BEGINNERS' CORNER
EDITED BY DMITRI THORO, SAN JOSE STATE COLLEGE
DIVISIBILITY II
We shall continue our investigation of some "background m a t e r i a l " for the b e ginning Fibonacci e x p l o r e r . Whenever n e c e s s a r y , we may a s s u m e that the i n t e g e r s
involved a r e not negative (or zero).
1.
DEFINITIONS
Two i n t e g e r s a and b a r e relatively p r i m e if their g r e a t e s t common divisor
(g. c. d. ) i s 1.
When convenient we will use the c u s t o m a r y abbreviation (a,b) to
designate the g. c. d. of a and b. Finally, as previously implied, we shall say that
n
* s composite if n has m o r e than two divisors,
2.
El.
ILLUSTRATIONS
If d = a and d| b , then d ( a ± b ) ; i. e. , a common divisor of two n u m b e r s is a
divisor of their sum or difference.
PROOF: d | a m e a n s t h e r e e x i s t s an integer a such that a = a'd. Similarly.
we may write b = b T d. Thus a ± b = d (a? ± b ? ) which p r o v e s that d[ (a ^ b). [ This
follows from the definition of divisibility. ]
E2.
If
d = (a,b).
then j -7 , -7 J = 1; i. e. , if two n u m b e r s a r e divided by their
g. c. d. , then the quotients a r e relatively p r i m e .
[This r e s u l t is a wide!}- used
"tool. "]
a
b
PROOF: Since d i s the g. c. d. of a and b. —, and -7 a r e certainly i n t e g e r s :
to
let us call them a'
show that
r
and b , respectively.
(a'. b ) = 1.
a' = a"d'
d
d
Thus a = a'd,
"
and b = b"d'.
Then (there exist i n t e g e r s
b = b'd. and we a r e to
a" and b " such that)
T
This implies a = a'd = a"d d and
f
i . e . . d d is a common divisor of a and b. But we a s s u m e d d'
T
dd
E3.
to
The trick is to use an indirect argument.
Suppose (a ? , bT) = df > 1.
T
•
1
b = b'd -• b"d'd;
1. which means
> d — c o n t r a r y to the fact that d is the g r e a t e s t common divisor of
a and b.
If a and b a r e relatively p r i m e , what can you say about the g. c. d. of a -r b
and a - b?
For example, (13 -f 8, 13 - 8) - 1. but (5 - 3. 5 - 3) - 2.
out, however, that these a r e the only possibilities.'
57
1
It t u r n s
58
BEGINNERS' CORNER
If (a,b) = 1, then
PROOF:
(a + b , a - b) = 1 or 2.
(a + b, a - b) = d.
Then by E l , dj [(a •+ b) ± (a - b)] ; i. e. ,
(I) If d i s even, set d = 2 K. Then from
2a = ' 2KaT, 2b = 2Kb' we have a = Ka r ,
d|2a
Let
[April
and d | 2 b .
b = Kb\
Therefore
K m u s t be 1 (why?), and hence d = 2.
(ii) Similarly if d is odd, then d would have to divide both a and b ,
whence
d = 1.
Can you see what objection a pedantic r e a d e r might have to this proof?
E4y
The following is a special c a s e of a r e s u l t due to Sophie Germain, a F r e n c h
mathematician (1776-1831).
n 4 + 4 is composite for n > 1 .
PROOF: Unlike the preceding i l l u s t r a t i o n s , h e r e one needs to stumble onto
a factorization,
n 4 + 4 = (n2 + 2)2 - (2n)2 = (n2 + 2 + 2n) (n2 + 2 - 2n) does the t r i c k ,
for this shows that N = n 4 + 4 is divisible by a number between 1 and N and
hence must be composite,
E5.
We now consider one of the s i m p l e s t p r o p e r t i e s of Fibonacci n u m b e r s .
Two consecutive Fibonacci n u m b e r s are always relatively p r i m e .
PROOF: Certainly this i s obvious for the first few n u m b e r s : 1. 1, 2 f 3, 5,
8,
,
Let us use an i n d i r e c t argument.
:
Suppose
to
II-
F
and
F
n
_, i s the f i r s t
n +
J
p1 a i r for which (F . F , ) = d > 1. Now examine the p a i r F _, and F . Since
ii' n+1
n-1
n
F ., - F = F 1 ?J d is a divisor of F 1 (for d| F , 1 , d! F and hence, byJ E l .
n+1
n
n-l
n-1
< n+V
n
'
their difference). This m e a n s that F - and F a r e not relatively p r i m e — a con'
n-1
n
tradiction to our assumption that F and F
i s the first such p a i r .
3.
Tl.
SOME USEFUL THEOREMS
Any composite integer n has at least one p r i m e factor.
PROOF:
(i) Since n i s composite, it must have at least one divisor g r e a t e r
than 1 and l e s s than n.
(ii) Let d be the s m a l l e s t divisor of n such that 1 < d < ii.
(iii) Suppose d is composite; let d r jd, 1 < d' < d.
(iv) Thus we have n = n t d = li^l'd"; i. e. , d' j n but d < d' —
to the definition of d.
