Download 20. The Fermat Equation v2.

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.
We will examine a somewhat modified and more general equation X 2 " DY 2 4, which
includes the original Fermat equation. Indeed, in order to solve x 2 " dy 2 1, we need only
solve 4x 2 " 4dy 2 4, i.e., Ÿ2x 2 " Ÿ4d y 2 4. The D in X 2 " DY 2 4 may be q 0 or 1 mod 4,
the case 0 mod 4 corresponding to Fermat’s equation.
The (simple) continued fraction ¡a 1 ; a 2 , a 3, . . . ¢ with integers a i , all positive, except for a 1 ,
which may be negative or zero, is the fraction
a1 a2 1
1
a 3 a 1...
4
Every finite continued fraction ¡a 1 ; a 2 , a 3 , . . . , 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 1 ; a 2 , a 3 , . . . , a k ¢ q kk
is called the k th convergent of ¡a 1 ; a 2 , a 3 , . . . ¢. In the sequel, we will be working with infinite
CFs. [Many people start CFs off with a 0 today.] Here are some basic properties of CFs.
1.
p1 a1
q1 1
p2 a2a1 1
q2 a2
p k a k p k"1 p k"2
q k a k q k"1 q k"2
for k 3, 4, . . . or as it is sometimes presented:
1
2.
3.
4.
5.
a1 a2 a3
C
a k"2 a k"1
ak
C
0
1
p1 p2 p3
C
p k"2 p k"1 p k a k p k"1 p k"2
C
1
0
q1 q2 q3
C
q k"2 q k"1 q k a k q k"1 q k"2
C
Let x 1 x be a positive number with x ¡a 1 ; a 2 , a 3 , . . . ¢, 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 1 ; a 2 , a 3 , . . . ¢ with
a calculator (or by hand): namely find the integer part of x, subtract it off, find the
reciprocal, repeat.
C 1 C 3 C 5 . . . x C C 6 C 4 C 2 .
p
If x ¡a 1 ; a 2 , a 3 , . . . ¢, and C k q kk , then
a. gcdŸp k , q k 1,
b. p k1 q k " p k q k1 Ÿ"1 k1 .
u is a purely periodic CF with period n, if a 1 0 and
u ¡a 1 ; a 2 , a 3 , . . . , a n , a 1 , a 2 , a 3 , . . . , a n , . . . ¢, or.,
u ¡a 1 ; a 2 , a 3 , . . . , a n , a n1 a 1 , . . . , a 2n, u ¢. If kn is even, and u is a purely periodic
CF,
a. p kn q kn"1 " p kn"1 q kn Ÿ"1 kn 1,
p up kn"1
for k 1, 2, 3, . . .
b. u q kn
kn uq kn"1
Definition A reduced (quadratic) equation is a quadratic equation au 2 bu c 0 with
integer coefficients, a non-square discriminant D b 2 " 4ac, one of whose roots
"b D
1 and the other one " 1 "b"2a D 0. The positive root is called a
2a
reduced number.
Example. 3u 2 " 10u " 1 0 is a reduced equation with roots u u 10" 112
6
X ". 0971.
10 112
6
1 and
The solution of X 2 " DY 2 4 with non-square D q 0, 1 mod 4 proceeds as follows:
1.
2.
3.
4.
5.
u is a reduced number if and only if the CF expansion of u is purely periodic.
For every non-square D q 0, 1 mod 4, there is a reduced number u with
discriminant D.
Every CF expansion of u (from 2) provides an infinite number of solutions to
X 2 " DY 2 4.
Every solution to X 2 " DY 2 4 is obtained in this way.
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, . . .
Now for the proofs.
1.
Let u ¡a 1 ; a 2 , a 3 , . . . , a n , a 1 , a 2 , a 3 , . . . , a n , . . . ¢ be purely periodic with period n, and
2
kn be even. Note that u a 1 u 1. Since u H p kn " q kn"1 . The discriminant
p kn up kn"1
q kn uq kn"1
, q kn u 2 " Hu " p kn"1 0 where
D H 2 4q kn p kn"1
H 2 4Ÿp kn q kn"1 " 1 H 2 4p kn q kn"1 " 4
Ÿp kn q kn"1 2 " 4
is a non-square, since it is 4 less than a square. r D is thus irrational, and
Hr
r H, since D H 2 4q kn p kn"1 . The two roots of q kn u 2 " Hu " p kn"1 0 are 2q
kn
H"r
Hr
H"r
and 2q
,
the
latter
being
negative.
Thus
u
.
u
is
the
negative
root.
