ON THE CONJECTURE OF JESMANOWICZ CONCERNING PYTHAGOREAN TRIPLES

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1 BULL. AUSTRAL. MATH. SOC. VOL. 57 (1998) [ ] 11D61 ON THE CONJECTURE OF JESMANOWICZ CONCERNING PYTHAGOREAN TRIPLES MOUJIE DENG AND G.L. COHEN Le a, b, c be relaively prime posiive inegers such ha a 2 + b 2 = c 2. Jesmanowicz conjecured in 1956 ha for any given posiive ineger n he only soluion of (an) x + (bn) v = (en)* in posiive inegers is x y = z = 2. Building on he work of earlier wriers for he case when n = 1 and c = b + 1, we prove he conjecure when n>l,c = b+l and cerain furher divisibiliy condiions are saisfied. This leads o he proof of he full conjecure for he five riples (o,6,c) = (3,4,5), (5,12,13), (7,24,25), (9,40,41) and (11,60,61). 1. INTRODUCTION Le a, b, c be relaively prime posiive inegers such ha a 2 + b 2 c 2, and le n be a posiive ineger. Clearly, he Diophanine equaion (1) (na) x + (nb) y = (nc) z, has he soluion x y = z = 2. Wheher here are oher soluions in posiive inegers when n 1 has been invesigaed by a number of wriers. Sierpinski [6] showed here were no oher soluions when n = 1 and (a, b, c) = (3,4,5), and Jesmanowicz [2] ha here were no ohers when n = 1 and (a, b, c) = (5,12,13), (7,24,25), (9,40,41) and (11,60,61). He conjecured ha he equaion (1) has no posiive ineger soluions for any n oher han x y = z = 2. The general soluion of a 2 + b 2 = c 2 in relaively prime posiive inegers is of course well known o be a = u 2 v 2, b = 2uv, c u 2 + v 2, where u > v > 0, gcd (u, v) 1 and one of u, v is even, he oher odd. A number of oher special cases of Jesmanowicz's conjecure have since been seled. Lu [5] proved i when v = n = 1. In 1965, Dem'janenko [1] exended earlier resuls in several papers by proving he conjecure o be rue whenever n I or 2 and u = v + 1. Takakuwa and Asaeda (see [7]) have proved he conjecure in a number of special cases in which, in paricular, Received 5h January, 1998 The firs auhor would like o hank he second auhor and UTS for heir hospialiy and excellen working condiions during his say a UTS. We are graeful o he referee for several suggesions which led o improvemens in he presenaion of hese proofs. Copyrigh Clearance Cenre, Inc. Serial-fee code: /98 SA

2 516 M. Deng and G.L. Cohen [2] n = 1 and v = 1 (mod 4), and Takakuwa [8] has proved i when u is exacly divisible by 2, v = 3, 7, 11 or 15, and n = 1. More recenly, Le has verified he conjecure if n = 1 and 2 exacly divides uv and, in [3], c is a prime power, in [4], v = 3 (mod 4) and u ^ 81u. A more general conjecure has been considered by Terai (see [9]). He asks wheher he equaion (1) wih n 1 and a? + b q = c r, has any posiive ineger soluions oher han (x, y, z) = (p, q, r). In paricular, he has considered (p, q, r) (2,2,3) and (p,q,r) = (2,2,5). Some auhors and reviewers have sipulaed ha n = 1 in (1), bu his is apparenly no par of Jesmanowicz's conjecure. Nor is i a paricularly easy case when n > 1. In his paper, we shall ake a = 2fc + 1, 6 = 2k(k + 1), c 2k(k + 1) + 1, where A; is a posiive ineger, and will obain by compleely elemenary means cerain condiions on n under which he only posiive ineger soluion of he equaion (1) is x = y = z = 2. This will lead us o prove Jesmanowicz's conjecure in full for his generalisaion of he original five