A Note on Je{ś}manowicz' Conjecture for Non-primitive Pythagorean Triples
aa r X i v : . [ m a t h . N T ] F e b A note on Je´smanowicz’ conjecture for non-primitivePythagorean triples
Van Thien Nguyen, Viet Kh. Nguyen, Pham Hung Quy
Abstract
Let ( a, b, c ) be a primitive Pythagorean triple parameterized as a = u − v , b =2 uv, c = u + v , where u > v > n , the Diophantine equation( an ) x + ( bn ) y = ( cn ) z has only the positive integer solution ( x, y, z ) = (2 , , x, y, z ) = (2 , ,
2) with n > v = 2 , u is an odd prime. Asan application we show the truth of the Je´smanowicz conjecture for all prime values u < MSC : 11D61, 11D41
Keywords : Diophantine equations, Non-primitive Pythagorean triples, Je´smanowicz’conjecture
1. Introduction
Let ( a, b, c ) be a primitive Pythagorean triple. Clearly for such a triple with 2 | b onehas the following parameterization a = u − v , b = 2 uv, c = u + v with(1 . u > v > , gcd( u, v ) = 1 , u + v ≡ Conjecture 1.1.
For any positive integer n , the Diophantine equation (1 .
2) ( an ) x + ( bn ) y = ( cn ) z has only the positive integer solution ( x, y, z ) = (2 , , . The primitive case of the conjecture ( n = 1) was investigated thoroughly. Although theconjecture is still open, many special cases are shown to be true. We refer to a recent survey[11] for a detailed account. 1uch less known about the non-primitive case ( n > x, y, z, n ) of (1 .
2) is called exceptional if ( x, y, z ) = (2 , ,
2) and n >
1. For a positiveinteger t let P ( t ) denote the set of distinct prime factors of t , and P ( t ) − their product. Thefirst known result in this direction was obtained in 1998 by M.-J. Deng and G.L. Cohen ([5]),namely if u = v + 1, a is a prime power, and either P ( b ) | n , or P ( n ) ∤ b , then (1 .
2) has onlypositive integer solution ( x, y, z ) = (2 , , .
2) to have exceptional solutions.
Theorem 1.2. ([10]) If ( x, y, z, n ) is an exceptional solution of (1 . , then one of thefollowing three conditions is satisfied: ( i ) max { x, y } > min { x, y } > z , P ( n ) $ P ( c );( ii ) x > z > y , P ( n ) ⊂ P ( b );( iii ) y > z > x , P ( n ) ⊂ P ( a ) . However, as noted in [15] by H. Yang and R.-Q. Fu, the case x = y > z is not completelyhandled by the arguments used in [10]. Furthermore they completed the unhandled case ([15],Theorem 1) based on a powerful result of Zsigmondy ([18], cf. [3], [4]). In fact one can givea unified simple proof of Theorem of Le-Yang-Fu (Theorem 1.2) by using a weaker versionof the Zsigmondy theorem as stated in Lemma 3 of [5] ( cf. § . v = 1 , u = 2 k , k = 1 , , . . . Most recently X.-W. Zhang and W.-P. Zhang ([17]), and T. Miyazaki ([12]) indepen-dently proved Conjecture 1.1 for the (infinite) family (1 . v = 2 , u is an odd prime which wasknown recently for few values u : u = 3 ([5]), u = 5 – by Z. Cheng, C.-F. Sun and X.-N. Du, u = 7 – by C.-F. Sun, Z. Cheng, and by G. Tang, u = 11 – by W.-Y. Lu, L. Gao and H.-F.Hao ( cf. [11] for references). Let’s formulate our main results. We rewrite (1 .
2) as(1 .