2q
2q
kn
kn
kn
Is it larger than "1? Since u u " qknkn"1 , " u kn"u1 kn , and q kn q kn"1 we see that
" u p kn"1u/q kn"1 C knu"1 . But C kn"1 u C kn , so " u uu 1, and "1 u 0.
Thus u is a reduced number.
p
p
/q
Now let u be a reduced number, the positive root of the reduced equation
au 2 bu c 0, where without loss of generality a 0, and a, b, c are relatively
br
b"r
prime. u "2a
and u "2a
where r D b 2 " 4ac . Since
2
c
b "D
"b
a u u 4a 2 0 and u u a , it follows that c 0 and b 0, the latter
because u 1 and "1 u 0.
Let u g u1U with g ¡u ¢. Then u U 1. In fact u U is a reduced number. To
see this, let a U "Ÿag 2 bg c , b U "Ÿ2ag b and c U "a. a U , b U and c U are
relatively prime, since a, b, c are. It’s somewhat tedious, but not hard to verify the
following:
a. ax 2 bx c 0 and a U x 2 b U x c U 0 have the same discriminant D,
b.
uU 1
u"g
U
U
2aŸr"b 2a
rb U
r 2 "b U2
2aŸb U r 2a
"b U r
2a U
1
u "g
, i.e., u U is a root of a U x 2 b U x c U 0.
b "r
b U2 "r 2 b U "r , and since "1 u 0, g u 1,
u U "2a
U
U
"1 u 0.
Thus only reduced numbers occur in the CF expansion of a reduced number.
Also note that "1U g " u , so that g "1U . Thus g is determined by u U as well
u
u
as by u.
c.
Now there are only a finite number of reduced numbers with discriminant D.
(D b 2 " 4ac and "ac 0 allows for only a finite number of bs, and then a finite
number of as and bs.) This means that in the CF expansion of a reduced number
u, some reduced number U must reappear after a finite number of steps, e.g.,
u ¡b 0 ; b 1 , b 2 , . . . , b k , U ¢, U ¡a 0 ; a 1 , . . . , a j , U ¢. From the conclusion of the last
R
paragraph, b k a j ; continue "backing up" to see that u is purely periodic.
2.
There are two cases to consider:
a. If D q 0 mod 4, D 4m for some positive integer m. Let g ¡ m ¢, a 1,
b "2g and c g 2 " m. The discriminant of ax 2 bx c 0 is
"b D
"b" D
b 2 " 4ac 4m D, and its roots are 2 and 2 .
g m 1 ´ 2g 2 m 2 ´
"b D
2
1. Next
3
b.
m " 1 g m ´ "1 g " m 0 ´ "2 "b " D 0, and thus
"1 "b"2 D 0.
If D q 1 mod 4, D 4m 1 for some positive integer m. Let g be the
largest integer for which g 2 g m (so Ÿg 1 2 Ÿg 1 m or
g 2 3g 2 m). Note that m is not of the form g 2 g for any g, since D
is not a square. Let a 1, b "Ÿ2g 1 and c g 2 g " m. The
discriminant of ax 2 bx c 0 is b 2 " 4ac 4m 1 D, and its roots
D "b
"b D
"b" D
are 2 and 2 . D u 5 so D 2g 1 2, i.e., 2 1. Next
g 2 g m ´ 4g 2 4g 1 4m 1 D
´ Ÿ2g 1 2 D
´ "b D
"b" D
´
2
0,
and
g 2 3g 2 m ´ 4g 2 12g 9 4m 1 D
´ Ÿ2g 3 2 D
´ 2g 3 D
´ D " 2g " 1 2
and thus "1 3.
"b" D
2
0.
´
D b
2
´
"b" D
1
"1
2
R
Let u ¡a 1 ; a 2 , a 3 , . . . , a n , a 1 , a 2 , a 3 , . . . , a n , . . . ¢ be the positive root of a reduced
equation
au 2 bu c 0
(1)
with discriminant D (and a 0, gcdŸa, b, c 1). Then from 1 above,
q kn u 2 " Hu " p kn"1 0
(2)
where kn is even and H p kn " q kn"1 . Thus (1) and (2) have the same roots.
Since gcdŸa, b, c 1, the coefficients of (2) are constant multiples, say y times the
coefficients of (1), i.e.,
(3)
y
q kn
a
p kn "q kn"1
"b
p kn"1
"c
.
Now let
(4)
x p kn q kn"1
Then x 2 " b 2 y 2 Ÿp kn q kn"1 2 " Ÿp kn " q kn"1 2 . From (3), 4acy 2 "4q kn p kn"1 .
