly
as:
Y1i .., Ym)= O?]s
<XI, .* sXn>E S-(3yjj,-*,Ym) [P(xjj, * sXni
or equivalently:
S = {<xl
**">
Xn | (3 y, * ,Ym) [P(xl
31*X
nsYlq.. 9Ym)-?]}*-
casesalwayswill)havenegativecoefficients.
NotethatP may(andinnon-trivial
in thearticleexceptwhere
shouldalwaysbe so construed
The word"polynomial"
stated.Alsoall numbers
isexplicitly
thecontrary
inthisarticleare positiveintegers
is stated.
unlessthecontrary
The mainquestionwhichwillbe discussed(and settled)in thisarticleis:
of theeventual
A vagueparaphrase
answeris: any
WhichsetsareDiophantine?
is Diophantine.What does thephrase
set whichcouldpossiblybe Diophantine
mean?Andhowisall thisrelatedto Hilbert's
beDiophantine"
"whichcouldpossibly
Thesequitereasonablequestionswillonlybe answered
muchlater.
tenthproblem?
forshowing
thetaskwillbedeveloping
thatvarioussets
In themeantime,
techniques
are indeedDiophantine.
A fewverysimpleexamples:
whichare notpowersof2:
(i) thenumbers
y,z)[x = y(2z + 1)],
xeS..(3
numbers:
(ii) thecomposite
xeS.S*(3y,z) [x = (y + 1)(z + 1)],
thatis thesets{<x,y> x < y},
onthepositive
relation
(iii) theordering
integers;
{<x,y> Ix _y}:
x < y (3 z) (x + z = y),
x:!gy.(3z)
(x +z-
I =y),
(iv) thedivisibility
relation;thatis {<x,y>Ix I y}:
xj y..(3
z) (xz = y).
as othersetsto consider,
thesetofpowersof2 and
Examples(i) and(ii) suggest,
Asweshalleventually
ofprimesrespectively.
see,thesesetsareDiophantine;butthe
proofis not at all easy.
Anotherexample:
(v) theset W of<x,y,z> forwhichxj y and x < z: Here
xl y.*.(3u)
(y
=
xu) and x <z.*.(3v)
(z
=
x + v).
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236
[March
MARTINDAVIS
Hence,
<x, y,z> EW (3it, v)[(y-xu)2 + (z-x-v)2
=O].
a Diophantine
So, in defining
general.
justusedisperfectly
Notethatthetechnique
=
set one mayuse a simultaneoussystemP1 = O, P2 = 0, *,Pk 0 of polynomial
singleequation:
can be replacedbytheequivalent
equationssincethissystem
2
P1 +P22k+ o+ Pk2=0?
integer
ofoneormorepositive
valuedfunction
a positiveinteger
Bya "function"
will alwaysbe understood.
arguments
if
is calledDiophantine
A function
f of n arguments
DEFINITION.
{<XI,
..
xn,Y>l
y =f(xi,
.
,xX)
is a Diophantine set, (i.e., f is Diophantineif its "graph" is Diophantine).
areDiophantine?
functions
hereis: which
questionthatwillbeanswered
Another
is associatedwiththe triangularnumbers,
function
Diophantine
Animportant
of theform:
thatis numbers
T(n) = 1 + 2 + *. +n
- (n + 1)
2
z, thereis a unique
foreachpositiveinteger
function,
SinceT(n) is an increasing
n ? 0 suchthat
T(n) < z < T(n + 1) = T(n) +n+ 1.
Hence each z is uniquely representableas:
z=T(n)+y;
y<n+1,
as:
uniquelyrepresentable
or equivalently,
z = T(x + y -2) + y.
In this case, one writesx = L(z), y = R(z); also one sets
P(x,y) = T(x + y -2)
+ y - 1.
since
functions
NotethatL(z), R(z) and P(x,y) are Diophantine
z = P(x,y)
2z=(x+y-2)(x+y-1)+2y
x = L(z)
(3y)[2z=(x+y-2)(x+y-1)+2y]
y = R(z)
(3x) [2z = (x + y-2) (x + y-1) + 2y].
P(x; y) mapsthesetoforderedpairsof positiveintegersone-one
The function
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1973]
HILBERTS TENTH PROBLEMIS UNSOLVABLE
237
pairwhichis mappedinto
And,foreachz,theordered
integers.
ontothesetofpositive
z byP(x, y) is (L(z), R(z)). ("P" is for"pair", "L" for"left", and "R" for"right".)
Note also thatL(z) < z, R(z) < z. To summarize:
THEOREM 1.1 (PairingFunctionTheorem').There are Diophantinefunctions
P(x,y), L(z), R(z) such that
(1) for all x, y, L(P(x,y)) = x, R(P(x,y)) = y, and
(2) for all z, P(L(z),R(z)) = z, L(z) < z, R(z) ? z.
Theorem,
Remainder
usefulDiophantine
function
isrelatedtotheChinese
Another
statedbelow:
of moduli
sequence
DEFINITION. The numbers
mln**, mN arecalledanadmissible
prime.
if i #j impliesthatmi and mj are relatively
THEOREM 1.2 (ChineseRemainderTheorem).Let a,, *,aN be any positive
integersand let ml * mNbe an admissiblesequenceof moduli.Then thereis an x
such that:
x
a, modml
x
a2 mod M2
x
aN mod MN.
is provedforexamplein [25],p. 33.(Thatx can
theorem
TheChineseremainder
stated.Butsincetheproductof themoduli
be assumedpositiveis notordinarily
thisis obvious.)
solution,
addedto a solutiongivesanother
Now let the funetionS(i, u) be definedas follows:
S(i, u) = wI
forwhich:
wherew is theuniquepositiveinteger
w-L(u)
mod 1 + iR(u)
w < 1 + i R(u).
whenL(u) is dividedby1 + i R(u).
remainder
Herewissimply
theleastpositive
THEOREM 1.3 (Sequence Number Theorem). There is a Diophantinefunction
S(i, u) such that
(1) S(i, u) < u, and
(2) for each sequencea,, ***,aN, thereis a numberu such that
S(i,u) = a1 for 1 < i < N.
Proof. The firsttaskis to showthatS(i, u) as definedjust above, is a Diophantine
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238
MARTIN DAVIS
[March
ofequations
function.
Theclaimisthatw S(i,u) ifandonlyifthefollowing
system
has a solution:
2u
X
=
(x+y-2)(x+y-1)+2y
w + z(1 + iy)
1+iy -w+v-1.
Thisis because(by thediscussion
leadingto thePairingFunctionTheorem),the
to:
first
equationis equivalent
x = L(u) and y = R(u).
Then(usinga technique
alreadynoted)oneneedsonlysumthesquaresofthethree
equationsto see thatS(i,u) is Diophantine.
let al, *- aN be givennumbers.
Choosey to
Now S(i, u) < L(u) ? u. So finally,
be somenumber
thaneach of at, * aN and divisible
greater
byeachof 1,2,**-,N.
1 + y, 1 + 2y,**,1 + Ny are an admissible
Thenthenumbers
sequenceofmoduli.
(For, if dj 1 + iy and dfI +jy, i <I, then dj [j(1 + iy) - i(1 +jy)], i.e., di -i
so thatd < N; butthisis impossible
unlessd = 1 because dl y.) Thisbeingthecase,
canbe appliedto obtaina number
Theorem
x suchthat
theChineseRemainder
x3
a
x
a2 mod l+ 2y
x
aN
mod l+y
modl+Ny.
Let u =P(x,y), so thatx =L(u) and y = R(u). Then, for i =1,2, .. N
a L=- L(u)
mod 1 + iR(u)
and ai < y = R(u) < I + iR(u). Butthenby definition,
ai = S(i, u).
A striking
of Diophantine
characterization
setsof positiveintegers(cf.[26]) is
given by:
THEOREM
1.4. A setS ofpositiveintegersis Diophantineif and only if there
isa polynomial
P suchthatS isprecisely
thesetofpositive
in therangeofP.
integers
Proof. If S is relatedto P(x1,** , xm)as in thetheoremthen
x E-S *(3 xl, ^*,xm)[x = p(xl , xm)].
let
Conversely,
PS(x,
X1,XXm) x[
Q(x x1
Let P(x,xl *** xm)= x[l _ Q2(s,
Xm) Tn, i-XSX=
)].
XI ,.*.*, Xm
Then,i'fx E-S, chooseX,, **>xmsuch
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1973]
239
HILBERT S TENTH PROBLEM IS UNSOLVABLE
' -,xm) = x; so x is in the rangeof P. On the
thatQ(x,xl, ,Xm) = 0. Then P(x, xl, *
xm)mustvanish(otherwise
x,I
otherhand,ifz = P(x, x, *., Xm),z > 0, thenQ(x,x
1-_Q2<_0) so thatz = x and x eS.
2. Twenty-four
easylemmas.Thefirst
majortaskis to provethattheexponential
Thisis thehardestthingwe shallhaveto do.
function
h(n,k) = nk is Diophantine.
The proofis in ?3. In thissectionwe developthemethodswe shallneed,usingthe
so-calledPell equation:
where
X2 -dy2=1,
X,Y>O,
d =a21,
a >1
M(*
a self-contained
Although
thisis a famousequationwitha considerable
literature,2
the
obvious
solutions
to
(*):
Note
is given.
treatment
LEMMA
x= 1
y=0
x=a
y=l.