T2. Given n > 1.
than a.
Therefore
a contradiction
d must be a p r i m e .
Suppose that the quotient q.
If n is not divisible by 2. 3, 4.
in the division of n by a, is l e s s
. (a - 1). a, then n is a p r i m e .
1963]
BEGINNERS' CORNER
59
PROOF: (i) If q (q < a ) is the quotient and r the r e m a i n d e r in the division
of n by a,
we may w r i t e
n
r
- = q + -,
0 <
r
<
a.
(ii) A s s u m e that n i s not divisible by 2, 3 , 9 9 ' , a but has a divisor d, 1 < d < n.
We shall show that this leads to a contradiction.
(iii) Since djn,
n = dd T ,
where 1 < dT < n.
(iv) By (ii), df > a and by (i) a > q; hence d? >q o r d? > q + 1.
Also
d>a;
T
multiplying d - q + 1 by d > a. we a r r i v e at
(v) n = ddf > aq + a.
But by (i), n = aq + r < a q + a
since r < a.
This i s the
d e s i r e d contradiction which p r o v e s that n cannot be divisible by d, 1 < d < n9
and hence m u s t be a p r i m e .
T3. If n > 1 i s not divisible by pt = 2, p 2 = 3, p 3 = 5, p 4 = 7, • • • , p k ? where
p^. i s the l a r g e s t p r i m e whose s q u a r e does not exceed n,
then n i s a p r i m e .
PROOF: A s s u m e a2 — n < (a + l ) 2 and that n i s not divisible by 2, 3, ••• ,'
(a - 1), a. Then n = dd? i m p l i e s d ^ a + 1, df — a + 1,
— a contradiction.
Thus n m u s t be p r i m e .
whence n = dd? ^
(a+1) 2
The r e a d e r should convince himself
that h e r e (as well as in T2) it suffices to consider only p r i m e divisors - Vn.
4.
PROBLEMS
1.1 Suppose that
(i) p i s the s m a l l e s t p r i m e factor of n and
V .
(ii) p > vn
What i n t e r e s t i n g conclusion can you draw?
1.2 P r o v e that two consecutive Fibonacci n u m b e r s a r e relatively p r i m e by
using one of the identities on p. 66 (Fibonacci Q u a r t e r l y , F e b r u a r y , 1963).
1.3
by 6.
P r o v e that if p and p + 2 a r e (twin) p r i m e s , then p + 1 i s divisible
(Assume p > 3. ) [This p r o b l e m was suggested by J a m e s Smart. ]
n
1.4 P r o v e that if n i s divisible by k, 1 < k < n, then 2 - 1 i s divisible
by 2 k - 1.
F o r example, 2 35 - 1 = 34 359 738 367 is divisible by 2 5 - 1 =31 and
2 7 - 1 = 127.
BEGINNERS' CORNER
60
[April 1963 ]
1.5 P r o v e that t h e r e a r e infinitely many p r i m e s .
Hint: Assuming that
p
i s the l a r g e s t p r i m e , Euclid considered the expression N = 1 + 2 - 3 - 5 - 7 - - - p .
Now either N i s p r i m e or N i s composite.
Complete h i s proof by investigating
the consequences of each alternative.
Additional hints may be found on p. 80.
FIBONACCI FORMULAS
Maxey Brooke 3 Sweeny, Texas
If you have a favorite Fibonacci formula, send it to us and we will try to publish
it.
Some historically i n t e r e s t i n g ones a r e shown below.
1.
P e r h a p s the f i r s t Fibonacci formula was developed by Simpson in 1753.
F
2.
- F 2 = v( - l ) n
± 1 F
n+1
n - 1n
n
'
A very i m p o r t a n t formula was developed in 1879 by an o b s c u r e F r e n c h m a t h e matician, Aurifeuille.
In fact, it is his one claim to fame,
L_ = L (L Q - 5F + 3 ) ( L 0 + 5F + 3)
5n
n 2n
n
' 2n
n
'
3.
The only formula involving cubes of Fibonacci n u m b e r s given in Dickson T s
"History of the Theory of N u m b e r s " i s due to L u c a s .
F3
n+1
+
F3
n
_
F3
n-1
=
F
3n
3
F* ,' - 3F + F 3 n = 3 F 0
.
n+2
n
n-1
3n
The r e c u r s i o n formula for sub-factorials i s s i m i l a r to the one for Fibonacci
The late Jekuthiel Ginsburg offers
te
4.
numbers:
5.
P , , = n(P + P J ; P 0 = 1 ,
v
n+1
n
n-17' u
*
Vx = 0 .
i
Fibonacci n u m b e r s have been r e l a t e d to a l m o s t every other kind of n u m b e r .
H e r e i s He S. Vandiver's relation with Bernoulli n u m b e r s .
Y
B
2kF2k
B
2k
S
hmod^
^ P = 5a+ 1
k=0
^ 3
)
F
2(k-l) ~
X (modp
)
i f P = 5a ± 2
k=0
p - 3 denotes the g r e a t e s t integer not exceeding (p - 3 ) / 2 .
I think that this is a good idea.
Ed.