4) [( u − n ] x + (4 un ) y = [( u + 4) n ] z An arithmetical argument (given in Lemma 3.5 below) shows that u − u − u u , gcd( u , u ) = 1, so that there are three possibilities toconsider: u ≡ ± , Theorem 1.3. If ( x, y, z, n ) is an exceptional solution of (1 . and u ≡ ± ,then y is even. In view of Theorem 1.3 the possibility u ≡ − x, y, z are even, which is in general impossible by an auxiliary argument (Lemma 3.6below).Let ν q ( t ), for a prime q , denote the exponent of q in the prime factorization of t , andlet (cid:16) m (cid:17) denote the Jacobi quadratic residue symbol.2 heorem 1.4. If ( x, y, z, n ) is an exceptional solution of (1 . , then one of the follow-ing cases is satisfied (1) ν ( u −
1) = 3: ( ν ( x ) , ν ( y ) , ν ( z )) = (0 , ≥ , u admits a proper decomposition u = t t , gcd( t , t ) = 1 and t , t ≡ satisfying certain special Diophantineequations ;(2) u ≡ , u = w s , where s = ν ( z − x ) − ν ( x ) and either of the following (2 . w ≡ ± ν ( x ) , ν ( y ) , ν ( z )) = (0 , ≥ , u ≡ (cid:16) u p (cid:17) = (cid:16) wp (cid:17) , ∀ p | ( u + 4) and (cid:16) wp (cid:17) = (cid:16) u + 4 p (cid:17) , ∀ p | u ;(2 . w ≡ ± ν ( x ) , ν ( y ) , ν ( z )) = ( β, , β ) , β ≥ u ≡ ± (cid:16) wp (cid:17) = 1 , ∀ p | ( u + 4) and (cid:16) wp (cid:17) = (cid:16) up (cid:17) , ∀ p | u . Moreover, if u ≡ , then w can not be a square. Corollary 1.5.
Conjecture 1.1 is true for v = 2 , u – an odd prime < . Let’s explain the ideas in proving our main results. As for Theorem 1.3 and Theo-rem 1.4 we exploit a total analysis of Jacobi quadratic and quartic residues. In the case u ≡ u = t t , which leads to certainspecial Diophantine equations. Theorem 1.4 helps us substantially in reducing the verifica-tion process, as the possibility u ≡ u <
100 in proving Corollary 1.5.The paper is organized as follows. In § §
4. The case u ≡ §
5. Theverification for u <
100 in Corollary 1.5 will be given in the final section.
Acknowledgement.
The authors would like to thank the referee for many valuablecomments and suggestions greatly improving the content of the paper.
2. A Simple Proof of Theorem 1.2
We shall use the following weaker version of Zsigmondy’s theorem.
Lemma 2.1 ( cf. [5], Lemma 3). For
X > Y > co-prime integers, (1) if q is a prime, then gcd (cid:0) X − Y, X q − Y q X − Y (cid:1) = 1 , or q ;(2) if q is an odd prime, then gcd (cid:0) X + Y, X q + Y q X + Y (cid:1) = 1 , or q. ℓ r is a common primepower divisor of X − Y and ( X q − Y q ) / ( X − Y ). Clearly(2 . X q − Y q X − Y ≡ ℓ r )On the other hand from the fact that X ≡ Y (mod ℓ r ) it follows(2 . X q − Y q X − Y = X q − + X q − Y + · · · + XY q − + Y q − ≡ qY q − (mod ℓ r )Since ℓ ∤ Y , (2 . − (2 .
2) imply that ℓ = q , and r = 1. (cid:3) Remark 2.2.
Part (1) is a special case of Theorem IV in [3].
Lemma 2.3.
For a prime divisor q of ( X − Y ) and positive integer β (2 . ν q ( X q β − Y q β ) = β + ν q ( X − Y ) Proof.
Applying part (1) of Lemma 2.1 β times one hasgcd (cid:0) X q β − − Y q β − , X q β − Y q β X q β − − Y q β − (cid:1) = q ; · · · gcd (cid:0) X − Y, X q − Y q X − Y (cid:1) = q. Hence the formula (2 . . (cid:3) In view of Lemma 2 of [5] there are no exceptional solutions with z ≥ max { x, y } , so asin [10] we have to eliminate the following three cases: (I) x > y = z ; (II) y > x = z ; (III) x = y > z. (I) x > y = z : dividing both sides of (1 .