2.1. There are no integersx , y, positive,negative,or zero,whichsatisfy
(*)for which1 <x+yld<
a +Id.
Proof. Let x, y satisfy(*). Since
1 = (a + Vd)(a-Nd)
=
(x + yd)(x-yfyd),
the inequalityimplies(takingnegativereciprocals)-1 <
-
x + yld <
-
a+
Id.
0 < 2yI/d< 2 I/d,i.e., 0 < y < 1, a contradiction.
Addingtheinequalities:
LEMMA 2.2.
satisfy(*). Let
Let x,y and x',y' be integers,positive,negative,or zero which
X"+ Y
(x +yV)
(x' + y'Vd).
Then,x",y"satisfies(k).
Proof.Taking conjugates:x" - y"Id = (x - y Id) (x' - y'Id).
gives:
(x")2
DEFINITION. xn(a),
-
d(y")2
=
(x2 - dy2) ((X')2 - d(y')2)
=
Multiplying
1.
forn > 0, a > 1, by setting
y,(a) are defined
xn(a) + yn(a)/d = (a + Vd)n.
on a is notexplicitly
thedependence
Wherethecontext
permits,
shown,writing
XnL .sYn(
LEMMA2.3.
xn,,Yn
sati'sfy()
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240
[March
MARTIN DAVIS
Proof.Thisfollowsat oncebyinduction
usingLemma2.2.
LEMMA
solutionof(*). Then for some n, x =
2.4. Let x, y be a non-negative
xn,
Y =Yn.
Proof.To beginwithx + yVd > 1. Oj theotherhandthesequence(a +
Henceforsomen > 0O
increasesto infinity.
(a +
I
<d)"? x + yl/d< (a + 1ydY'+
theresultis proved;so supposeotherwise:
If thereis equality,
Xn + Yn,fdi < x + yV/d < (xn + Yn,/d) (a +
V-d).
Since (xn + ynV/d) (xn- yn Jd) = 1, the number xn- yn /d is positive. H4ence,
< a + >/d. But thiscontradicts
Lemmas2.1 and 2.2.
1 < (x + yld) (Xn-Yn
y,d)
The defining
relation:
Xn+ Yn,Id = (a + >d)"
formula:
is a formalanalogueof thefamiliar
(cosu) + (sinu)y1- 1 = eiU= (cos1+ (sin1)1
-
1)u,
theroleofsinandd playingtheroleof -1.
withxnplaying
theroleofcos,Ynplaying
identities
haveanaloguesin which-1 is replaced
Thus,thefamiliar
trigonometric
by d at appropriate
places.For examplethePell equationitself
x-
dY2
=
1
Nextanaloguesof thefamiliar
is just the analogueof the Pythagorean
identity.
are obtained.
additionformulas
LEMMA 2.5. xm?n= XmXn ? dYnymand Ym?n = XnYm? XmYn.
Proof.
Xm+n+Ym+nvJd
=
(a + Vd)m+n
=
(Xm + Ym V/d)(Xi + Yn>/d)
-
(Xmxn+ dYnYm) + (XnYm+ XmYn)Id.
Hence,
Similarly,
(Xm.n +
Xm+n =
XmXn + dYnYm
Ym+n =
XnYm+ XmYn.
Ym.n Id) (Xn+ YnId)
= Xm+ ym-/d. So
Xm_n+ Ym-nn/d = (Xm + ym
-/d) (xn
d)-
.as above.
and one proceeds
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1973]
LEMMA
HILBERT S TENTH PROBLEM IS UNSOLVABLE
241
2.6. Ym?i = a Ym? Xm,and xm?1 = ax, + dym.
Proof.Take n = 1 in Lemma2.5.
Thefamiliar
notation(x,y) is usedto symbolize
theg.c.d.ofx and y.
LEMMA
2.7. (x.,,yR)= 1.
Proof. IfdIx and djYn, then dfx 2-dy2 i.e., dj1.
LEMMA
2.8. YnJYnk
Proof.Thisis obviouswhenk = 1. Proceeding
byinduction,
usingtheaddition
formula(Lemma2.5),
Yn(m+1) = XnYnm+ XnmYn
By theinductionhypothesis
Ynj Ynm.Hence,YnIYn(m+1)LEMMA
2.9. Yn|IYtifand onlyifnj t.
Proof.Lemma2.8 givesthe implication
in one direction.For the converse
supposeYnjYtbutntt. So one can writet = nq + r, 0 < r < n. Then,
+ XnqYr.
Yt = XrYnq
= 1- (If df Yn,
Since (byLemma2.8) ynj Ynq,it followsthatYnJ
XnqYr.But (yn,xnq)
d Xnq,
byLemma2.7,impliesd = 1.) Henceynj Yr.
thenbyLemma2.$ dfYnqwhich,
But,sincer < n,wehaveYr< Yn(e.g.,byLemma2.6). Thisis a contradiction.
LEMMA2.10. Ynk k xyn
mod (yM,)3.
Proof.
(a + Idyk
V/d
Xnk+ Y,,k
=
(Xn+ YnQd)k
(
k-iy
x
?W
j=
nJ
Jdj12
n
So,
k
j(Jxkydjh-h)/2
Ynk =
j odd
forwhich]> I are
Butall termsofthisexpansion
LEMMA
2.11.
0 mod(yR)3.
O
Yn I YnYn
Proof. Set k = YR in Lemma 2.10.
LEMMA2.12. If YnI Yf,thenYn t.
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242
[March
MARTIN DAVIS
Proof. By Lemma 2.9, n I t. Set t = nk. Using Lemma 2.10, y2 k xn-7y., i.e.,
ynI kx'-71.But byLemma2.7,(Yn,
Xn)= 1. SO,Ynk and henceYnI t.
and Yn+I = 2ayn-yn.
LEMMA 2.13. xn+ = 2axn-xn_
Proof.ByLemma2.6,
Xn+ = axn+ dyn,
Ynl+ = aYn+ Xn
= aXn - dyn,
Yn-1 = aYn - Xn,
Xn
So, Xn+
Jr
Xn-I = 2axw Yn+l + Yn-l = 2ayn.
together
withtheinitialvaluesxo = 1,
equations,
Thesesecondorderdifference
x1 = a, Yo = 0, Yi = 1, determinethe values of all the Xn,Yn.Various propertiesof
themforn = 0, 1 andusingthese
thesesequencesareeasilyestablished
bychecking
show
for
fromits
difference
thattheproperty n + 1 can be inferred
equationsto
examplesfollow:
holdingforn and n - 1. Somesimple(butimportant)
LEMMA2.14. yn-n mod a
1.
-
usinga -1, mod
inductively,
Proof.For n = 0,1 equalityholds.Proceeding
a-1:
2ayn-Yn-y
Yn+1-
mod a-1.
2n-(n-1)
LEMMA2.15. If a _ b mod c, thenfor all n,
mod c.
yn(a) yj(b)
xn(a)-x.(b),
is an equality.
Proceeding
byinduction:
Proof.Againforn = 0,1 thecongruence
yn+1(a)
=
2ay,(a) - yl1(a)
-
2by.(b)-y,,-1(b)
mod c
= 7,.+ Il(b).
LEMMA2.16. Whenn is eveny,,is evenand whenn is odd y, is odd.
Proof Yn
+ = 2aYnYn - I Yn-, mod 2. So whenn is even,Yn Yo = 0 mod 2,
and when n is odd, Yn-Yi = 1 mod 2.
LEMMA
2.17. xn(a) - yn(a)(a
Proof. xo - yo(a
-
-
y) Y" mod 2ay _ y2
y) = 1 and xl
yl(a
-
_ 1.
y) = y, so theresultholds for n1= 0
-
and 1. UsingLemma2.13 and proceeding
by induction:
Xn+ - Yn+(a
-
y) = 2a[Xn
-
yn(a
-
Y)]
-
[Xn-
Yn-1(a
= 2ayn _ yn-I
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-
y)]
1973]
243
HILBERT'S TENTH PROBLEM IS UNSOLVABLE
=
y"nI(2ay-1)
=
yn-I
y2
yn+1
LEMMA
2.18. For all n, Yn+1
Proof. By Lemma 2.6, Yn+1
yn> n for all n.
LEMMA
>
>
Yn_ n.
Yn.Since yo = 0 > 0, it followsby inductionthat
2.19. For all n, xn+1(a) > x"(a) > an;
xn(a) < (2a)".
By Lemmas 2.6 and 2.13 a x"(a) < xn+ 1(a) < (2a)xn(a).
by induction.
Proof.
Next some periodicity properties of the sequence
LEMMA
2.20. X2"n=-
The result follows
Xk are obtained.
xj mod xn.
Proof. By the addition formulas (Lemma 2.5)
X2n j
=
Xnxn j + dYnYn j
-
dYn(YnXi ?XnYi)
-
dy 2xj
=
(x2- _)xj
-
LEMMA
2.21.
x; mod
X4"+J
mod xn
mod xn
- x;
mod xn.
xn.
Proof. By Lemma 2.20
x4n"j
LEMMA
n=1,
i=0
-X2n"
mod xn.
ijxX
n > 0. Then i=j,
2.22. Let xi- xj mod xn, i<J<2n,
and j=2.
unless a =2,
Proof. First suppose xn is odd and let q = (xn - 1)/2. Then the numbers
-q+2, ., -1,0 , 1, ,q - 1, q are a complete set of mutually
-q, -q+1,
incongruent
residues modulo
xn. Now by Lemma
1 =xo
Using Lemma 2.6, x - I<-
< X < ...
