21. The Fermat-Gauss Impossibility Theorem.
Prove that the sum of two cubes cannot be a cube, i.e.
x3 y3 z3
is not true for any integers x, y, z 0.
This is a special case of Fermat’s Last Theorem, which was expressed by Fermat in the
following way in the Arithmetic of Diophantus, edited by Fermat’s son, and published in
1670: "It is impossible to divide a cube into two cubes, a fourth power into two fourth
powers, and in general any power except the square into two powers with the same
exponents." Fermat added: "I have discovered a truly wonderful proof of this, but the
margin (of the notebook) is too narrow to hold it." Unfortunately, Fermat neglected to
disclose this "wonderful proof."
Fermat’s Last Theorem became very famous since many of the greatest
mathematicians since Fermat, including Euler, Legendre, Gauss, Dirichlet, Kummer, and
others tried unsuccessfully to prove it in general. Andrew Wiles finally proved Fermat’s Last
Theorem in 1994.
Euler proved the impossibility of x 3 y 3 z 3 (1770) as did Gauss (Complete Works, vol.
II) Euler’s method was relatively complicated; Gauss on the other hand, proved simply and
clearly the impossibility of the more general equation
)3 *3 +3
for any complex numbers ), *, + of the form xJ yO, where x and y are integers and
J
1"i 3
1i 3
and O 2
2
are cube roots of "1. (O is the conjugate of J. ) Let
R Z¡J ¢ xJ yO | x, y are integers
For convenience, we call members of R G-numbers. R — Z, since every integer
n nJ nO. G-numbers have many similar properties to common integers. Many of these
are listed in the supplement. (The reader is advised to at least read through the statements
in the supplement, if not the proofs.) Property X is especially important for our purposes.
Gauss’ Proof of the Impossibility of the Equation
)3 *3 +3
We will use Greek letters to designate G-numbers and lower case Roman letters to
denote usual integers. First we make the substitutions ) 8, * 1 and + "0 to write the
1
equation in the form
8 3 1 3 0 3 0,
(1)
in which no two of the "bases" 8, 1, 0 have a common divisor. (Any G-prime divisor of two of
them would divide the third, and could then be divided out.)
Under the assumption that (1) is true for some non-zero G-numbers, we will prove
I. Exactly one of the bases 8, 1, 0 has the G-prime divisor = J " O i 3 .
II. There is another equation of the same form as (1) in which the base with prime factor
= contains the divisor = fewer times that the base in (1).
These two theorems, however, contradict each other. By repeated application of II., it is
possible to obtain an equation like (1) which has no base divisible by =, which contradicts I.
Proof of I.
If none of the three bases 8, 1, 0 were divisible by =, then by X.b. from the supplement
8 3 q e mod 9, 1 3 q f mod 9, 0 3 q g mod 9 with e 2 f 2 g 2 1,
and consequently, by (1), e f g q 0 mod 9, which is, however, impossible. (Each of e, f, g
is either 1 or "1. ) Without loss of generality, we assume that
=|0,
= 4 8,
= 4 1.
Proof of II.
=|0 ´ 0 3 q 0 mod = 3 , and then from (1) 8 3 1 3 q 0 mod = 3 . Since 8 3 q e mod 9 and
1 3 q f mod 9, e f q 0 mod = 3 . = 3 "3i 3 , and thus e f q 0 mod 3, and it follows that
f "e. 8 3 1 3 q e f 0 mod 9, = 4 9, and thus 0 3 "Ÿ8 3 1 3 q 0 mod = 4 . Since = is a
G-prime,
0 q 0 mod = 2 .
From 8 3 1 3 q 0 mod = 3 and the (easily verified) identity
8 3 1 3 IED,
where
I 8J 1O, E 1J 8O, D 8 1
it follows that at least one of the factors I, E, D is divisible by =. In fact, each one is divisible
by =, because I " E Ÿ8 " 1 = and I E D. Let
I =I U , E =E U , D =D U .
2
I U , E U , D U are pairwise relatively prime. If for instance - divided both I U and E U , then - would
also divide I U " E U 8 " 1 and =ŸI U E U 8 1, and thus - would divide 28 and 21; since 8
and 1 are relatively prime, it follows that - 2 (up to a unit factor). By the division algorithm
in R (see VI. below), we would have either
8 25 / and 1 26 / or
8 25 / and 1 26 " / with
/ 3 o1. (Note that / p 0, since that implies that 2|8 and 2|1. ) Then I 8J 1O 2Ÿ5J 6O /
or
2Ÿ5J 6O /=,
neither of which, however, is divisible by - 2.
