ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Seria I : PRACE MATEMATYCZNE X I I I (1969) ANNALES SOCIETATIS MATHEMATICAE POLONAE Series I : COMMENTATIONES MATHEMATICAE X I I I (1969)
W. Ze l ic h o w ic z (Poznań)
O n the equation {со2 — 1)(г2—1) = {у2—I )2
The question whether the equation
(1) (x2- l ) { z 2- l ) = (y2—l ) 2
has integral solutions, with all three of x , y , z distinct and even, was raised by Schinzel and Sierpiński in [1]. We shall prove the following
Th e o r e m. Equation (1) has no integral solutions with x , y , z distinct and even.
P ro o f. Suppose that there exist three integeres x , y , z all distinct and even satisfying equation (1). We can restrict ourselves to the case 0 < x < у < z.
Write
(2) z2 — x2-\-r.
Then (1) becomes
(3) ж4 + (r—2)x2—t (у2—l ) 2 — У+1 = 0.
Denoting by A the discriminant of equation (3), we get
(4) A = r2+ 4 (y 2- l ) 2.
I t is obvious that A has to be square of an integer 8 so write
(5) r2 + 4(y2- l ) 2 = S2.
I t is well known (see [2], p. 234) that all the solutions of (5) in positive integers are given either by
(6) r = (ul + Z u ^ ) ^ ,
(7) 2(y2—1) = (2v\ + 2u1v1)Jcl ,
(8) s = (u\-{-2vlJr 2u1vl)Jc1
or by
(9) r = {2х1 + 2и2и2)к2,
(10) 2(у2—1) = (u22 + 2u2v2)Tc2,
(11) s = (u22 + 2v22-jr 2u2v2)Tc2,
Roczniki PTM — P ra ce M atem atyczne X III 8
114 W. Ze l ic h owi cz
where u1,v 1,Tc1, resp. u2, v 2, &2 are positive integers, chosen in snch a way that (7) or (10) are satisfied.
Putting u\hx = mx, 2v\lcx = nx, 2uxvxk x = tx and likewise 2v\lc2 — и\Ъ2 = n2, 2u2v2~k2 = t2 we get in the both cases, all the solutions of (5) in the form
(12) r = m + Z,
(13) 2(y2- l ) = n + t,
(14) s — mĄ-n + t,
with the condition
(15) 2mn = Z2,
where m , n , t are any positive integers, satisfying (12)-(15).
As t is even, we can write
(16) Z = 2%
where Z is odd and i > 1
Then condition (15) becomes mti = 22l~1l2. Let dxd2 to be any pos
itive integers snch that dxd2 and assume
(17) m — 21 dxi
where j > 0, and
(18) n = 22i~j~xd2.
Then owing to (13) and (16) n is even and consequently 2 i—j — 1 > 1 Substituting (16), (17) and (18) to (12), (13) and (14) we get
(19) r — 21 dx-\-2i l 1
2 = 2^-1 - 2 ^ 1Z+ 1 ,
(20) . . . ,
8 = 2 Ч + 22г' ?_Ч + 2гг.
Obviously we restrict ourselves to positive value for x2 in equation (3).
Hence
(21) ' x2 = 22i- j- 2d2+1.
Substituting (21) and (19) to (2) we get (22) z2 = 2 4 + 2*Z + 22W- 2d2+ l .
Because x is even we have j = 2i — 2. Then equations (20), (21) and (22) become
(23) ®* = d (.+ l,
(24) y2 ^ t f - ' l + d . + l , (25) г* = 22i' 2d1 + 2i l + d2+ l .
E q u ation (x2 — 1 ) (s2 — 1 ) = (у2 — l ) 2 115
We claim that i ^ 3.
The expression c?2+ l , as the square of x, where x is even, is divis
ible by 4, but 21-1Z + d2+ l is also the square of y, where у is even, and so is divisible by 4. Consequently 2г~11 is divisible by 4 and so we have i > 3.
Since x, у , z are distinct and even, we may write (26) x = 2aa, y = 2 pb, z = 2Yc,
where a, /3, у > 1 and а,Ъ , c are all odd. Owing to (23), (24) and (25) we have
(27) z2- y 2 = 2 2vc2- 2 wb2 = 2i~1(2i~1d1+ l ), (28) z2- ^ 2 = 22V - 2 2V = 2i (2i~2d1+ l) , (29) y2~ x 2 = 2 2V - 2 2V = 2*” 1?.
As i > 2, the numbers 2l-1d1 + Z, 2t_2d1 + Z, I are odd.
First we claim that all three of a , ft, у in (26) cannot be distinct.
Suppose that they are distinct. Then from (27) and (29) we would have min(2/?, 2y) = min(2a, 2/3) = i —1 = 2/?
and so the integer i would be odd. This is impossible because from (28) we have г — min(2a, 2y) = 2k.
The case a = /3 Ф у is also impossible, since from (27) and (28) it would follow i = %—1.
The case a — у Ф fi is impossible too, since from (29) we would get i —1 = min(2a, 2/3) and from (28), it would follow 2a < 7 —3 as c2—a2
= 8&.
Similarly the assumption fi = у Ф a would lead also to a contra
diction. Thus we have to consider the only possibility
(30) a = /3 = у =% .
I t is clear that for the existence of an integral solution of (1), with all three of x , y , z distinct and even, it is necessary that there exist positive integers x, у , z, D , X and y. satisfying all the equations
(31) x2—m 2 = 1, z * - D y 2 = 1, у2 = Ш у ф 1,
where D > 1 is a prime or a product of distinct and odd primes and Я Ф у. Therefore we define Я, у for given x, у , z in such a way that the above conditions are satisfied.