2) by n y one gets(2 . a x n x − y = c y − b y By considering mod c + b , and taking into account ( c + b )( c − b ) = a , one sees that y must be even, say y = 2 y . Now put X = c , Y = b , so X ≡ Y (mod a ) , gcd( Y, a ) = 1.Taking mod a and in view of (2 . ≡ X y − Y y X − Y = X y − + X y − Y + · · · + XY y − + Y y − ≡ y Y y − (mod a )4ne concludes that a | y .For any q ∈ P ( a ) let β = ν q ( y ), so that y = q β y with q ∤ y . Putting U = X q β , V = Y q β for short, we have(2 . X y − Y y = ( U − V )( U y − + U y − V + · · · + U V y − + V y − )and(2 . U y − + U y − V + · · · + U V y − + V y − ≡ y V y − q )Lemma 2.3 and (2 . , (2 .
6) imply that(2 . ν q ( X y − Y y ) = ν q ( U − V ) = β + 2 ν q ( a )In view of (2 .
4) the equality (2 .
7) means that a x − | y in contradiction with y = y/ < a x − as x > y, a > (II) y > x = z : Similarly dividing both sides of (1 .
2) by n z one gets(2 . b y n y − x = c x − a x Arguing as above with mod c + a , one sees that x must be even, say x = 2 x . Put X = c , Y = a . Considering mod b and from (2 .
8) it follows that b | x . So ν q ( X x − Y x ) = ν q ( x ) + 2 ν q ( b ) for any q ∈ P ( b ), therefore b y − | x in contradiction with x = x/ < b y − as y > x, b > (III) x = y > z : dividing both sides of (1 .
2) by n z one gets(2 .
9) ( a x + b x ) n x − z = c z First we claim that x must be even . Indeed, if x is odd, then from (2 .
9) it follows thatthere is an odd prime q ∈ P ( a + b ) ∩ P ( c ), so q ∈ P ( ab ), as c = a + b . A contradictionwith gcd( a, b ) = 1.Writing now x = 2 x one sees that x must be odd . Since otherwise for an odd prime q ∈ P ( a x + b x ) ∩ P ( c ) taking mod q and by (2 . ≡ a x + b x = a x + ( c − a ) x ≡ a x (mod q )one gets a contradiction with gcd( a, c ) = 1.Now from (2 .
9) we see that(2 .
10) ( a ) x + ( b ) x a + b = c z − n x − z > x > z ≥
2. So there is an odd prime q ∈ P ( c ) dividing (( a ) x + ( b ) x ) / ( a + b ).Considering mod q and taking into account a ≡ − b mod q, q ∤ a one has0 ≡ ( a ) x + ( b ) x a + b = ( a ) x − − ( a ) x − b + · · · − a ( b ) x − + ( b ) x − ≡ x a x − (mod q )5ence q | x , and so (( a ) q + ( b ) q ) | (( a ) x + ( b ) x ). Applying part (1) of Lemma 2.1 weget(2 .
11) gcd (cid:0) a + b , ( a ) q + ( b ) q a + b (cid:1) = q On the other hand from (2 .
10) one knows that (( a ) q + ( b ) q ) / ( a + b ) is a product ofprimes in P ( c ). It is easy to see that (( a ) q + ( b ) q ) / ( a + b ) > q . So either ν q (cid:0) (( a ) q +( b ) q ) / ( a + b ) (cid:1) ≥ ν q ( a + b ) ≥
2, or both of them must have another common primefactor in P ( c ), a contradiction with (2 .
3. Preliminary reduction
We need some reduction of the problem. The following result is due to N. Terai.
Lemma 3.1 ([13]).
Conjecture 1.1 is true for n = 1 , v = 2 . Because of Lemma 3.1 we will assume henceforth n > i ) in Theorem 1.2. Lemma 3.2 If ( x, y, z, n ) is an exceptional solution, then either x > z > y , or y >z > x. Note that the proof of Lemma 3.2 relies essentially on the condition n >
1. It could beinteresting to find a proof of this result for the case n = 1.Furthermore, in the case when u is an odd prime and v = 2, H. Yang, R.-Q. Fu [16]succeeded to eliminate the possibility ( ii ) in Theorem 1.2. Lemma 3.3.