Addition of these last two equations yields
x 2 " Dy 2 4Ÿp kn q kn"1 " q kn p kn"1 4 1 4. (See property 5a above.) Thus every
4
pair of convergents C kn"1 and C kn with kn even provides solutions to X 2 " DY 2 4
by (3) and (4).
4.
Let Ÿx, y be a solution of X 2 " DY 2 4 with x and y positive integers (and
"b D
non-square D q 0, 1 mod 4 . Let u 2a be the positive root of the reduced
equation au 2 bu c 0; recall that c 0 by 1. Also let
P x"by
2
Q ay,
, p "cy,
q xby
2
.
Q and p are clearly positive integers. Since
Ÿx
by Ÿx " by x2 " b2y2
4 Dy 2 " b 2 y 2
4 " 4acy 2
4Ÿ1 " acy 2 4Ÿ1 Qp is a multiple of 4, and Ÿx by Ÿx " 2y 2x is even, x by and x " by are each
even, and thus P and q are positive integers too. It is easy to check that
Pq " Qp 1
"ŸP"q "p
because x 2 " Dy 2 4. And from a y , b , c y , we get
y
Q 2
P"q
p
2
y u " y u " y 0, Qu qu Pu p and
Pu p
u
.
Qu q
p
These last two relations among p, q, P, Q and u, suggest that q and QP might be
consecutive convergents of the CF expansion of u, from which x and y can be
Q
obtained: x P q and y a . That’s what we proceed to show. First
"b" D
b"r
of the reduced equation
2ŸQ " q Ÿ2a " b y " x. Since the root 2a "2a
2
au bu c 0 is between "1 and 0, b r 2a or 2a " b r. Thus
r2y2 " x2
Dy 2 " x 2
2ŸQ " q ry " x ry x ry x ry"4 x
and Q " q ry"2x . However, since D r 2 b 2 " 4ac is at least 5 (D is a
non-square q 0, 1 mod 4), y u 1 and x u 3, and it follows that ry x 5 so
Q " q "52 . Hence Q u q.
Q
Let u ¡a 1 ; a 2 , . . . , a n ¢ where n is the period of the CF, and expand QP into a
C.F. ¡b 1 ; b 2 , . . . , b 2m ¢ with an even number of terms. (This is no loss of generality
since ¡b 1 ; b 2 , . . . , b k , 1 ¢ ¡b 1 ; b 2 , . . . , b k 1 ¢ and
th
¡b 1 ; b 2 , . . . , b k , b k1 ¢ ¡b 1 ; b 2 , . . . , b k , b k1 " 1, 1 ¢ if b k1 1. ) Let C k be the k
U
p
convergent of ¡b 1 ; b 2 , . . . , b 2m ¢. Then C 2m QP , C 2m"1 q U , and Pq U " Qp U 1.
Since Pq " Qp 1 too, we have PŸq U " q QŸp U " p , and we can then conclude
5
that q U q because Q and P are relatively prime, whence Q divides q U " q but q
and q U are positive numbers t Q. Then p U p too, and we get
¡b 1 ; b 2 , . . . , b 2m , u ¢
i.e.,
u
Pu p U
Pu p
U
Qu q
Qu q
Pu p
¡b 1 ; b 2 , . . . , b 2m , u ¢
Qu q
but u ¡a 1 ; a 2 , . . . , a n ¢, from which we conclude that 2m is a multiple of n. Thus all
solutions of X 2 " DY 2 4 are obtained from the C.F. expansion of u using an
n, 2n, 3n, . . . periods if n is even, and 2n, 4n, 6n, . . . periods if n is odd.
5.
Let ŸX 1 , Y 1 be the smallest positive solution of X 2 " DY 2 4 found by this C.F.
xy D
method, and let Ÿx, y be any other (positive) solution. Let z 2 and
X Y
xy D
2
Z 1 1 21 . Then we must show that z is some power of Z 1 . Dörrie calls
a Lagrange number if Ÿx, y is a positive integral solution of X 2 " DY 2 4. He
proves the following
D
x y
Lemma If z 1 1 21
latter if z 2 z 1 .