cases seled by Sierpinski and Jes"manowicz, ha is, for k {1,2,3,4,5}. For any ineger N > 1 wih prime facorisaion [J p *, we wrie C(N) = II P» i=i»=i All Greek and Roman leers in his paper denoe posiive inegers unless specified oherwise. The following wo heorems will be proved. THEOREM 1. Le a- 2Jfc + l, b = 2k(k+ 1), c= 2k(k+ 1) + 1, for some posiive ineger k. Suppose ha a is a prime power, and ha he posiive ineger n is such ha eiher C(b) \ n or C(n) \ b. Then he only soluion of he Diophanine equaion (an) x + (bn) y = (cn) z is x = y = z = 2. THEOREM 2. For each of he Pyhagorean riples (a, b, c) - (3,4,5), (5,12,13), (7,24,25), (9,40,41), (11,60,61), and for any posiive ineger n, he only soluion of he Diophanine equaion (an) x + (bn) v = (cn) z isx = y = z = 2. Theorem 2 is he confirmaion of Jesmanowicz's conjecure in he five cases saed, corresponding o k = 1, 2, 3, 4, 5, respecively, in Theorem 1. The firs case, when k = 1, is an immediae corollary of Theorem 1. The remainder of he proof of Theorem 2 uses Theorem 1 and special argumens in each of he cases k = 2, 3, 4, 5, wih no paern apparen. I is plausible ha similar approaches will be successful for he nex permissible cases k = 6, 8, 9, 11,..., bu he deails have no been carried ou. Three lemmas will be required. LEMMA 1. Le a = 2k + 1, b = 2k(k + 1), c = 2k(k + 1) + 1, for some posiive ineger k. The only soluion of he Diophanine equaion a x +b y = c* isx = y = z 2. This is Dem'janenko's resul, menioned above, when u v + l and n = 1.

3 [3] On he conjecure of Jesmanowicz 517 LEMMA 2. If z ^ max{x, y}, hen he Diophanine equaion a x + b y c*, where a, b and c are any posiive inegers (no necessarily relaively prime) such ha a 2 +b 2 = c 2, has no soluion oher han x y z = 2. PROOF: If z 1 hen z = y = 1 and (a + b) 2 > a 2 + b 2 = c 2, so a + b > c. Suppose z ^ 2 and, wihou loss of generaliy, ha x ^ y. If y 1, hen a x + b y = a + b<a 2 +b 2 = c 2 ^ c*. If y ^ 2, hen a x + &» < (a 2 ) y/2 + {b 2 ) y/2 ^ (a 2 + b 2 ) y ' 2 = c"^ c\ and here is sric inequaliy unless x y = z 2. LEMMA 3. If p is an odd prime and gcd (a, b) = 1, hen D Then PROOF: Le q be a prime divisor of a + b, so ha q \ a and b = a (mod q). ap a + b = a""1 - a p - 2 b b p ~ 1 = pa p - x (mod g). I follows, if q is a divisor of (a p + b p )/(a + 6), ha q p and ha p is an exac divisor of (a? + W)/{a + b). D 2. PROOF OF THEOREM 1 By Lemma 1, we may suppose n > 1, and by Lemma 2 ha z < max{x,y}. Of course, a 2 + b 2 = c 2. Noice also ha a 2 b + c, c = b + 1, b = k(a + 1), c=a;(a l) +o, and a, b, c are relaively prime in pairs. We also suppose ha equaion (1) holds, and will show ha his leads o a conradicion. There are wo main cases o he proof, depending on wheher gcd (n, c) = 1 or gcd (n, c) > 1, and numerous subcases in each case, indexed by a decimal numbering sysem. 1 Suppose gcd (n, c) = 1. We canno have x = y, since hen z < x and, from (1), we may wrie n x ~ z (a x + b x ) <?. Then gcd (n, c) > 1, a conradicion. 1.1 Suppose x > y, so ha we may wrie n v (n x ~ y a x + b v ) = rfc*. Then clearly z ^ y, so ha also z < x Suppose n\b v. Since we may wrie n x ~ y a x + b v n z ~ v c*, hen we canno have z > y, so z = y in his case, and n x ~ z a x + b z c*. Modulo a, we have k z = (-k) z, and gcd (o, k) = 1, so z is even. Wrie z = 2z\, so ha - b Zl ).