Suppose that u is an odd prime and v = 2 . Then equation (1 . has noexceptional solutions ( x, y, z, n ) with x > z > y . Lemma 3.4.
For a positive integer w (1) if ν ( w ) ≥ , then ν [(1 + w ) x −
1] = ν ( w ) + ν ( x );(2) if ν ( w ) = 1 and x is odd, then ν [(1 + w ) x −
1] = 1;(3) if ν ( w ) = 1 and x is even, then ν [(1 + w ) x −
1] = ν (2 + w ) + ν ( x ) . In particular ν [(1 + w ) x −
1] = 2 + ν ( x ) , if w ≡ or if w ≡ and x is even.Proof. (1) The conclusions of Lemma 3.4 are true trivially for x = 1. Assuming now x ≥ .
1) (1 + w ) x − w ( C x + C x w + · · · + C x − x w x − + C xx w x − )Clearly ν ( j ) ≤ j − j = 2 , · · · , x, and so ν ( C jx w j − ) = ν (cid:16) xj C j − x − w j − (cid:17) ≥ ν ( x ) + j − > ν ( x ) , ν ( w ) ≥
2. Hence the conclusion follows from taking ν ( · ) on both sides of (3 . . (2) Obvious from (3 . C x + C x w + · · · + C x − x w x − + C xx w x − is odd in this case.(3) Writing x = 2 x we have(3 .
2) (1 + w ) x − w ) x − w ) x + 1]If x is odd , i.e. ν ( x ) = 1, then ν [(1 + w ) x −
1] = 1 by the part (2) above, and ν [(1 + w ) x + 1] = ν (2 + w ), as(1 + w ) x + 1 = (2 + w )[(1 + w ) x − − (1 + w ) x − + · · · − (1 + w ) + 1]and (1 + w ) x − − (1 + w ) x − + · · · − (1 + w ) + 1 is odd .If x is even , then ν [(1 + w ) x + 1] = 1, since(1 + w ) x + 1 = 2 + C x w + C x w + · · · + C x − x w x − + w x . Therefore ν [(1 + w ) x −
1] = ν [(1 + w ) x −
1] + 1 by (3 . (cid:3) The following claims play a central role in the next sections.
Lemma 3.5. If ( x, y, z, n ) is an exceptional solution of (1 . , then u − admits aproper decomposition u − u u , gcd( u , u ) = 1 and with one of the following conditionssatisfied: (1) u ≡ and ν ( z ) = ν ( u −
1) + ν ( x ) − u ≡ , ν ( z ) = ν ( u + 1) + ν ( x ) − , and ν ( x ) ≥ u ≡ , u is a square and and ν ( z ) = ν ( x ) . Proof.
In view of Lemmas 3.2, 3.3 we may assume the existence of an exceptionalsolution with y > z > x (the case ( iii ) of Theorem 1.2). Dividing both sides of (1 .
4) by n x one gets(3 .
3) ( u − x = [( u + 4) z − (4 u ) y n y − z ] n z − x It is easy to see that gcd( u + 4 , n ) = 1. So (3 .
3) is equivalent to the following system(3 . ( ( u + 4) z − (4 u ) y n y − z = u x n z − x = u x with u − u u , gcd( u , u ) = 1 . The system (3 .
4) can be rewritten as(3 .
5) ( u + 4) z − y u y n y − z = u x or equivalently(3 . ′ ) [( u + 4) z − − ( u x −
1) = 2 y u y n y − z k ( z − x ) = mx , and n m = u k .Clearly u >
1. Assume now u = 1. As u ≡ ν ( · ) both sidesof (3 .
5) and by (1) of Lemma 3.4 we have ν [( u + 4) z −
1] = 2 + ν ( z ) < y . So (3 . ′ ) isinconsistent. So u > . ν ( z ) = ν ( u x − − u ≡ ν ( z ) = ν ( u −
1) + ν ( x ) − u ≡ x is odd , then by (2) of Lemma 3.4: ν ( u x −
1) = 1, impossible by(3 . . ′ ) is inconsistent.If u ≡ x is even , then by (3) of Lemma 3.4: ν ( u x −
1) = ν ( u + 1) + ν ( x ).Hence by (3 .