2.19,
<xn-1.
xnla < i xn; so xn_1 < q. Also by Lemma 2.20, the numbers
Xn+1,Xn+29 ...X2n-1,X2n
are congruent modulo
xn respectively to:
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244
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MARTIN DAVIS
moduloxn.Thisgives
incongruent
x0,x1,x2,**. , x2 are mutually
Thusthenumbers
the result.
Nextsupposexnis evenand let q = xn/2. In thiscase,it is thenumbers
- q +1,
- q +2, ... - 1,O,1,... ~q - 1,q
residuesmodulox".(For, - q = q
incongruent
setofmutually
whicharea complete
<
So
the
will
follow
as above,unless xn-1 = q
result
mod xn.)As above,xn1 q.
= x"/2, so thatx"+ --q mod x", in whichcase i = n-1, =n + 1 would
ourresult.But,byLemma2.6,
contradict
axn_l+
xn=
dYn- 1,
so thatxn= 2xn_1impliesa = 2 andyn-i= 0, i.e.,n = 1. So theresultcanfailonly
fora =2, n = 1 and i =0, j =2.
LEMMA
2.23. Let xj xi mod xn, n > O,O< i < n, O < j < 4n, then either
i or j =4n-i.
j=
Proof. Firstsupposej < 2n. ThenbyLemma2.22, = ia unlesstheexceptional
caseoccurs.Sincei > 0, thiscan onlyhappenifj = 0. Butthen
i = 2 > 1 = n.
letj > 2n and set j = 4n -j so 0 < j < 2n. By Lemma2.21,X _ Xj
Otherwise,
case ofLemma2.22occurs.Butthis
-xi modxn.Againj = i unlessthe exceptional
lastis out of thequestionbecausei, j > 0.
LEMMA
2.24. If O < i < n and xj
xi mod xn, thenj
+i mod4n.
Proof. Write]= 4nq + j, 0 _ j < 4n. ByLemma2.21,
X_Xj_X.
By Lemma2.23i=
j or i
=
4n - j. So, ]s
mod xn.
j-
+i mod4n.
ofDiophantine
equations:
function.
Considerthesystem
3. Theexponential
(I)
x2 - (a2-1)y2 = 1
(II)
u2 - (a2-1)v2 = 1
(III)
s2
-
(IV)
v
= ry2
(V)
b = 1+4py=a+qu
(VI)
s
= x+cu
(VII)
t
= k + 4(d-1)y
(b2-1)t2 = 1
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HILBERT S TENTH PROBLEM IS UNSOLVABLE
245
y = k+e-1.
(VIII)
Then it is possibleto prove:
3.1. For givena, x, k, a > 1, thesystemI-VIII has a solutionin the
remainingargumentsy,u, v,s, t,b, r,p, q, c, d, e ifand onlyifx = xk(a).
THEOREM
Proof.Firstlettherebe givena solutionofI-VIII.ByV, b > a > 1. ThenI, II,
III imply(byLemma2.4) thatthereare i,j, n > 0 suchthat
x = xi(a), y = yi(a), u = x"(a), v = y(a),
s = xj(b), t = yj(b).
ByIV, y < v so thati < n. V andVI yieldthecongruences
b-a
mod x"(a);
xj(b)-xi(a)
mod x"(a)
and by Lemma2.15 one getsalso
Thus,
mod x"(a).
xj(b)
xj(a)
xi(a)
xj(a) mod x"(a).
By Lemma2.24,
mod 4n.
ij?+i
(1)
Next,equationIV' yields
(y, (a))'
y(a).
y|
so thatbyLemma2.12,
yi(a)n 7
and (1) yields:
j -+
(2)
i mod 4yi(a).
By equationV
b -1
mod 4yi(a),
so by Lemma2.14,
(3)
yj(b)
j
yj(b)
k mod 4y,(a).
k-+
i mod 4y,(a).
mod 4yi(a).
By equationVII,
(4)
Combining
(2), (3), (4),
(5)
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246
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MARTIN DAVIS
Equation VIII yields
k < yi(a)
and by Lemma 2.18,
i < yi(a).
Since the numbers
- 2y+ 1, -2y +2,
,
1,0, 1,
2y
forma complete set of mutuallyincongruentresidues modulo 4y
inequalitiesshow that (5) impliesk = i. Hence
x ix(a)
=
4y,(a), these
Xk(a).
Conversely,let x = xk(a). Set y yk(a) so that I holds. Let m = 2kyk(a) and let
u = x"(a), v =yr(a). Then II is satisfied.By Lemmas2.9 and 2.11 y2' v. Hence one
can choose r satisfying
IV. Moreoverby Lemma 2.16, v is even so thatu is odd. By
Lemma2.7, (u,v) = 1. Hence (u,v 4y) = 1. (If p is a primedivisorof u and of4y,then
p jy because u is odd, and hence p Iv since y v.) So by the Chinese Remainder
Theorem(Theorem 1.2), one can findbo such that
bo-1
mod 4y
bo-a
mod u.
Since bo + 4juy will also satisfythesecongruences,b,p, q satisfying
V can be found.
III is satisfiedby settings = xk(b), t = yk(b). Since b > a, s = xk(b)> xk(a) = x.
By Lemma 2.15 (usingV), s -x mod u. So c can be chosen to satisfyVI. By Lemma
2.18, t> k and by Lemma 2.14, t= k mod b-I and hence usingV, t = k mod 4y.
So d can be chosen to satisfyVIT. By Lemma 2.18 again, y > k, so VIII can be
satisfiedby settinge = y - k + 1.
COROLLARY
3.2. The function
is Diophantine.
g(z,k) =
Xk(Z + 1)
Proof. Adjoin to the systemI-VIII:
a
(A)
=
z +1.
By thetheorem,thesystem(A), 1-VIII has a solutionifand onlyifx = xk(a) = g(z, k).
Thus a Diophantinedefinitionof g can be obtainedin the usual way by summing
the squares of 9 polynomials.
Now at last it is possibleto prove:
THEOREM
3.3. The exponentialfunction
h(n,k) =-nk
is Diophantine.
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HILBERT S TENTH PROBLEM IS UNSOLVABLE
247
First,a simpleinequality:
LEMMA
3.4. If a >
yk,
then2ay - y- - 1 >
yk.
Proof.Set g(y) =2ay - y2 - 1. Then (since a _2) g(1) =2a-2 > a. For
So g(y)> a for 1< y<a. Then for a>y y y,
1<y<a, g'(y)=2a-2y>0.
2ay _ y2 1 > a > yk
Now, adjoin to equationsI-VIII:
IX
(x-y(a-n)-m)2
X
m + g = 2an-n2 2
XI
w-n + h = k + l
XII
a2 -(w2-1)
-
=(f-1)2(2an-n2
1)2
(w-1)2z2 = 1.
at oncefrom:
Theorem3.3thenfollows
LEMMA 3.5. m =
ing argumen?ts.
nk
ifand onlyif equationsI-XII havea solutionin the remain-
Proof. Suppose I-XII hold. By XI, w > 1. Hence (w - 1)z > 0 and so by XII
a > 1. So Theorem3.1 appliesand it followsthatx = xk(a), y = yk(a). By IX and
Lemma2.17,
mod 2an-n2
m-nk
1.
XI yields
k,n <w.
By XII (usingLemma 2.4),forsomej, a = xj(w), (w - 1)z = yj(w). By Lemma 2.14,
j
0 mod w - 1
so thatj > w - 1. So by Lemma 2.19,
a ? ww1 > n
Now by X, m <
2an-n2 -
1, and by Lemma3.4,
nk< 2an-n2
_
1.
and bothlessthanthemodulus,theymustbe equal.
Sincem and nk are congruent
Conversely,
supposethatm = nk. Solutionsmustbe foundforI-XII.Chooseany
numberw such that w > n and w > k. Set a = xw_I(w)so that a > 1. By Lemma
2.14,
Yw-l(w)
0 mod w - 1.
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248
[March
MARTIN DAVIS
So one can write
(W) = z(w
Yw-
-1);
XI can be satisfied
thusXII is satisfied.
bysetting
h=w-n,
As before,a
> nk
I=w-k.
so thatagainbyLemma3.4,
m=
nk
<
2an -
1
n
and X can be satisfied.Settingx = xk(a), y = yk(a), Lemma 2.17 permitsone to
definef suchthat
(f- 1)(2an -n2
x - y(a - n) - m
so thatIX is satisfied.
Finally,I-VIII can be satisfied
byTheorem3.1.
ofDiophantine
Now thatit has beenprovedthatthe
4. Thelanguage
predicates.
is
other
exponentialfunction
functions
andsetscan be handled.
Diophantine,
many
As an example,let
h(u,v,w) = uv .
The claim is thath is a Diophantinefunction.For:
y
uvW
(3z)
(y =
z&
vw)
where"&" is the logician'ssymbolfor "and". UsingTheorem3.3, thereis a
polynomialP such that:
z~~~~~~~~~~~~
y = u4Z
(3rj, *,rn)[P(y, u,z, rl,
Z
=
Vw
3S
1,
*sSO
VI ,Ws
[P(Z,
r
I SSn
0,7n
]9
O=
]
Then,
y = uVW (3(z,ri,*,rn,si,
*
+
,sn) [Pp2(y,u,z, ri,.*,r)
P2(Z,
V,W, S1, *',
S)
=
0].