4 518 M. Deng and G.L. Cohen [4] The facors on he righ canno boh be divisible by a. Since a x > a z = a 2 * 1 = (c + b) Zl ^ c Zl 4- b Zl > c Zl b Zl, we have a conradicion Suppose n i". Then i is no he case ha C(n) \ b, so necessarily 8 8 C(b) I n. We may ake b = f\ r?* (prime facorisaion), so n = JJ r i' wih Vi ~ 1 i=l =l for each i 1,..., s. By he division algorihm, we wrie ^y = Uvi + Zj, say, where i ^ 1 and 0 ^ li < i>i, for i = 1,..., s If x > y + i for all i = 1,..., s, hen we may wrie (2) II r^+i) (n & -*> a* + f[ r'a = f{ r?' cf. =l ^i=l i=l ' i=l Since we canno have r^ \ c for any i 1,..., s, because gcd(n,c) 1, and since li < Vi for each i, i follows ha we mus have z y + i y + a, so x = = s =, say, and (2) reduces o f] r v i i{ - x ~ v ~ ) =i apparen ha li = 0 for i 1,..., s, so ha a x + Yl rj* = <?. I is hen =i say, where gcd (y',') = 1. Also, (2) furher reduces o (4) n x - z a x + l=c*. If z is even, hen, wriing z 2zi, we have n x ~ z a x (c Zl + l)(c zi - 1). However, a canno divide boh facors on he righ, and so his is impossible. a x >a z = a 2 ' 1 - (6 + c)* 1 > c zi + 1 > c* 1-1, Suppose now ha z is odd. Using (3), we have n = b v /* so ha, from (4), 6y'(a:-z) o i' = (^ _ f' = ^f, + y* _ f' Since f, ig eve]lj henfc Z) so (f + 1)* _ 1 is divisible by b exacly. Hence y'(x - z) = '. Since gcd (y',') = 1, hen y' = 1, x = z + ' and, from (3), y'. Since z = y + = y(l + '), hen ' is even and x is odd. Wrie x = 2x\ + 1. We have n x ~ z = ra 4 ' = 6"' = b, so ha, from (4), c z - 1 = 6a 1 = a(c - 1)(6 + c) Xl = a(c - l)(2c - I) 11. Modulo c, we have a( 1) X1 impossible, since c > a + 1. = 1, from which c (a + 1) or c (a 1). Bu his is

5 [5] On he conjecure of Jesmanowicz If x ^ y + i for a leas one i = 1,..., s, hen we can quickly obain a conradicion. The approach may be illusraed by aking x ^ y + i and x > y + U for i = 2,..., s (if s ^ 2). Then, adjusing (2), we may wrie r? X f[r? {v+i) fflr^1^^ a x + r? {v+lx) flr^ = J^rf z c*. =2 N=2 =l ' =l s Bu since x > z, his implies ha r\ PJ r" Z c z, which is he desired conradicion. =2 1.2 Suppose x < y and wrie (1) as n x (a x + n y ~ x b y ) = n z c z. Then clearly y > z ^ x If n\a x, hen we canno have z > x,so z = x and we have n v ~ z b v c z a z. Consider his equaion modulo 4 if k = 1, in which case a = 3, 6 = 4 and c = 5, and modulo k +1 if k > 1. In boh cases, we conclude ha z mus be even. Wrie z = 2z\. If k = 1, hen n»-*4» = Z = (5*i i)(5 Zl )- The facors on he righ are boh even bu canno boh be divisible by 4. Hence one of hem is divisible by 2 2 "- 1. Bu 22j/-l > 22z-l = 24Z!-1 ^ 23Z! = (5 + 3)*1 -. 5*1 + 3*1 > 5*1 _ 3*1. We have a conradicion. Suppose k > 1. We have n y ~ z b v (c Zl +a Zl )(c Z1 - a zi ), and we observe ha b = 2A;(fc + 1), fc (c - a) (c*! - a z i) and gcd(c zi + a z i,c z i - a z i) = 2. If Z! is even, or if z\ is odd and k is even (in which case, a = c = 1 (mod 4)), hen if 1 + a Zl is divisible by 2 bu no by 4, so ha 2»~ 1 J!/ (c* 1 - a Zl ). However, Zl > (2fe 2 + 2fe + 1) Z1 = c z > > c z i - a z S which is also a conradicion. If z\ and k are boh odd, hen, since c = -a = 1 (mod (fc + 1)), we have (k + 1) (c 2^ + a z >) and 4f(c*i -a z i)- Hence 2"- 1 (A; + l) y (c 2 i +a z i). Bu 2«" 1 (ifc + l) v > (2(* + l)) z = i(4fc 2 + 8k + 4)* 1 ^ (2Jfc 2 + 4fc + 2) 21 = (c + a) Zl ><?!+ a z i, our final conradicion in his case Suppose n \ a x. Wrie a = p a, where p is prime, and n p". Also, wrie ax = f +1, where 0 ^ / < v.