6) one gets ν ( z ) = ν ( u + 1) + ν ( x ) − u ≡ ν ( u x −
1) = 1, if x is odd (by (2) of Lemma 3.4), and ν ( u x −
1) = 2 + ν ( x ), if x is even (by (3) of Lemma 3.4). Hence for (3 .
5) to be consistentone has necessarily ν ( z ) = ν ( x ), which implies ν ( z − x ) ≥ ν ( x ) + 1. So from the secondequation of (3 . n z − x = u x it follows that u must be a square, hence u ≡ u u ≡ u u = u − ≡ u ≡ ν ( u x −
1) = 2 + ν ( x ),and by the same reason ν ( z ) = ν ( x ). Hence the system (3 .
4) is inconsistent, if u is not asquare. (cid:3) Lemma 3.6.
In the notations above if x, y, z are even, then (3 . is inconsistent.Proof . In this case we can rewrite (3 .
5) in the form of Pythagorian equation (cid:0) u x/ (cid:1) + (cid:2) y u y/ n ( y − z ) / (cid:3) = (cid:2) ( u + 4) z/ (cid:3) . Hence ( cf. (1 . X, Y , say with 2 | Y such that(3 .
7) ( u + 4) z/ = X + Y (3 .
8) 2 y u y/ n ( y − z ) / = 2 XY In view of [17], Lemma 2.2, equation (3 .
7) has solutions(3 . u + 4 = A + B , | B (3 . ν ( Y ) = ν ( z/
2) + ν ( B )Since u + 4 ≡ .
9) that ν ( B ) = 1. From (3 .
8) we have ν ( Y ) = y − .
10) implies y = ν ( z ) + 18 contradiction with y > z . (cid:3) Corollary 3.7.
In the notations above if y, z are even and (3 . is consistent, then x is odd and u ≡ . Moreover u admits a proper decomposition u = t t such that gcd( t , t ) = 1 and (3 . t x + t x = 2( u + 4) z/ (3 . t x − t x = 2 y +1 u y/ n ( y − z ) / (3 . ν ( t x −
1) = ν ( t x −
1) = ν ( u x − − Proof . By Lemma 3.6 x is odd. In fact one can rewrite (3 .
5) as A · B = u x with gcd( A, B ) = 1 , where A = ( u + 4) z/ − y u y/ n ( y − z ) / , B = ( u + 4) z/ + 2 y u y/ n ( y − z ) / . Hence(3 . A = t x , B = t x with u = t t and gcd( t , t ) = 1If t = 1, then by (1) of Lemma 3.4: ν [( u + 4) z/ −
1] = 2 + ν ( z/ < y = ν (2 y u y/ n ( y − z ) / ). So A = 1 is impossible.Now from (3 .
14) we have two possibilities:(1) z/ t ≡ t ≡ z/ t ≡ t ≡ u ≡ . − (3 .
13) follow immediately from (3 . (cid:3) Corollary 3.8.
In the situation of Corollary 3.7 we have t , t ≡ and ν ( u −
1) = 3 . Proof . We will show that z/ ν ( u −
1) = ν ( u x −
1) = ν ( A −
1) + 1 = 3.Assume on the contrary that ν ( z ) ≥
2. In view of (3 .
14) one has x ≥
3, as t < t
4. We claim that x >
3. Indeed, if x = 3, then n = u by (3 . z = 4 by B = t x of (3 . y = 6 as A = t x >
0. Now from the equation t x = A in (3 .
14) we seethat ( t , uu , u + 4) is a primitive solution of(3 . X + Y = Z Euler ([7], pp. 578–579) indicated a primitive parameterization for the Diophantineequation (3 .
15) with 3 ∤ Z, | Y as follows X = ( s − t )(3 s − t )(3 s + t ) , Y = 4 st (3 s − st + t )9ith s, t co-prime, 3 ∤ t and s t (mod 2). Hence 8 | Y which shows that t x = A in (3 . x ≥
4, then by Theorem 1.1 of [1], (3 .
