Download 20. The Fermat Equation v1.

20. The Fermat Equation
Find all integer solution of x 2 " dy 2 1, where d is a positive whole number but
not a square.
This extremely important problem in number theory was posed by Pierre Fermat in
1657, first to his friend Frénicle and then to all contemporary mathematicians. The first,
very complicated solution, was obtained by the Englishmen Lord Brouckner and John
Wallis. The simplest and best solutions to this problem were discovered by Euler, Lagrange
and Gauss. [Euler: "De usu novi algorithmi...", Novi commentarii Academiae Petropolitanae
ad annum 1765. Lagrange: "Solution d’un problème d’arithmétique", Miscellanea
Taurinensia, vol. IV, 1768. Gauss: Disquisitiones arithmeticae, 1801] They are all based on
the properties of continued fractions.
Dörrie’s "solution" is not entirely correct. See Note 2 below. The following outline of the
solution comes from Elementary Number Theory by Burton, McGraw Hill, 2007.
(Most proofs are based on property 1 below, and can be found in Burton’s book.)
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¢
A (simple) continued fraction a 0 ; a 1 , a 2 , . . . where the a i are integers, all positive, except
for a 0 , which may be negative or zero, is the fraction
a0 a1 ¡
1
1
a 2 a 1...
3
¢
Every finite continued fraction a 0 ; a 1 , a 2 , . . . , a n is a rational number, and conversely, every
rational number can be written as a finite continued fraction. Irrational numbers have infinite
continued fraction expansions.
¡
¢
p
C k a 0 ; a 1 , a 2 , . . . , a k q kk
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¢
is called the k th convergent of a 0 ; a 1 , a 2 , . . . .
1.
p0 a0
q0 1
p1 a1a0 1
q1 a1
p k a k p k"1 p k"2
q k a k q k"1 q k"2
for k 2, 3, . . . or as it is sometimes presented:
1
a0 a1 a2
2.
3.
0
1
p0 p1 p2
1
0
q0 q1 q2
a k"2 a k"1
ak
p k"2 p k"1 p k a k p k"1 p k"2
q k"2 q k"1 q k a k q k"1 q k"2
¡
¢
C
C
C
Let x 0 x be a positive number with x a 0 ; a 1 , a 2 , . . . , and x k1 x k "1¡x k ¢ (provided
that x k p x k ). Then a k x k . This gives an easy way to find a 0 ; a 1 , a 2 , . . . with
a calculator (or by hand): namely find the integer part of x, subtract it off, find the
reciprocal, repeat.
If d 0 is not a square, then d a 0 ; a 1 , a 2 , a 3 . , , , . a 3 , a 2 , a 1 , 2a 0 where the bar
gives the repeating or periodic part of d . For example
a.
28 5; 3, 2, 3, 10 with period 4, and
b.
29 5; 2, 1, 1, 2, 10 with period 5.
C 0 C 2 C 4 . . . C 5 C 3 C 1 .
p
If x a 0 ; a 1 , a 2 , . . . , and C n q nn , then
a. gcd p n , q n 1,
p
b.
x " q nn q n11 q n t q12 .
¡ ¢
¡ ¢
¡
¡
4.
5.
C
C
C
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Ÿ
¢
¡
¡
¢
¢
¢
¢
n
If 1 t b t q n , then x " q nn t x " ab , i.e., q nn is the best approximation
to x with denominator of q n or less.
d. If x is any irrational number, and x " ab 1 2 , with positive relatively
2b
prime integers a and b, then ab is one of the convergents of x.
p
If p, q is a positive solution of x 2 " dy 2 1, then q is a convergent of the continued
fraction expansion of d . (Proof: p " q d
p q d 1 so
p
c.
6.
p
q
" d and 0 1
q pq d
p
q
p
" d 1
q q d q d
2
2
d
2q 2 d
1
2q 2
, and 5d
establishes the result.) For example x " 28y 1 has the the positive solution
p 127, q 24 and 28 5; 3, 2, 3, 10 with convergents:
¡
¢
5 3
¡
2
10
0
1
5 16 37 127 1307
1
0
1 3
7
¢
Theorem. Let d a 0 ; a 1 , a 2 , . . . , a n , and let
1.
3
24
pk
qk
247
C
C
C
be its convergents.
If n is even, then all (positive) solutions to x 2 " dy 2 1 are
x p kn"1 , y q kn"1 for k 1, 2, 3, . . .
2.
If n is odd, then all (positive) solutions to x 2 " dy 2 1 are
x p 2kn"1 , y q 2kn"1 for k 1, 2, 3, . . .
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¢
Example 1. x 2 " 28y 2 1. Here 28 5; 3, 2, 3, 10 has period 4, and the first two
2
solutions are x p 3 127, y q 3 24 and x p 7 32257, y q 7 6096.
5 3
2
3
10
3
2
0
1
5 16 37 127 1307 4048 9403 32257
1
0
1 3
7
24
247
765
¡
C
C
C
3
1777 6096
¢
Example 2. x 2 " 29y 2 1. Here 29 5; 2, 1, 1, 2, 10 has period 5, and the first
solutions are x p 9 9801, y q 9 1820.
5 2
1
1
2
10
2
1
1
2
10
0
1
5 11 16 27 70 727 1524 2251 3775 9801 101785
1
0
1 2
3
5
13 135 283
418
701
1820 18901
C
C
C
Note 1. If x 1 , y 1 is the smallest positive solution of x 2 " dy 2 1, then every positive
solution is given by x n , y n where
xn yn d x1 y1 d
n
for n 1, 2, 3, . . .
2
For example, 127 24 28
32 257 6096 28 so x 32257, y 6096 is also a
2
2
solution to x " 28y 1 (which we found above in Example 1).
Note 2. Dörrie claims to solve the equation X 2 " DY 2 4, where D q 0, 1 mod 4 is not a
square. He points out that in order to solve x 2 " dy 2 1, we need only solve
4x 2 " 4dy 2 4, i.e., 2x 2 " 4d y 2 4. He finds a "reduced equation"
au 2 bu c 0 with discriminant D, expands the positive root in a purely periodic
continued fraction a 1 ; a 2 , . . . , a n , a 1 , a 2 , . . . , a n , a 1 , . . . , and asserts that every solution
can be found from a 1 ; a 2 , . . . , a n , a n1 , . . . , a N where N is an even multiple of n. This is
not true as his example shows: x 2 " 112y 2 4. A reduced equation with discriminant
32 7
112 is 3u 2 " 10u " 1 0 and 3 3; 2, 3, 10 with n 4. A solution x 254, y 24
can already be derived from a 1 ; a 2 , . . . , a n , a n1 , . . . , a N with N n, i.e., N an odd
multiple of n. All that is required is that N kn be even, not that k be even. This
error is continued in his "proof" and other versions of it found on the web. So reader
beware!
Ÿ Ÿ ¡
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Note 3. Equations of type X 2 " DY 2 4 thus "include" equations of type x 2 " dy 2 1, but
are more general, e.g., X 2 " 5Y 2 4. If x 1 , y 1 is the smallest positive solution of
X 2 " DY 2 4, then every positive solution is given by X n , Y n where
Xn Yn D 2
X1 Y1 D
2
n
for n 1, 2, 3, . . .
3