Now recall that 0 q 0 mod =, and set
F3 0
=
F. Then
"Ÿ8 3 1 3 "IED
03
"I U E U D U with I U E U D U .
3
3
=
=
=3
Since I U , E U and "D U are pairwise relatively prime, to within unit factors ), * and +, I U , E U and
"D U must be cubes of pairwise relatively prime G-numbers >, @ and A :
I U )> 3 , E U *@ 3 , "D U +A 3 with ) 6 * 6 + 6 1,
so that
F 3 )*+> 3 @ 3 A 3 , )> 3 *@ 3 +A 3 0.
3
F
)*+ a G-unit, and since 4 is a G-unit (since its norm is 1)
This makes 4 3 >@A
n
n 3
we conclude that )*+ ŸJ ŸJ 3 Ÿ"1 n o1. Set )*+ E. Recall next that
0
0 q 0 mod = 2 so that =| = F. Since = is a G-prime, = divides >, @ or A (and just one of
them since I U , E U and "D U are pairwise relatively prime), say
(2)
=|A but = 4 > and = 4 @.
Then, however (see X. below), > 3 q e mod 9 and @ 3 q f mod 9 with e 2 f 2 1, and by (2),
)e *f q 0 mod 3. (Note that =|A ´ 3| "3 3 i = 3 |A 3 . ) Since
|)e *f| t |) ||e| |* ||f| 1 1 2, it follows that )e *f 0. Thus, with F "fe Ÿ o1 , we
get
* F),
F) 2 + F))+ *)+ E
and from (2)
3
)> 3 *@ 3 +A 3
0
3
3
3
)> F)@ +A
0
F) 2 Ÿ)> 3 F)@ 3 +A 3 0
or
F) 3 > 3 ) 3 @ 3 EA 3 0.
If we set 8 1 F)>, 1 1 )@ and 0 1 EA, this can be written as
8 31 1 31 0 31 0
(3)
an equation of the same type as (1) with =|0 1 EA, since =|A, but to a lower power than =
0
F divides 0, because A >@4
=>@4 and = is relatively prime to E. This finishes the proofs of
I and II, and thus the Fermat-Gauss Impossibility Theorem.
Supplement on R Z¡J ¢
I.
Recall J 1i 3
2
.
iY
2πi
J2=e 3 =-O
πi
1+i
J=e 3 =
2
J3=-1
1=J+O
4πi
J4=e 3 =-J
3
X
1-i 3 5
O=
=J =e
2
5πi
3
=J=-J2
Then
J O 1,
JO 1,
J 2 O 0,
O 2 J 0, J 3 "1, O 3 "1.
II.
R is a ring. It suffices to check that R is closed under addition, subtraction and
multiplication. Addition and subtraction are easy. For multiplication, note that
4
ŸaJ
bO ŸcJ dO acJ 2 adJO bcOJ bdO 2
"acO ad bc " bdJ
Ÿad bc " bd J Ÿad bc " ac O
III. The norm of the complex number z
a bi is z 0 NŸz z z a 2 b 2 u 0. It is
well known, and easy to show that the norm of the product of two numbers equals
the product of the norms, i.e., NŸwz NŸw NŸz . The norm of the G-number
Ÿab Ÿa"b 3 i is easily shown to be a 2 b 2 " ab. This is a non-negative
aJ bO 2
integer, since a and b are integers. It equals zero if and only if a b 0. The
smallest possible positive norms are 1, 2 and 3.
€ There are exactly six G-numbers with norm 1 : J, J 2 , J 3 "1, J 4 , J 5 , J 6 1.
€ There are no G-numbers with norm 2. (Think about the possibility of
a 2 b 2 " ab 2 when a or b is odd, or when both are even.)
€ There are exactly six G-numbers with norm 3 : = J " O i 3 and =J n for
n 1, 2, 3, 4, 5.
IV.
/ in R is a unit, or a G-unit, if there is some 1 in Z¡J ¢ such that /1 1. This
implies that NŸ/1 1 or NŸ/ NŸ1 1. Thus NŸ/ 1, and from III, / J n ,
n 1, 2, 3, 4, 5, 6. There are six units in R : o1, oJ and oO. (Compare with Z, which
has two units 1 and "1. )
V.
Let 0 R. The associated numbers of 0 are 0J n , n 1, 2, 3, 4, 5, 6, i.e., 0 times a
unit. The six associated number of = J " O i 3 are
=J "1 " O, =J 2 "1 " J, =J 3 "=,
=J 4 1 O, =J 5 1 J, =J 6 =.