D
x 2 y 2 D
2
and z 2 are Lagrange numbers, so are z 1 z 2 and
z1
z2
, the
Proof of the Lemma
z1z2 1
2
1
2
z1
z2
x 1 x 2 y 1 y 2 D
2
x 1 x 2 "y 1 y 2 D
2
x x y y D
x 2 y 1 x 1 y 2
2
x 2 y 1 "x 1 y 2
2
D , and
D .
x x "y y D
x y x y
x y "x y
We must show that x 3 1 2 2 1 2 , y 3 2 1 2 1 2 , x 4 1 2 2 1 2 and y 4 2 1 2 1 2 are
positive integers and x 2i " Dy 2i 4 for i 3, 4. It is an easy algebraic exercise to
show that x 2i " Dy 2i 14 Ÿx 21 " Dy 21 Ÿx 22 " Dy 22 4. Now recall that D q 0, 1 mod 4;
when D q 0 mod 4, both x 1 and x 2 are even, and x 3 , y 3 , x 4 , y 4 are integers; when
D q 1 mod 4, x i q y i mod 2, so both are even or both are odd, and it also follows that
x 3 , y 3 , x 4 , y 4 are integers. Clearly x 3 , y 3 0. Since x 21 " Dy 21 4, x 1 D y 1 and
similarly x 2 D y 2 . Thus x 1 x 2 Dy 1 y 2 and x 4 0. Finally
z2 z1 ´ x2 y2 D x1 y1 D
´
Ÿy 2
" y1 D x1 " x2
but also x 1 y 1 D
x1 " y1 D x2 y2 D
z 1 x 1 " y 1 D z 2 x 2 " y 2 D and thus
x2 " y2 D
4 , i.e.
z2 z1 ´ x1 " y1 D x2 " y2 D
´
Ÿy 2
" y1 D x2 " x1.
It follows that y 2 " y 1 0 or y 1 y 2 . Then
x 21
y 21
" D y 21 x 22
y 22
" D y 22 implies that
6
x1
y1
x2
y2
and y 4 x 2 y 1 " x 1 y 2 0.
R
Proof of 5. Suppose that z is not a power of Z 1 . Since D u 5, Z 1 1, and for some
integer n, Z n1 z Z n11 . Then 1 Zzn Z 1 . By the Lemma, Zzn is a Lagrange
1
1
number, but it’s smaller than Z 1 , the smallest Lagrange number. This contradiction
R
shows that every z is a power of Z 1 .
Example 1. x 2 " 112y 2 4.
Solution A reduced equation with discriminant D 112 is 3u 2 " 10u " 1 0. Here
10 112
¡3; 2, 3, 10 ¢ and we find the convergents of u to be
u
6
3 2 3
10
C
0
1
3 7 24 247
C
1
0
1 2 7
C
72
and the smallest positive solution (from the proof of 4) is x p 4 q 3 247 7 254,
q
24. All solutions can be found from
y a4 72
3
2
25424 112
2
n
2 127 12 112
n
n
. For example
x
y
2 64, 514
6, 096
3 16, 386, 302 1, 548, 360
Example 2. x 2 " 28y 2 1.
Solution. Solve Ÿ2x 2 " 112y 2 4. From example 1, the solutions are
254
127, 24 , 64514
32257, 6096 , . . .
Ÿx, y 2
2
Example 3. x 2 " 13y 2 1.
Solution. Solve Ÿ2x 2 " 52y 2 1. A reduced equation with discriminant 52 (by the proof
6 52
of 2) is u 2 " 6u " 4 0 with positive root 2 ¡6; 1, 1, 1, 1 ¢. Since the period is
odd, we must use two periods to find the first solution:
6 1
1
1
1
6
1
1
1
1
0
1
6 7 13 20 33 218 251 469 720 1189
1
0
1 1 2
3
5
33
38
71
109 180
2x p 10 q 9 1189 109 1298, y a10 180, and x 649, y 180 is the smallest
n
1298180 52
positive solution. All solutions can be found from 2
2
q
n
n
2 649 90 52
: if u n v n 52 2 649 90 52
and
n
1
x n y n 13 649 180 13 , then x n 2 u n and y n v n . (This is an easy proof
n
by induction.) Thus x n y n 13 649 180 13
gives all the solutions to
7
x 2 " 13y 2 1 (in positive integers).
Example 4. x 2 " 41y 2 4.
Solution. A reduced equation with discriminant 41 (by the proof of 2) is u 2 " 5u " 4 0
5 41
with positive root 2 ¡5; 1, 2, 2, 1 ¢. Since the period is odd, we must use two
periods to find the first solution:
5 1 2
2
1
5
1
2
2
1
0
1
5 6 17 40 57 325 382 1089 2560 3649
1
0
1 1 3
7
10 57
67
191
449
640
x p 10 q 9 3649 449 4098, y a10 640 is the smallest positive solution. All
n
n
4098640 41
solutions are Ÿx n , y n where x n y n 41 2
2 2049 320 41 . For
2
q
example,
x 2 y 2 41
x 3 y 3 41
2 2049 320 41
2
16 793 602 2622 720 41
2 2049 320 41
3
68 820 176 898 10 747 905 920 41
8