6 520 M. Deng and G.L. Cohen [6] Suppose y > x +, and wrie (1) as n x+ (p' + n y ~ x ~*b y ) = n z c*. From his, i follows ha z = x + and I = 0, so ha n v ~ z b v = <? - 1. If z is odd, hen, as in he las paragraph of , c* 1 is exacly divisible by b. Bu y > z, so y > 2 and b 2 (c* - 1). Then z mus be even. Wrie z -2z x. We have c* = 2 (mod 6), from which (c Zl +1,6) = 2. Since n y ~ z b y (c* 1 + l^c* 1-1), we mus hen have / (c*i - 1). Bu hv h! 1 - > -y- = 2<c- a) Zl (c + a) zi ^ c»i + a^ > c^ - 1. This is a conradicion. If / ^ x +, hen wrie (1) as n y (n x+ ~ y p l + b y ) = n'c". Since y > z, we have n c z, a conradicion. 2 In he second main case, we suppose gcd (n, c) > 1. Wrie c = Yl Q?* (prime facorisaion). * =1 2.1 Suppose firs ha C(n) c, so ha we may wrie n = flqf', say, wih s ^ i=1 Pi ^ 1 for i = 1,..., s Suppose x y, so z < x. Prom (1), we have f so ha «=1 i=l =l i=s+l (5) a* + V- _ i=l =s+l I is clear from his ha (*iz 0i(x z) ^ 0 for each i 1,..., s. We shall show ha a\z j3\{x z) > a\. Suppose his is no rue. If = 1 hen s = 1 and g i z -0i( x - z > j- q i = C < a x + b x, conradicing (5). If > 1 hen, since Qi 1 ^ c/i2 < c 1 = b and Y\ if* ^ c/<li < c 1 = 6, we have i=2 =l»=»+! i=2 anoher conradicion. Thus o\z /?i(x z) > a\, and similarly aiz - /3<(i - z) > oii for i = 2,..., s. I follows from (5) ha (6) a x + b x = 0 (mod c).