Now this procedureis perfectlygeneral: Expressionswhichare alreadyknown
setsmaybe combinedfreelyusingthelogicaloperations
to yieldDiophantine
of
"&" and "(3)"; theresulting
willagaindefinea Diophantine
set.(Such
expression
expressionsare sometimescalled Diophantinepredicates.)In this "language" it is
to use thelogician's"V" for"or", since:
also permissible
(3 rl, -,rn)[PI = O] V(OSI,
',Sm) EP2 = ?]
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HILBERT S TENTH PROBLEM IS UNSOLVABLE
249
Threeimportant
are givenby:
Diophantinefunctions
THEOREM
4.1. Thefollowingfunctionsare Diophantine:
(1)
f(n,k) =(
g(n) = n!
(2)
(3)
y
H (a + bk).
h(a, b,y) =
k=1
thefamiliar
thistheorem
notation[oc],wherea is a realnumber,
will
In proving
be usedto meantheuniqueintegersuchthat
[ac]? a < Lx]+ 1.
LEMMA
4.1. For 0 < k ! n, u > 2n
+
[(u
1)n/uk]
=
i= k
u
I',
Proof.
?(I
(u+1)n/uk=
.u
=S + R
where
S=
j
n-ikk
3
R
nI-
'7
=
( i- .)uik
k
and
ThenS is an integer
R < u '
i=0
(.
Ivi
i-
<
ui 1
i0O
-
u(I.
<
1.
+
(
1)n
So,
S ?
(u + 1)n/uk < S + I
whichgivesthe result.
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250
[March
MARTIN DAVIS
LEMMA
4.2. For 0 < k < n, u > 2 ,
[(U +
kn
1)nlUk]-
odu
ofthesumforwhichi > k aredivisiblebyu.
Proof. In Lemma4.1 all terms
LEMMA
4.3. f(n, k)
=
()
is Diophantine.
Proof. Since
=2n <u,
(k--)
Lemma 4.2 determines(n) as the unique positive integercongruentto
[(u + 1)n/uk]modulou and < u. Thus,
z-(
k
)
(3u, v,w) (v = 2n& u
>
v
&w = [(u + I)n!uk] &z
W
mod u&z<u).
to note that each of the above
To see that(k) is Diophantine,it then suffices
v=2" isofcourseDiophanpredicates;
by"&" areDiophantine
separated
expressions
u > v is of courseDiophantinesinceu > v
tineby Theorem3. The inequality
(3x)(u = v + x). Also,
z3w
mod u & z<u*(3x,y)(w=z+(x-1)u&u=z+y).
Finally
w = [(u
+
1)n /uk]
(3x,y,t) (t = u + 1 &x = tn&y =u k&w ?xIy < w + 1),
and w < x/y< w + 1 wy < x < (w + l)y.
LEMMA
4.4. If r > (2x)x+1 then
x=
[r (rx)l]
Proof. Let r > (2x)x+l. Then,
~~~rxx!
X/ r
x
=
r(r - 1) ..(r-x
= x! -,
4\
+1)
,
x
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HILBERT'S TENTH PROBLEM IS UNSOLVABLE
< xx
x
Now,
= 1+ x
1
1 x--+
r
r~
x
2
r
+
<1+-{1
r
2
r
~~~ ll
r
+ 3: +
+
r
2x
r
And,
r~~~ j
2x
+2x
2x
2
< 1+<
I +
=l r
(x)
2x x
r
So,
rxi(r)
< x! + <
+
x!
.x!2x
2x+1 xx+'
r
_
< X! + 1.
LEMMA
4.5. n! is a Diophantinefunction.
Proof: m = n! *
(3r,s,t,u,v) {s = 2x + 1& t = x + 1 &r
&u=
r&v =(r)
n
= St
& mv < u < (m + 1)v}.
LEMMA4.6. Let bq -a mod M. Then,
I (a + bk) bYy!(y Y)
mod M.
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251
252
[March
MARTINDAVIS
Proof
byY!
q+
= by(q+y) (q +y-l)-.
(q+l1)
(bq + yb) (bq + (y-1)b) *.. (bq + b)
=
3 (a + yb) (a + (y-I)b) **(a
LEMMA 4.7. h(a, b, y)
=
fkj=
+ b)
(mod M).
(a + bk) is a Diophantine
1
function.
Proof. In Lemma 4.6 choose M = b(a + by)' + 1. Thein, (M, b) = 1 and
bq _ a modM is solvableforq and then
fJ7k1 (a + bk). Hencethecongruence
as theuniquenumber
whichis congruent
moduloM
fk=1 (a + bk) is determined
M
>
to byy! q + Y) and is also < M. I.e.,
y
Z
HI
k=1
(a
p,q, r,s,t, u, v,w, x)
+ bk) (M,
r = a + by&s
&bq
= rY & M =
bs + 1
a + Mt & u- by&v = y! &z < M
& w =q?y
&x=
(w)&z+MP=uvx}
forthe exponential
forv = y! and for
function,
Usingthe previousexpressions
x -(), we obtaintheresult.
in Lemmas4.3,4.5,and 4.7.
ofTheorem4.1 is contained
The assertion
The languageof Diophantinepredicates
use of
permits
quantifiers.
5. Bounded
usedbylogiciansare:
&, V, and 3. Otheroperations
for"not"
(Vx) for"forall x"
for"if*.., then ..."
can lead to
as willbe clearlater,theuseofanyoftheseotheroperations
However,
Thereare also thebounded
whichdefinesetsthatarenotDiophantine.
expressions
existentialquantifiers:
,
'6(3y)s
"
which means "(3y) (y < x &
andtheboundeduniversalquantifiers:
"(Vy)x
X
whichmeans "(Vy) (y > xV
P)
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19731
253
HILBERT S TENTH PROBLEM IS UNSOLVABLE
It turnsoutthattheseoperations
maybe adjoinedto thelanguageof Diophantine
thatis, thesetsdefinedby expressions
of thisextendedlanguagewill
predicates;
stillbe Diophantine.
I.e.,
THEOREM
R
5.1. If P is a polynomial,
= {<y,x1, **,xn>|
Z,X1, **,Xn,Y1, * ,Ym) =
(3z)s5(3y1 **,Ym)[P(Y
0]}
and
S
{<y, XI, *,
xn> I(Vz)5(y(3y1
, Ym)[P(Y, z, xI, .*,
Xn,Y,1, , Ym)= OD}
thenR and S are Diophantine.
is trivial.Namely,
ThatR is Diophantine
<Y, XI,
,ym)(z < y&P = O).
IXn> c-R.t(3z, yl,
is farmorecomplicated.
The proofoftheotherhalfofthetheorem
LEMMA
5.1.
(Vk)gY(3y1,
,Ym) [I(Y,k, XI,
Ym) =
Y,,
,Xn,
0]
*I*Ym)-u[P(y,k,xl1,** , xn, Y I,* Ym)= ?]
(3u) (Vk):,Y(3yI,
Proof.The fightside of theequivalencetrivially
impliestheleftside. For the
converse,
supposetheleftsideistrueforgiveny,x1,* xn. Thenforeach k = 1,2, . ,y
thereare definitenumbers (k)**,
y(k)
forwhich:
P(y,k,x1 **,,(k)
,
, y(k))
0.
of the mynumbers
Takingu to be themaximum
{(k)|
m; k =
j = J, **,
1,2,**,y},
is likewisetrue.
it followsthattherightsideoftheequivalence
LEMMA 5.2.
(1)
Let Q(y,Ux1 ...X)
be a polynomialwiththe properties:
~~Q(Y'U,X1,*,Xn) > U, (2)
Q(Y,U,X1,*" Xn)> Y,
,
<! Q(Y,U,X1,***,Xn)*
IYm)|
(3) k < y and yl,**,Ym_ u imply|P(y,k,xl-,*,xn,Yl,
Then,
"' Xns
Y,I ..* ~Ym)= 0]
(Vk)5Sy(3yI *y,Y5)u [P(y, k,xl **
y
(3c,t,a,,
...$am)
[1 +ct=
k=1
(1 +
kt)
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254
[March
MARTIN DAVIS
u
f
I
&t = QU3o8eX13 ... 3,x)! & 1 + ct
&... &1 +Ct
f
j =1
(amJ-(j)
mod 1+ct].
*,am)-=O
&P(y,c,-x,,xn,al,
(a,l-j)
j=i
Thepointofthislemmais thatwhiletherightsideoftheequivalence
seemsthe
morecomplicated
ofthetwo,itis freeofboundeduniversal
quantifiers.
in the - direction:
Proof.Firsttheimplication
Foreachk = 1,2, **, y,letPkbea primefactorof 1 + kt.Lety,k) be theremainder
whena' is dividedbypk(k = 1,2,***,y;
i = 1,2,**,m).It willfollowthatforeach
k,i:
(a)
1< Yk) < u
(b)
P(y,k,x,
...xnyk)
mk))
-
0.
To demonstrate(a),
notethatPk i + kt,1 + kt| 1 + ct and 1 + ct f[J>
=1(a, -j). I.e.,
.~~~~~~~~~~~~~~~~~~~~~~~~~~~~
pkI1iJ=l(ai -j). Since Pk is a prime,Pkj ai -j forsome1j2,
j
*.*?u. That is
Since t = Q(y,u,x1,
,x)!,
mod Pk.