VI. There is a division algorithm for G-numbers. If ), *
a quotient 4 and remainder + both in Z¡J ¢ such that
) 4* +,
where NŸ+ t 34 NŸ* .
Proof.
)
*
)*
**
)*
*0
hJkO
*0
J
*0
h
*0
J
k
*0
R with * p 0, then there is
O. (Recall that * 0 NŸ* . ) By the
division algorithm for the integers, h m* 0 r and k n* 0 s for some
*
integers r, s with 0 t r, s * 0 . If r 20 , let h Ÿm 1 * 0 Ÿr " * 0 where
"* 0
r " * 0 0. Similarly for s. Thus without loss of generality we can
2
assume that |r|, |s| t 12 * 0 . For example 19 4 5 " 1 with |"1| 1 t 5.
Then )* m *r0 J n *s0 O ŸmJ nO *r0 J *s0 O and
) ŸmJ nO * r
*0
s
O * 4* +. Since + ) " 4*, + Z¡J ¢,
and
5
NŸ+ N
r
*0
2
r
*0
t
J
1
4
3
4
1
4
s
*0
O NŸ* 2
s
*0
1
4
"
r
*0
NŸ* s
*0
NŸ* NŸ* .
R
If * 2, then we can always pick 4 R so that ) 24 + where + 0, J, O
or J O 1. Indeed if ) aJ bO with a and b even integers,
4 a2 J b2 O and + 0; if just one of a and b is even, we can get + J or
O, and if both are odd, + J O 1.
VII. Repeated application of the division algorithm
)
41* +1
with NŸ+ 1 t
3
4
NŸ* *
4 2 + 1 + 2 with NŸ+ 2 t
3
4
NŸ+ 1 t
3
4
2
NŸ* + 1 4 3 + 2 + 3 with NŸ+ 3 t
3
4
NŸ+ 2 t
3
4
3
NŸ* B
produces the well-known Euclidean algorithm for finding the greatest common
divisor of two integers. It must eventually terminate with a remainder of 0, since
n
3
NŸ* v 0 The last non-zero remainder + is the greatest common divisor (up
4
to a unit factor) of ) and *. If the GCD is a unit, then ) and * are said to be
relatively prime.
VIII. Just as with the (usual) integers, the theorems about divisibility, and unique
factorization follow from the Euclidean algorithm, and we have
€ If gcdŸ), * is a unit, and *|)6, then *|6.
€ If gcdŸ), + and gcdŸ*, + are units, so is gcdŸ)*, + .
€ A G-prime is a G-number that has no divisor other than its six associated
numbers and the six units.
€ = J " O i 3 is prime. (If = 56, then NŸ= NŸ5 NŸ6 , i.e.,
3 NŸ5 NŸ6 , and then 5 or 6 has norm 1, making it a unit.)
€ 2 is prime. (If 2 56, then NŸ2 NŸ5 NŸ6 , i.e., 4 NŸ5 NŸ6 .
Since there are no G-numbers with norm 2, either 5 or 6 has norm
1, making it a unit.)
€ Every G-number can be written as a product of G-primes in only one way,
up to the order in which the factors are written, and in which associated
numbers are considered the same. For instance ) * + and )J * +O
are considered to be the same.
IX.
) q * mod 6 if and only if 6|Ÿ) " * .
X.
Two results about divisibility by = i 3 in R :
a.
4 aJ bO is divisible by = if and only if 3|Ÿa b .
6
b.
If = 4 4, then 4 is congruent to a G-unit / mod 3, and 4 3 q o1 mod 9.
Proof of a. If =|4, then aJ bO ŸmJ nO ŸJ " O Ÿ2n " m J Ÿn " 2m O and
a 2n " m, b n " 2m. Then a b 3Ÿn " m . Conversely, if a and b
satisfy a b 3g, solve the system n " m g, 2n " m a for m a " 2g
and n a " g to get 4 ŸmJ nO ŸJ " O .
Proof of b. If = 4 4, then 3 4 Ÿa b , and therefore one of the following pairs
of equations is true:
a 3h
and b 3k e or
a 3h e and b 3k
or
a 3h e and b 3k e
with e o1. Set 5 hJ kO to get
4 35 eO or
4 35 eJ or
4 35 e.
Thus 4 35 / for some G-unit /, and 4 is congruent to a G-unit mod 3.
Next 4 3 9Ÿ35 3 35 2 / 5/ 2 / 3 , and since / 3 ŸJ n 3 J 3n o1 or
R
/ 6 1, 4 3 96 o 1, i.e., 4 3 q o1 mod 9.
7