7 [7] On he conjecure of Jesmanowicz 521 If x is odd, say x = 2xi + 1, hen a x + b x = aa 2x i xi = a(-l)* 1-1 (mod c). By (6), hen c (a 1) or c (a + 1), which is impossible since c> If x is even, say x = 2x\, hen, from (6), ( l) Xl +1 = 0 (mod c), so xi is odd. In ha case, a x + b x = (a 2 ) xi + (f 2 )* 1 is divisible by a 2 + b 2, and, since z < x implies x > 2, he quoien mus exceed 1. Furhermore, by (5), (a x + b x )/(a 2 + b 2 ) is divisible by qj, say, for some j = 1,...,. Since a 2 = -1 = b 2 (mod c), we have a+1. ± 2)j jad-i) = Xi = 0 (mod qj)j ha is, qj \ x x. Then a? q i+b 2q i divides a 2x i+b 2x i. Furhermore, (a 2q i + b 2g i)/(a 2 + b 2 ) divides a 2xi +b 2xi, and, from (5), mus be a produc of primes in {qi,..., q\ I follows hen from Lemma 3 ha gcd( a , (a 29 -* + b 2q i)/(a ) J = qj. However, i is clear ha (a 2q 3 + b 2q i)/{a 2 + b 2 ) > qj, and f] q 2 \ (a 2 + b 2 ), so we have a conradicion. =i Now suppose x > y. From (1), we may wrie i=l =l i=l If z ^ y hen q\ \ b, conradicing gcd (6, c) = 1, so z < y and we wrie (7) n x -y a x + by = f[q^z-^y- z) - J] <7 Q ' 2. i=l «=»+l Again we have a conradicion if ajz 0j (y z) > 0 for some j 1,..., s, since hen q } : 6, so f[ q iz ~ Pi{v ~ z) = 1. I follows ha s < * bu, since f] iv < c /ii <b,v/e «=1 1=8+1 have JJ «?" < 6 Z < 6 y < n*-^1 + 6", i=s+l which is hen a conradicion of (7). Similarly, we canno have x < y. 2.2 If C{n) \ c, hen we may wrie n = TI1TI2, where ri\ > 1 and gcd(ni,ri2) = gcd(ni,c) = If x = y hen (1) becomes n x n x (a x + b x ) = n\n\c*. Since z < x, his implies ha n\ \ n^c?, a conradicion.

8 522 M. Deng and G.L. Cohen [8] Suppose x > y, and wrie (1) as n\n y 2{n x - y a x + V) = nln^c*. If z ^ y hen gcd (6, c) > 1, since gcd (n, c) > 1, and his is a conradicion. If z < y hen ni I c z, and his is also impossible. Similarly, we canno have x < y. This complees he proof of Theorem 1. D 3. PROOF OF THEOREM 2 In he noaion of Theorem 1, we mus exend ha proof for he special cases k = 2, 3, 4 and 5, wihou he resricion ha C(b) \ n or C(n) \ b. In effec, we need only look o he case in he proof of Theorem 1, so ha we may assume gcd (n, c) = 1, x > y, n \ b y and C(b) \ n. The four values of k mus be considered in urn. For his purpose, we coninue he previous decimal indexing. 3 Take k = 2, so (a,b,c) = (5,12,13). The relevan assumpions are: x > y, n 12 y and 6 \ n. Then n is eiher a power of 3 or a power of Le n = 3 r, and wrie y = r +1 where 0 ^ I < r. If x > y +, hen we may wrie (1) as nv + (n x -v- 5 x + 3'4«) = n z 13 z. I follows ha z = y +, so ha nx-z 5 x + 3i 4 y = 13z and hen ha i = 0 since x > z Tneil) modulo 5, {-l) y = 3 Z, from which z mus be even. Wrie z = 2z x, so ha n x ~ z 5 x = (13* 1 + 2! ')( y ). The facors on he righ canno boh be divisible by 5, and, noing ha z = y +1 > y, 5 X > 5 Z = 25 Z1 > 13 Z1 + 4* 1 > y > 13* 1-2 y, so we have a conradicion. If x ^ y +, hen we may wrie (1) as n x (5 x + 3 l n y+ - x 4 y ) = n z 13 z. This is clearly impossible, since x > z. 3.2 Now le n = 2 s, and wrie 2j/ = s + I, where 0 ^ / < s. As in 3.1, we easily show ha we canno have x ^ y +, so x > y + and we may wrie ny+( n x-y- 5 x + 2/ 3 y) _ nz l3 z This i mp i ies na z = y +, and hen ha I = 0, so (8) n x ~ z 5 x + 3 y = 13 Z. Then 3^ = 3 Z (mod 5), so y and z are boh even or boh odd. If 4 n x ~ z, hen (8), considered modulo 4, shows ha y is even. If n x ~ z = 2, hen (8), considered modulo 3, shows ha x is odd so ha z = x - 1 is even. Thus we may pu z = 2zi and y = 2/i, and hen n x - z 5 x = (13 Zl + 3 s 'i)(13 Zl - 3 yi ). As in 3.1, we may show his o be impossible. 4 Now ake k 3, so (a,b,c) = (7,24,25). We are assuming ha x > y, n \ 2\ y and 6 \ n, so ha again n is a power of 3 or a power of 2.