)
jaai=
(2) implies that every divisor of 1 + kt must be
> Q(y, u, xi, ***,xn").So Pk > Q(y,,u, xi, ..** x") and by (1), pk > u. Hence j<u
Sincey(k)is theremainderwhenai is dividedbyPk, also )(5c<Pk. So,
yik) = j
To demonstrate
(b), first
notethat
mod
1+ct=1+kt=0
Pk.
Hence
k+kct-=c+kct
i.e., k-c
mod Pk,
mod Pk. Wehavealreadyobtained
yk)-=ai
mod p,.
Thus,
Y(f k,x1,*
",y)-P(y,
c, xi
,
Xn.a,, X,
0 modpk.
Finally
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am)
<
Pk.
1973]
255
HILBERT'S TENTH PROBLEM IS UNSOLVABLE
,n,(
|P(y, k,xl,...Y
)| _
(Y(ksx)
This proves(b) and completesthe proofof the
To prove the => implication,let
P(Y'k>
=
<Pk..'
implication.
(k) ... Ymk)=
?~
for each k = 1,2, .*, t, whereeach yj(k)< u . We sett = Q(y,u,x1, x")!, and since
Hk= (1 + kt) =1 mod t, we can findc suchthat
...,
1+ Ct =
y
H
(1 + kt).
k= 1
Now, it is claimed that for 1 ? k < I< y,
(1 + kt,1 + It) = 1.
For, let p 11+ kt,p| l + It. Then P I-k, so p < y. But since Q(y,u,x1, ,xn) > y
thisimpliesp I t whichis impossible.Thus the numbers1 + kt forman admissible
sequence of moduli and the Chinese RemainderTheorem (Theorem 1.2) may be
applied to yield,foreach i, 1 < i < m, a numberai such that
a,
_yek)
modI + kt,
k-=1, 2, -*,y.
As above, k _ c mod I + kt. So
P(y c,x
-X,n aj, - am)
P(y, k,xl, **, X",
-
k)
...
y(k))
mod 1 +
kt,
0.
Since the numbers 1 + kt are relativelyprime in pairs and each divides
., x", a1, **, am) so does theirproduct. I.e.,
P(y, c,x1,
P(y, c, x
1, ..
x.,
xna
am) -0
mod 1 + ct.
Finally,
i.e.,
ai y=k) mod 1 + kt,
1 +kt az _
(k)
Since1 <y(k)<u
1kt|IIf
j=1
(ai-J)
Andagain sincethe 1 + kt'sare relativelyprimeto one another,
I+ ct
Hrl (ai -j).
u
j=1
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256
[March
MARTIN DAVIS
Nowitis easyto completetheproofofTheorem5.1using Lemmas5.1 and 5.2.
Firstfinda polynomial
Q satisfying
(1), (2), (3) of Lemma5.2. Thisis easyto do:
Write
N
k,xj,. sX.,Yi ..sYm) Ya tr
P(yu
r=1
whereeach trhas theform
tr=c'k
S=|cy
Xq,Xq,2.
qnyl
S2 .Sm
2
y
2
Set Ur =
forc an integer
positiveor negative.
Y
cya+bX 1XX2...xqnus1?S2+--+Sm
N
Q(Y'
=
sxn)
UsX,,..
u
+
Y
E
+
r=1
Ur.
Thus:
Then(1), (2), and(3) ofLemma5.2holdtrivially.
1,*,Ym)[P(ysk,x X'n*,Y"Y1,**,Ym)= ?
(Vk)<Y(3y
+ ct
(3u, c, t,a,***,am)
Y
k =1
,x n)!&1 +ctI
&t=Q(y,uxI X
u
& ...&1 + ctlHi
j=l
mod I + ct]
*a+ , ejh,g&1,
*, gm,h1, ,e hn )
,
&a
171(a1 -j)
j=1
(a,n-])
&P(Y'c,xj,-,x,,aj,-,am)--0
(hu,c,t,a
( + kt)
Y
t =f!
=iao-aU &e2=ya2-Uh
&gD
&e&4gm.am1u
u
& h,=
f
k=1
& .. &hm=
ti
(gI+ k) &h2
t
rI
k=1
=
fl
k=1
+k)
(92
k) &e| hj&e| h2&}&e|
(gmn+
& I= P(y' , xl,s,, x, a,,,,",an)
&e
hm
1
and this is Diophantine by Theorem 4.1.
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and let
1973]
HILBERT S TENTH PROBLEM IS UNSOLVABLE
257
6. Recursive
functions.
So farone trickafteranotherhas been used to show
thatvarioussetsare Diophantine.
Butnowverypowerful
methodsareavailable:it
turnsoutthatthe expandedversionof the languageof Diophantinepredicates,
theuseofboundedquantifiers
permitting
(sanctioned
byTheorem5.1)together
with
theSequenceNumber
Theorem
(Theorem
1.3)enablesonetoshowinquitea straightforward
waythatalmostanysetwe pleaseis Diophantine.
Someexamplesare in order:
(i) theset P of primenumbers:
x > I &(Vy,z)<x[yz <xv yz > x V y = 1 V z = 1].
xeP
oftheprimesis:
Another
Diophantine
definition
X GP-=x
> I &((x -1)! ,x) = I
x > 1 &(3y,z,u,v) [y = x -1 &z = y! &(uz
-
vx)2 = 1];
is themorenaturalone.
butthefirst
definition
From Theorem 1.4 it followsthat thereis a "prime-representing"
polynomial
is primeif and onlyif it is in the rangeof P. For an
P, i.e.,a positiveinteger
of sucha polynomial
explicitconstruction
P, cf. [23a].
(ii) thefunctiong(y) = ,7ly=1 (I + k2). Here we use the SequenceNumber
Theoremto "encode" the sequence g(1), g(2),
.
so that
S(i,u) = g(i),
,g(y) into a singlenumberu, i.e.,
i = 1,2, .,y.
Thus, z = g(y)
>(3u)
{S(1,u)
=
2 &(Vk)<Y[k = 1 V(S(k,u)
(3u) {S(l,u)
& b-S(a,
=
=
(1 + k2)S(k - 1,u))] & z = S(y,u)}
2 & (Vk)?y[k = 1 V (3a,b,c) (a
u) & c = S(k, u) & c = (1 + k2)b)] & z
k- 1
S(y,-u)}.
By now it is clearthattheavailablemethodsare quitegeneral.Theyare so
thatthe questionbecomes:how can any "reasonable"set or function
powerful
escapethesemethods,
i.e.,notbe Diophantine?
Thestrength
canbe testedbyconsidering
ofthemethods
theclass ofall compuTheseare thefunctions
whichcan be computed
functions.
tableor recursive
bya
or
machinehavingarbitrarily
finite
program computing
largeamountsof timeand
ofthisclass(all ofthemequivalent)
definitions
atitsdisposal.Manyrigorous
memory
are available.One of thesimplest
is as follows:
The recursive functions3 are all those functionsobtainable from the initial
functions
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258
[March
MARTIN DAVIS
c(x) = ,s(X) =x + 1; U"(x1',
**x)
i
t < i <n;
S(i,u) (Thesequencenumber
function)4
iterativelyapplyingthe three operations: composition,primitiverecursion,and
minimalizationdefinedbelow:
COMPOSITION
yieldsthefunction
,
h(X 1, , "XJ)fg
1(x 1,* ,X.),*
g(x,
,X.))
fromthe given functionsg1, , gn and f(t1,* ,r, .)
PRIMITIVE RECURSION
equations:
yields the functionh(xl,
h(xl,
x., t) =f(xl,
**,x",z)
whichsatisfiesthe
**x.)
h(xl,** x" t + 1) = g(t,h(x1,***,Xn,t) xl, ** ,xJ,
romthegivenfunctions
f, g.
Whenn = 0,f becomesa constant
so thath is obtaineddirectly
fromg.
MINIMALIZATION
yieldsthefunction:
h (x 1, *, xj=
mi'n,U(x
x,y)
1, *.*1 X,
= g(X,
XnY)]
X,
fromthegivenfunctionsf,
g assumingthatf,g are such that foreach x1,
,x. there
. ,x",y) = g(x1,.,x",y); (i.e.,h must
is at leastoney satisfying
theequationf(x1,
be everywhere
defined).
The mainresultof thisarticleis:
TEILOREM6.1. AfunctionisDDiophantine
ifand onlyifit is recursive.
To beginwith,consider
thefollowing
shortlistofrecursive
functions:
since
(1) x + y is recursive
x + I = s(x),
x + (t + 1) = s(x + t) = g(t,x + t,x),
whereg(u, V,w) = S(U3(U, V,W)).
since
(2) x *y is recursive
x l
x (t+ 1)=(x
whereg(u, v,w) =
=
U1(x)
t)+x=g(t,x
tx),
U3(u, V,W) + U3(u, V,W).
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1973]
259
HILBERT'S TENTH PROBLEM IS UNSOLVABLE
(3) For eachfixedk,theconstant
function
Ck(X) = k is recursive,
since c1(x) is
and Ck+I(X) = Ck(X) + c(x).
one of theinitialfunctions
(4) Any polynomialP(x1,
***,
x") with positiveintegercoefficients
is recursive,
of additionsand mulcan be expressed
bya finite
iteration
sinceanysuchfunction
of variablesand c(x). E.g.,
tiplications
2x2y+ 3xz3 + 5 =
C2(X) * X * X * y + C3(X) * X * Z * Z * Z + C5(X).