Taking into account (23), (24) and (25) we have d% — DX2,
d2 + 2i~1l = BXy, 22i~2d1 + 2i l + d2 = B y 2.
(32) (33) (34)
116 W. Z e l ic h owi cz
Substituting (32) into (33) we get
(35) 2<_1Z = ВЦ/л — Л).
Similarly from (32), (35) and (36) we get 22l~2dx — В (у — Л)2.
I t follows that exist such odd integer d that dx — Dd2 and
(36) /и-Л = 2l- l d.
Substituting (33) to (35) we have l = Did. Now substituting (36) with (30) to (31) we get
(37) 22V = DA2+ 1 ,
(38) 22V = J V + 1 ,
(39) 22’ 62 = DA^+1.
Before completing the proof of our theorem we have to establisch the following two lemmas.
Le m m a 1. Let x0, A0 be the smallest solution o f the Pell equation
(40) х2- В Л 2 = 1
(excluding x0 = 1, A0 = 0), where both В and x0 are odd, then in any solu
tion, x is odd.
P ro o f. If x and В are both odd, then A has to be even and from the general formula (see [2], p. 257)
(41) $п+ 1 == Лп_^_ i — Aq xn -I- Xq Aji
we get by induction that then хп+х is also odd.
Le m m a 2. Let x0, A0 be the smallest solution of (40), where D is odd, and let
(42) x0 = 2v6,
where 6 is odd and v > 1. Then any other even solution for x defined by (41) has an even index and is o f the form
(43) x,k = 2 vdNk
and
m = (22’ д \ к+1>кх1к) ъ ,
where Nk is odd and qk is even, for any even h, and qk is odd for any odd h.
P ro o f. If x0 is even and В is odd, then A0 is odd. Putting n = 0 in (41) we get from (42):
(45) хй = 22vó2+-DA2, Aj = 2’’+1<5A0.
E q u a tio n (x2 — 1 ) (z2 — 1 ) = (y2 — l ) 2 117
Similarly
(46) a?2 = 2V 6NX, = 22vd2 + 3BX20, (47) X2 = (22v d2ql -{-DAl)A0, qx = 3.
Thus we proved the lemma in the case ft = 1.
Suppose the lemma is true for some k. Then we have for 2ft+ 1 owing to (41)-(44):
(48) xik+i = 21’ d2Nk + 22’ d2m l q k + D k+1Xl’1+2, (49) X2k+1 = 2 'S N kX0 + 23’ d3qkXk + 2’’dDkXf‘+l.
(60) and
(51)
Similarly for 2 (ft+ l) we have from (41), (42), (48), (49)
*^2fc+2 = 2 6Nk_
Nk+1 = 22’’d2Nk + 22,+1d2m l q k + 2 Dk+lX?+2+ N kD%,
X-*k+‘l — (‘X2*’ Я?+2)Я0 , к <]2 к qk+1 = 2Nk + B K q k + 22Vd2qk+ D 4 -
I t is easy to verify, because of (45) and (48), that the solutions for x with odd indices are odd and from (46) and (48), that the solutions for x with even indices are even and of form (43). Similarly the solutions for Л with even indices are odd and of form (44).
From (51) it follows that if qk is odd, then qk+1 is even and conversely.
As qx = 3 thus for each odd ft, qk is odd, and for each even ft, qk is even.
This completes the proof of Lemma 2.
Because of (37)-(39) and both lemmas, for the existence of the solutions of (1) in distinct even integeres, it is necessary that there exists Pell equation with such two distinct even solutions for x that the re
spective values for A satisfy (39).
Suppose X2k and X2i are solutions of this kind. Substituting such solutions to (39) with (44), we get
22v b2 = [24v ó4 qk qz-f 22v d2 (qiBk X? + qkB l Xf)]BX20+ B k+l+1 X f+2l+2 + 1 . From (40) and (42) we have
(52) BX20 = 22v b2-
Substituting (52) into (50) we get
-1.
(53) 22’ b2 = 22’ d2l22’ $2qkql+ (q lDkX20k + qkD‘tZ)]D% +
k + l - i
i2Z\
+ ( —l ) s |fc+*+ 1 )(2 2,,<52)fc+z+1- sT 22"<52(ft + Z + l ) ± l + l .
118 W. Ze l ic h owi cz
If & + Z+1 is even, then the sum of the last two membres is equal to 2 the right-hand side of (53) is divisible only by 2 and the left-hand side is divisible at least by 4; thus we get a contradiction. If ft + Z - j - 1
is odd, then k-\-l is even and the numbers Jc and l are both even or both odd. Substituting to (36) the values у = /\2i, A — A2k defined by (44) we get, assuming Jc-\-l is even
The right-hand side is divisible only by 2 and so the left-hand side must be divisible only by 2 but this is impossible, as we consider the case i ^ 3.
So the assumption of the existence of a solution x , y , z of (1), where x, у , z are distinct and even, leads to contradiction.
This completes the proof of the theorem.
[1] A. S ch in z e l i W . S ie rp iń sk i, O równaniu се2— 2у2 = к, Wiadom. Mat. 7 (1964), рр. 229-232.
[2] W . S ie rp iń sk i, T eoria liczb, Warszawa 1950.
[3] — T eoria liczb, Część II, Warszawa 1959.
l-i
2l- ld = 22’ d2Xn(ql- q k) + X0^ (-1 )" (22"ó2/ 's =
R eferences