9 [9] On he conjecure of Jesmanowicz Suppose n 3 r, and y r +1 where 0 < I < r. As in 3.1, we see ha we mus have x > y +, and, as before, ha z ~ y + and 1 = 0. Then n x ~ z 7 x +8 y = 25 Z. Considering his equaion modulo 3, his implies ha we may wrie y = 2yi so ha nx-z 7 x _ ( 5 z + 8 yi )( 5 * - 8 yi ). However, 7 canno divide boh facors on he righ and 7* > V = 7 49» 1 > 5*(25 W + 8" 1 ) ^ 5 y+ + 8 yi = 5 Z + 8" 1 > 5 Z - 8 yi, so we have a conradicion. 4.2 If n = 2 s, hen, very much as in 3.2, we again obain a conradicion. 5 Nex, ake k = 4, so (a,b,c) = (9,40,41). We are assuming ha x > y, n \ 40 y and 10 \ n, so ha n is a power of 5 or a power of Suppose n = 5 r, and y = r + I where 0 ^ I < r. Again, we mus have x > y +, so ha, from (1), n y+ (n x -y- 9 x + 5 l 8 v ) = n z 41 z, and his implies ha z = y +, and hen ha 1 = 0. The equaion n x ~ z 9 x + 8 V = 41 Z, considered modulo 5, shows ha y is even, and hen, considered modulo 3, ha z is even. Wrie y = 2j/i and z = 2zi, so ha we have n x ~ z 9 x = (41 z i + 8! ' 1 )(41 il - 8 Vl )- The facors on he righ canno boh be divisible by 3, and 9 X > 9 Z = 81 Z1 > 41*1 +g*i > 41 z i +8^ > 41 z i -8 yi, so we have a conradicion. 5.2 If n = 2 s, hen, again as in 3.2, we obain a conradicion. 6 In our final case, ake k = 5, so (a,b,c) = (11,60,61). We assume ha x > y, n 60 y and 30 \ n. We need o consider he possibiliies n = 3 ri, n = 5 r2, n = 2 s, n = 3 ri 5 r2, n = 2 s 3 ri, n = r2, and he proofs in each of hese cases follow lines similar o he above. This complees he proof of Theorem 2. D REFERENCES [1] V.A. Dem'janenko, 'On Jesmanowicz' problem for Pyhagorean numbers', (in Russian), Izv. Vyssh. Ucebn. Zaved. Ma. 48 (1965), [2] L. Jesmanowicz, 'Several remarks on Pyhagorean numbers', (in Polish), Wiadom. Ma. 1 (1955/56), [3] L. Maohua, 'A noe on Jesmanowicz' conjecure', Colloq. Mah. 69 (1995), [4] L. Maohua, 'On JeSmanowicz' conjecure concerning Pyhagorean numbers', Proc. Japan Acad. Ser. A Mah. Sci. 72 (1996), [5] W.T. Lu, 'On he Pyhagorean numbers An 2-1, 4n and An 2 + 1\ (in Chinese), Ada Sci. Naur. Univ. Szechuan 2 (1959), [6] W. Sierpinski, 'On he equaion 3* + 4" = 5", (in Polish), Wiadom. Ma. 1 (1955/56), [7] K. Takakuwa, 'On a conjecure on Pyhagorean numbers. IIP, Proc. Japan Acad. Ser. A Mah. Sci. 69 (1993),

10 524 M. Deng and G.L. Cohen [10] [8] K. Takakuwa, 'A remark on Je^manowicz' conjecure', Proc. Japan Acad. Ser. A Mah. Sci. 72 (1996), [9] N. Terai, 'The Diophanine equaion a x + V = c z. II', Proc. Japan Acad. Ser. A Mah. Sci. 71 (1995), Heilongjiang Nongken Teachers' College A Cheng Ciy People's Republic of China PO Box 123 Broadway NSW 2007 School of Mahemaical Sciences Universiy of Technology, Sydney Ausralia g.cohen@mahs.us.edu.au