So (1), (2), (3), and composition
givestheresult.
is recursive:
Nowit is easyto see thateveryDiophantine
function
Letf be Diophantine,
and write:
y =f(xj,
*w,x.)<*(3tls
*ss tm) [P(X1,
=
Q(XD,
..,Xng
*' * Xn
Y,
Y9 tIg
tl,
..
...
9 tm)]
tm)
9
withpositiveinteger
coefficients.
whereP, Q arepolynomials
Then,by thesequence
numbertheorem:
f(x1, x..,
x)
= S(1, min,,[P(x1, X.,
x",S(1, u), S(2, u), ... , S(m + 1,u))
= Q(xl, ..., x,, S(1, u), S(2, u), ..., S(m + 1,u))]).
andminimalization).
so isf (usingcomposition
SinceP, Q, S(i, u) arerecursive,
To obtaintheconverse:S(i, u) is knownto be Diophantine;the otherinitial
to provethattheDiophantine
Henceit suffices
aretrivially
functions
Diophantine.
are closed undercomposition,
recursion
and minimalization.
functions
primitive
Composition:If h(xl,**,Xn)=w(gj(xj9**Ixn), **
9m
g are Diophantine,thenso is h since
..
y = h(xl **, Xn) ->(3t19 ** tm) [tj
& tm
I
g(XI,*
,
gm(Xlx*
* , X.)),
m(x
where f,g ,
...
Xn) &
Xn) & y =f(tl,
... tm)]
PrimitiveRecursion: If
h(xls
*-*9Xn,1)
= f(XI,
-qssXn)
h(xl **,Xn, t + 1) = g(t,h(xl, **X*,n t)9xl
...
,Xn),
to "code" the
then(usingthesequencenumber
theorem
andf,g are Diophantine,
numbers
h(xl, ... ,xn,1), h(xl
Xn,2),*,h(xj,***,x z):
y = h (x, I 9..,
Xn9 Z) <
(3u) {(3v) [v) = S(, u) & v =f(x1,I,
(Vt) z [(t=z)
xn)]
V (3v) (v = S(t + 1,u)
& v = g(t, S(t,u), xI, ..,x))]
& y = S(z, u)}
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260
[March
MARTIN DAVIS
so that(usingTheorem5.1) h is Diophantine.
Minimalization:If
h(x1,
--x
= g(xl,
mnU[(x1, j..,y)
...,x" y)],
wheref,g are Diophantine,thenso is h since,
y=
h(x-,X*>
(3Iz)[z
=
---Xn,Y)&Z
=A(XI,
XXI,
**?,Xn-Y)]
? (Vt):5y
[(t = Y) V (3u,v) (u f(x,, ... 9Xn9t)
&v=g(xl,..X
ot)&(
<
v Vv< u)].
of all theDiophantine
set.An explicitenumeration
7. A universal
Diophantine
withpositiveinteger
willnowbe described.
Anypolynomial
setsofpositiveintegers
canbebuiltupfrom1 andvariables
additions
and multiplicabysuccessi've
coefficients
tions.We fixthealphabet
Xo,X1,X2,X3,'
of all such polynomials
enumeration
of variablesand thenset up thefollowing
(usingthepairingfunctions):
P1I
=
P3i1
=
P3
- PL(i) + PR(i)
P3i+ I
WritePi = Pi(x0, x,
*
x),
XiPL(i) PR(i)
wheren is largeenoughso thatall variablesoccurring
in Pi are included.(Of coursePi will notin generaldepend on all of thesevariables.)
Finally,let
n= {xoj (3x1, ...,x") [PL(fn)(X0o * I,
Xn)
l
PR(n)(x0 X 1
.,xX)}.
Here,PL(,)andPR(n)do notactuallyinvolveall ofthevariablesxo, x1,
xn-but
I
clearlycannotinvolveanyothers.(RecallthatL(n), R(n) ? n.) Bythe way the seitis seenthatthesequenceofsets:
quencePi hasbeenconstructed,
DI1, 23,D39D4, *
includesall Diophantinesets. Moreover:
7.1 (Universality
THEOREM
Theorem5).
{<n, x> I x Dn} is Diophantine.
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1973]
261
S TENTHPROBLEM
IS UNSOLVABLE
HILBERT
it is claimedthat:
theorem,
Proof.Onceagainusingthesequencenumber
xe Dn. (3u) {S(1, u) = 1 & S(2, u) = x
& (Vi):5[S(3i, u) = S(L(i), u) + S(R(i), u)]
&(Vi)sn[S(3i
+ 1,u) = S(L(i), u) S(R(i), u)]
& S(L(n), u) = S(R(n), u)}.
side of thisequivalence
is
It is clearenoughthatthepredicateon theright-hand
to verify
theclaim:
so itis onlynecessary
Diophantine,
Let x E Dn for given x, n. Then there are numberst1,***, tn such that
number
theorem)
PL(n) (X, tl, ***, tn) = QL(n) (X, tl, **., ta). Chooseu (bythesequence
so that
(*)
j =
u) =Pj(x1, t,I *. , tn)3
~~~~SOS,
3n + 2.
,2
Then in particularS(2, u) = x and S(3i - 1,u) = ti-1, i = 2,3,
right-hand
side of theequivalenceis true.
sideholdforgivenn,x. Set
lettheright-hand
Conversely,
*,n
+ 1. Thus the
t, = S(5, u), t2 = S(8, u), ... , t = S(3n + 2, u).
Then, (*) mustbe true.Since S(L(n), u) = S(R(n), u), it mustbe the case that
PL(n)(X4 t3, *s*,tn) = pR(n)(X3, tI
.
tn)9
so thatxE Dn*
of all Diophantine
sets,it is easyto
SinceDI, D2,D3, .*', givesan enumeration
Thatis,define:
from
all ofthemandhencenon-Diophantine.
construct
a setdifferent
f
v = {n n
Dnl}
THEOREM7.2. V is notDiophantine.
Proof. This is a simple application of Cantor's diagonal method. If V were
Diophantine,thenforsome fixedi, V = Di. Does i E V? We have:
i c-V.>i c-Di;
i e-V
i 0 Di.
This is a contradiction.
THEOREM 7.3. Thefunctiong(n,x) definedby:
g(n,x)=1
if xq Dn,
g(n,x)=2
if xeDn,
is not recursive.
(Theorem6.1),say:
thenit wouldbe Diophantine
Proof.If g wererecursive
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262
[March
MARTIN DAVIS
y = g(n,x) *(3y1,* *,Ym)[P(n, xy,yl
Y,Y,**,Ym) =0?]
Butthen,it wouldfollowthat
V-={x I (3y,,**, Ym)[P(x,x,11Yll,... Ym)= ?]}
Theorem7.2.
whichcontradicts
UsingTheorem7.1,write:
"' Zk) [P(n,x,z,
Xc-Dn*(3zl **
Zk) =
O].
whereP is somedefinite
(thoughcomplicated)
polynomial.
Supposetherewerean
algorithmfor testingDiophantineequationsfor possessionof positiveinteger
solutions;i.e., an algorithmfor Hilbert's tenthproblem!Then forgiven n, x this
ornottheequation
couldbe usedtotestwhether
algorithm
P(n, x,ZI, -
z Zk)
=
has a solution,i.e., whether
or not x E Dn. Thusthe algorithm
could be used to
compute
thefunction
g(n,x). Sincetherecursive
functions
arejustthoseforwhicha
computing
algorithm
exists,g wouldhaveto be recursive.
This would contradict
Theorem7.3, and thiscontradiction
proves:
THEOREM7.4. Hilbert's tenthproblemis unsolvable!
Naturally
thisresultgivesno information
aboutthe existenceof solutionsfor
anyspecificDiophantineequation;it merelyguarantees
thatthereis no single
algorithm
fortesting
theclassofall Diophantine
equations.
Alsonotethat:
X E-v
- (3z,,
J(3z
(VZ1,**
**' Zk)
", ,ZO
[P(X,
X, ZI, ", ZJ) = 0]
[P(X, X, Z1s
Zk) [P(X, X, Z 1,
...
Zk) = 01 -+ I = 0}
9)Zk > ?
V P(x, X, Z, ..., Zk) < 0]
whichshowsthatifeither- orunbounded
universal
quantifiers
(Vz)or implication
( ) arepermitted
inthelanguage
ofDiophantine
thennon-Diophantine
predicates,
setswill be produced.
It is naturalto associatewitheach Diophantine
seta dimension
and a degree;
i.e.,thedimension
ofS is theleastn forwhicha polynomial
P existsforwhich:
(*)
S-{xl (3y1, Yn) [P(X,Y
Yn = ?]}
andthedegreeofS is theleastdegreeofa polynomial
P satisfying
n
(*) (permitting
to be as largeas onelikes).Nowitis easyto see:
THEOREM
7.5. EveryDiophantineset has degree ? 4.
Proof The degreeofP satisfying
(*) maybe reducedbyintroducing
additional
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1973]
HILBERT'S TENTH PROBLEM IS UNSOLVABLE
263
variablesZj satisfying
equationsof theforim
YiYk
Zi
zj zi= Yi
Zj =
zj
=
XYi
x
ofthezj's intoP itsdegreecan be broughtdownto 2.
substitutions
Bysuccessive
equationseach of
Hencetheequationis equivalentto a systemof simultaneous
thesquaresgivesan equationofdegree4.
degree2. Summing
A lesstrivial(andmoresurprising)
factis:
THEOREM7.6. There is an integerm such that every Diophantine set has
dimension< m.
Proof. Write
Dn= {xl (3YI,.. Ym)[P(x n,yl, .. IYm) = 0]}s
of Dnis <m for
theorem.Thenthe dimension
whichis possiblebytheuniversality
all n.
sets:
Aninteresting
exampleis givenbythesequenceofDiophantine
Sq = {X I (3y1 , yq) [X = (Yt + 1) .. (Yq + 1)]}.
numbers.
HereS2isthesetofcomposite
numbers;
Sq is thesetof"q-fold"composite
definition
of Sq (for
thatit is possibleto givea Diophantine
It is surelysurprising
large q) requiringfewerthan q parameters(cf. [19]).
in the universalDiophantineset?
How largeis m, thenumberof parameters
A directcalculation
usingthearguments
givenherewouldyielda numberaround50.
and JuliaRobinsonhave veryrecently
shownthat m = 14
ActuallyMatiyacevic
will suffice!
ofHilbert's
Theunsolvability
tenthproblem
canbe usedto obtaina strengthened
theorem:
formof Godel's famousincompleteness
Correspondingto any given axiomatizationof numbertheory,
thereis a Diophantineequation whichhas no positiveintegersolutions,but such
that thisfact cannot be provedwithinthegivenaxiomatization.
THEOREM 7.7.
of"axiomatization
of number
A rigorous
proofwouldinvolvea precisedefinition
heuristic
argument
theory"whichis outsidethescopeof thisarticle.An informal
follows:
all ofthetheorems
to systematically
generate
One usesthegivenaxiomatization
willbe someasserting
of the axioms).Amongthesetheorems
(i.e., consequences
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264
[March
MARTIN DAVIS
suchis encountered
thatsomeDiophantine
equationhas no solution.Whenever
it
on
a
list
called
LISTA.
At
thesametimea list,LIST B, is made of
isplaced
special
Diophantineequationswhichhavesolutions.LIST B is constructed
by a search
e.g.,atthenthstageofthesearchlookatthefirst
procedure,
nDiophantine
equations
inwhicheachargument
is < n. Thus every
(in a suitablelist)andtestforsolutions
willeventually
Diophantine
equationwhichhaspositive
solutions
integer
beplacedin
LIST B. If likewiseeachDiophantine
equationwithno solutionswouldeventually
forHilbert'stenthproblem.
appearin LIST A, thenone wouldhavean algorithm
ofa solutionsimplybegingenerating
Namely,
totesta givenequationforpossession
LIST A andLIST B untilthegivenequationappearsin onelistor the other.Since
some equationwithno solutionmustbe
Hilbert'stenthproblemis unsolvable,
fromLIST A. Butthisis justtheassertion
ofthetheorem.
omitted
sets. It is nowtimeto settlethe questionraisedat the
8. Recursively
enumerable
which
sets
are
beginning:
Diophantine?
DEFINITION. 8.1. A set S of n-tuplesof positiveintegersis called recursively
enumnerable
if thereare recursive
functionsf(x,x1,
1 ., x), g(x,gx1. -qXn) suchthat:
S
=
{<xl, ..*sxn>I(3x)
[AX,Xi,
xn)=g(x,x
-,
xJ)]}
THEOREM 8.1. A set S isDiophantineifand only if it is recursively
enumerable.
Proof.If S is Diophantinethereare polynomials
P, Q withpositivecoefficients
suchthat:
<x
4
x,, >c-S(3y,1,
(3u)[P(x1,... x ,
-,YM) [P(X I X..
",yi,"
Ym) -= Q(X i .. ~Xn Yi1.., Ym)]
**S(m,
SO1 u), u..,S(M,U)) = Q(X* ,* xu),
SO
)]
enumerable.
so thatS is recursively
if S is recursively
enumerablethere are recursivefunctions
Conversely
such
that
f(x,x1, .. ,x"), g(x,xI,--,xn)
<xi,
Xn>
x">S.*S (3x) [J(x,1,
XI,.
xn) =
g(x, x1,
.* (3x,z) [z =Xf(x,x1,.*X,xn)&z
Xn)]
= (X,x,
,x")].n
Thus by Theorem6.1, S is Diophantine.
The presentexpositionhas ignoredthe chronological
9. Historical
appendix.
orderin whichtheideas weredeveloped.The first
contribution
was by Godel in
his celebrated1931paper[16]. The mainpointof Godel's investigation
was the
in formalsystems.
ofundecidable
The undecidablestatements
existence
statements
and in orderto exhibitthesimple
Godel obtainedinvolvedrecursive
functions,
character
number-theoretic
of thesestatements,
Godel usedtheChineseremainder
to reducethemto "arithmetic"
form.The techniqueused is justwhatis
theorem
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1973]
265
HI'LBERT'S TENTH PROB3LEMIS UNSOLVABLE
usedhereinproving
Theorem1.3(thesequencenumber
theorem)and Theorem6.1
(in the direction:everyrecursive
function
is Diophantine).Howeverwithout
the
fordealingwithboundeduniversal
as discussedinthispaper,
techniques
quantifiers
the bestresultyieldedby Godel's methodsis thateveryrecursive
function(and
indeedeveryrecursively
enumerable
set)can be defined
bya Diophantine
equation
precededbya finite
number
ofexistential
andboundeduniversal
quantifiers6.
In my
doctoraldissertation
(cf.[5], [6]), I showedthatall butoneoftheboundeduniversal
so thateveryrecursively
couldbe eliminated,
enumerable
setS couldbe
quantifiers
definedas
S = {x I (3y) (Vk)<y(3y1 " Ym)[P(k, x, y,y1, ,Ym)= O]}.
Thisrepresentation
becameknownas theDavisnormal
form.(LaterR. M. Robinson
[31], [32] showedthatin thisnormalformone couldtakem = 4. More recently
Matiyacevic
has shownthatone can eventakem= 2. It is knownthatone cannot
one can alwaysgetm = 1 is open.)
alwayshavem= 0; whether
ofmyworkandat aboutthesametime,JuliaRobinsonbeganher
Independent
centeredaboutthe question:
study[27] of Diophantinesets.Her investigations
Is theexponentialfunctionDiophantine?The mainresultwas thata certainhypoth-
Thehypothesis,
esis impliedthattheexponential
was Diophantine.
which
function
becameknownas theJuliaRobinsonhypothesis,
has playeda keyrolein workon
is simply:
Hilbert'stenthproblem.Its statement
There existsa Diophantineset D such that:
(1) <u, v> EDD impliesv ? u'.
(2) For each k, thereis <u, v> E D such thatV >
Uk.
Thehypothesis
an openquestionforabout2 decades.(Actually
remained
theset
D
=
{<u,v> Iv=xx(2)&u>3}
satisfies
(1) and(2) byLemma2.19andis Diophantine
byCorollary
3.2, so thetruth
followsat once fromthe resultsin thisarticle.)
of JuliaRobinson'shypothesis
is
function
JuliaRobinson'sproofthatthishypothesis
impliesthattheexponential
is
usedthePell equation.And,theproofthattheexponential
Diophantine
function
indeedDiophantine
givenhereis closelyrelatedto a morerecentproof[28] by
herof thissameimplication.
whichwereexponential
In [27], JuliaRobinsonstudiedalso setsand functions
definablein termsof exponentiation)
thatis which
Diophantine(or existentially
oftheform:
possessdefinitions
(3u,,
..X
..
..
Ung VI, * * Vng Wl, ** Wn) EP(X 1
&u1
=vI
'&.
**.*,XWsU19
&u"
=
-
'
Un1V19-,** 1Vni WD
-9%*W)
=
O
Vln].
In particular,
the functions
(k) and n! were shown by her to be exponential
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266
MARTIN
DAVIS
[March
Diophantine.
This is reallywhatis shownin proving(1) and (2) of Theorem4.1.
The present
proofof (2) is justhers;theproofof(1) givenhereis a simplified
variantof thatin [27]. (It is due independently
to JuliaRobinsonand Matiyasevic.)
Theidea ofusingtheChineseremainder
theorem
to codetheeffect
ofa bounded
in theworkofmyself
universal
quantifier
first
occurred
andPutnam[7]. In [8], we
refined
ourmethods
andwereable to show,beginning
withtheDavis normalform,
thatIF thereare arbitrarily
long arithmetic
progressions
consistingentirelyof
primes
(stillan openquestion),
theneveryrecursively
enumerable
set is exponential
Diophantine.In our proofwe needed to establishthath(a, b,y) = fly= (a + bk) is
whichwe did extending
exponential
Diophantine,
JuliaRobinson'smethods.
(The
proofgivenhereof (3) ofTheorem4.1 is a muchsimplified
argument
foundmuch
laterbyJuliaRobinson-cf.[29].)JuliaRobinson
thenshowedfirst
howto eliminate
the hypothesis
about primesin arithmetic
then
how to greatly
and
progression,
theproofalongthelinesofLemma5.2ofthisarticle.Thusweobtainedthe
simplify
theorem
of[9] thateveryrecursively
setis exponential
enumerable
Diophantine.
Attention
was nowfocusedon theJuliaRobinsonhypothesis
sinceit was plain
thatit wouldimplythatHilbert'stenthproblemwas unsolvable.
werefoundto implytheJuliaRobinsonhypothManyintelesting
propositions
seemedimplausible
esis.7. Howeverthe hypothesis
to many,especially
becauseit
andsurprising
wasrealizedthatan immediate
wouldbe theexistence
consequence
of
an absoluteupperboundforthedimensions
ofDiophantine
sets(cf.Theorem7.6).
Thus in his review[19] Kreiselsaid concerningtheresultsof [9]: "... it is likelythe
present
resultis notcloselyconnected
withHilbert'stenthproblem.Alsoit is not
are uniformly
altogether
plausiblethatall (ordinary)
reducible
Diophantine
problems
ofvariablesoffixeddegree...."
to thoseina fixednumber
The JuliaRobinsonhypothesis
was finally
provedby Matiyasevic
[23], [24].
he
showed
if
we
define
that
Specifically
d-=a2
= 1,
an+1 = an + an-1
so thatan is thenthFibonaccinumber,
thenthefunction
Then
a2n is diophantine.
since,forn > 3,as is easilyseenbyinduction,
)<
an <2n2
the set
D = {<u,v>j v = a2,,&u > 2}
theJuliaRobinsonhypothesis.
satisfies
directdiophantine
definitions
Subsequently,
of the exponential
weregivenby a numberof investigators,
function
severalof
themusingthePellequationas inthisarticle(cf.[3], [4], [14], [18a]). The treatment
in ?2,3 isbasedon Matiyasevic's
thedetailsare JuliaRobinson's.
methods,
although
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1973]
HILBERT'STENTH PROBLEMIS UNSOLVABLE
267
In particular,
itwasMatiyasevic
whotaught
ushowtouse resultslikeLemmas2.11,
usedanalogousresults
himself
2.12,and2.22ofthepresent
exposition.
(Matiyasevic
fortheFibonaccinumbers.)
It wassoonnoticed(byS. Kochen)thatbya simpleinductive
argument
theuse
oftheDavis normalformcouldnowbe entirely
avoided,as has been done in the
presentexposition.
Let #(P) be thenumber
of solutions
oftheDiophantineequationP = 0. Thus
O? #(P) ? 8R. Hilbert'stenthproblemseeksan algorithm
fordecidingofa given
P whetheror not #(P) = 0. But there are many related questions: Is there an
algorithmfortestingwhether#(P) = N, or #(P) = 1, or #(P) is even? I was able
ofHilbert'stenthproblem)thatall
withtheunsolvability
to showeasily(beginning
areunsolvable.
Infactif
oftheseproblems
A
=
{0,1,2,3, 4o}
for
andB ' A, B # 0, B # A,thenonecanreadilyshowthatthereis no algorithm
ornot#(P) E B (cf.[15]).
whether
determining
suchas Hilbertdemandedwillbe forthcoming
Thefactthatnogeneralalgorithm
for
of
interest
adds to the
algorithms dealingwithspecialclassesof Diophantine
equations.AlanBakerand his coworkers
[1], [2] havein recentyearsmadeconin thisdirection.
siderableprogress
Notes
(but ofcoursenottheirbaingDiophantine)wereusedby Cantorin
1. Thesepairingfunctions
an
J.Robertsand D. Siefkeseach corrected
of therationalnumbers.
his proofof thecountability
They,as wellas W. Emerson,M. Hausner,Y. Matiyasevic,
ofthesefunctions.
errorin thedefinition
and JuliaRobinsonmade helpfulsuggestions.
= 1,
usedinsteadtheequationsx2 - xy2. For example,cf.[25],pp. 175-180.Matiyasevic
U2-
muv + V2
1.
Thiscreatesa minor
integers.
on thenonnegative
areusuallydefined
functions
3. The recursive
(e.g.,
withone in theliterature
technicalproblemin comparingthepresentdefinition
butannoying
in
cf. [6],p. 41; also Theorem4.2 on p. 51).Thusone can simplynotethatf(xl,..., x") is recursive
in theusual sense.Fromthe
thepresentsenseif and onlyiff(t1+ 1,. *,tn+ 1) - 1 is recursive
of thefunctions
involvedthisdoesn'tmatterat all;
"computability"
pointof viewof theintuitive
integers
as a "code" forthenonnegative
oneis simplyinthepositionofusingthepositiveintegers
using n + 1 to representn.
Thatis, S (i, u) can bs obtainedusingour three
4. InclusionofS (i, u) in thislistis redundant.
initialfunctions.
operationsfromtheremaining
to use theenumera5. Themethodofproofis JuliaRobinson's,[28],[30].If onewerepermitted
Theoremwould
theory([6],p. 67.Theorem1.4),theUniversality
function
tiontheoremin recursive
followat once fromTheorem6.1.
6. ActuallytheresultwhichG6delstated(as opposedto whatcan be obtainedat oncebyuse of
functions
oftheclass ofrecursive
weaker.Indeed,theverydefinition
was somewhat
histechniques)
came severalyearslaterin theworkof G6del,Church,and
of theirsignificance
and theperception
was a precise equivalentof theintuitive
Turing.In particularths sugg-stionthatrecursiveness
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268
MARTIN DAVIS
[March
by Churchand by
was madeindependently
notionof beingcomputableby an explicitalgorithm
whichis essentialin regardingthe technicalresults
Turing.And ofcourseit is thisidentification
a negativesolutionofHilbert'stenthproblem.(For further
discussedin thisaccountas constituting
cf. [6].)
discussionand references,
wouldfollowfromthenon7. For example,I showed([13]) thattheJuliaRobinsonhypothesis
solutionsof theequation
existenceof nontrivial
9 (U2 -!-7i2)2 __ 7(x2 + 7y2)2 - 2.
The methodsused readilyshowthatthesameconclusionfollowsiftheequationhas onlyfinitely
(and hencethe Julia
manysolutions.Cudnovskii[4] claimsto haveprovedthat2x is diophantine
thereis a possibility
thatsomeof Cudnovskii's
usingthisequation.Apparently
Robinsonhypothesis)
- butI havenotbeenable to obtaindefinite
ofMatiyasevic
workmayhavebeendoneindependently
aboutthis.
information
References
to thetheoryofDiophantineequations:I. On therepresentation
1. Alan Baker,Contributions
ofintegers
bybinaryforms,II. The Diophantineequationy2 xX3 + k, Philos.Trans. Roy. Soc.
London Ser. A, 263 (1968) 173-208.
2. AlanBaker,The Diophantineequationy2 == ax3 F bx2 + cx + d, J.LondonMath.Soc., 43
(1968) 1-9.
3. G. V. Cudnovskii,Diophantinepredicates(Russian),UspehiMat. Nauk, 25 (1970) no. 4
(154), pp. 185-186.
, Certain
problems
(Russian),OrdenaLeninaAkad.Ukrains.SSR, Preprint
arithmetic
4.
IM-71-3.
enumerablepredicates,J. Symbolic
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5. MartinDavis, Arithmetical
Logic, 18 (1953)33-41.
, Computability
6.
McGrawHill, New York, 1958.
and Unsolvability,
7. MartinDavis and HilaryPutnam,Reductionof Hilbert'stenthproblem,J.SymbolicLogic,
23 (1958) 183-187.
-, On Hilbert's
-and
tenthproblem,
U. S. AirForce0. S. R. ReportAFOSR TR
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Diophantineequations,Ann. Math.,74 (1961) 425-436.
-Proc.Symp.
10. MartinDavis, Applicationsof recursivefunctiontheoryto numbertheory,
PureMath.,5 (1962) 135-138.
, Extensions
and corollariesof recentworkon Hilbert'stenthproblem,IllinoisJ.
11.
Math.,7 (1963) 246-250.
12. MartinDavis and HilaryPutnam,Diophantinesetsoverpolynomial
rings,IllinoisJ.Math.,
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13. MartinDavis, One equationto rulethemall, Trans.New York Acad. Sci., SeriesII, 30
(1968) 766-773.
oftheexponential
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14.
function,
Diophantinedefinition
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35 (1972) 552-554.
undverwandter
16. KurtGodel,Uberformalunentscheidbare
SatzederPrincipiaMathematica
(1) Kurt,Godel,
SystemeI, Monatsh.Math.und Physik,38 (1931) 173-198.Englishtranslations:
On FormallyUndecidablePropositionsof PrincipiaMathematicaand Related Systems,Basic
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1973]
HILBERTS TENTH PROBLEMIS UNSOLVABLE
269
Books, 1962.(2) MartinDavis (editor),The Undecidable,Raven Press,1965,pp. 5-38. (3) Jean
Press,1967,pp. 596-616.
Van Heijenoort(editor),FromFregeto Gddel,HarvardUniversity
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ofthesolutionsofPell's equation(Rus18a. N. K. Kosovskii,On Diophantinerepresentations
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19. GeorgKreisel,Reviewof[9].Mathematical
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to Hilbert's
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po Konstruktivnoi
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,
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23b. -,
enumerablepredicates,Proc. Second
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ScandinavianLogic Symp.,editor,J.E. Fenstad,North-Holland,
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enumerablepredicates,Proc. 1970
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Intern.CongressMath.,pp. 234-238.
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to theTheoryof Numbers,2nd ed.,
25. Ivan Nivenand HerbertZuckerman,
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26. HilaryPutnam,
J.Symb.Logic,25 (1960)220-232.
An unsolvable
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27. JuliaRobinson,Existential
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6 (1969) [Studies
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in NumberTheory,editedby W. J.LeVeque,pp. 76-116].
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31. RaphaelM. Robinson,Arithmetical
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32. -,
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