11) is again impossible. (cid:3)
4. Proof of Theorem 1.3
The aim of this section is to show that the case u ≡ (cid:16) m (cid:17) , (cid:16) m (cid:17) we shall use in the following lemmas. Lemma 4.1.
For a prime p | ( u + 4) one has p ≡ and (cid:16) up (cid:17) = 1 .Proof . Since u ≡ − p ), so (cid:16) − p (cid:17) = 1, i.e. p ≡ (cid:16) up (cid:17) = (cid:16) up (cid:17) = (cid:16) u + u + 4 p (cid:17) = (cid:16) ( u + 2) p (cid:17) = 1 (cid:3) Lemma 4.2. If (3 . is consistent and u ≡ ± , then (cid:16) np (cid:17) = (cid:16) u p (cid:17) for anyprime p .Proof . Indeed, in this case by Lemma 3.5 ν ( z ) > ν ( x ). Hence ν ( z − x ) = ν ( x ), sowe have in (3 . ′ ) n m = u k with k, m odd, and therefore the conclusion of Lemma 4.2. (cid:3) We are ready now to prove Theorem 1.3. Let p | ( u + 4). By taking (cid:16) p (cid:17) on (3 .
5) andusing Lemmas 4.1, 4.2 one sees that(4 . (cid:16) u p (cid:17) x = (cid:16) np (cid:17) y − z = (cid:16) u p (cid:17) y − z = (cid:16) u p (cid:17) y (as z is even). Now taking the product of (4 .
1) over all (not necessarily distinct) primedivisors p | ( u + 4) we have(4 . (cid:16) u u + 4 (cid:17) x = Y p | ( u +4) (cid:16) u p (cid:17) x = Y p | ( u +4) (cid:16) u p (cid:17) y = (cid:16) u u + 4 (cid:17) y By the quadratic reciprocity law(4 . (cid:16) u u + 4 (cid:17) = (cid:16) u + 4 u (cid:17) = (cid:16) u (cid:17) = 1(4 . (cid:16) u u + 4 (cid:17) = (cid:16) u + 4 u (cid:17) = (cid:16) u (cid:17) = − u ≡ ± u ≡ ± . − (4 .
4) imply that (cid:16) u p (cid:17) y = ( − y =1, i.e. y must be even. (cid:3) Corollary 4.3.
The possibility u ≡ in Lemma 3.5 is not realized.Proof . Indeed, in this case ν ( z ) > ν ( x ) ≥
1, so (3 .
5) is inconsistent by Lemma 3.6. (cid:3)
Corollary 4.4.
In the case u ≡ of Lemma 3.5 we have ( ν ( x ) , ν ( y ) , ν ( z )) = (0 , ≥ , . Proof . By Lemma 3.5 and Theorem 1.3: y, z are even, hence x is odd by Lemma 3.6.From the proof of Corollary 3.8 it follows that ν ( z ) = 1. For a prime p | ( u + 4) by taking (cid:16) p (cid:17) on A = t x of (3 .
14) and using Lemma 4.1 one gets(4 . (cid:16) t p (cid:17) = (cid:16) np (cid:17) ( y − z ) / By the same reason of (4 .
4) we have (cid:16) t u + 4 (cid:17) = −
1, as t ≡ p | ( u + 4) such that(4 . (cid:16) t p (cid:17) = − . , (4 .
6) one concludes that ( y − z ) / (cid:16) np (cid:17) = − (cid:3) Remark 4.5.
One can have another proof of Lemma 3.6 as shown in several stepsbelow. Assuming y, z even, and arguing as in the proof of Corollary 3.7 one gets equations(3 .
14) together with (3 . − (3 . u ≡ t , t ):( i ) t ≡ , t ≡ ii ) t ≡ , t ≡ iii ) t ≡ , t ≡ iv ) t ≡ , t ≡ . u ≡ ± x even , hence ν ( z ) ≥ ν ( y ) = 1. Indeed, considering p | ( u + 4) and taking (cid:16) p (cid:17) on (3 .
5) one has byusing Lemmas 4.1, 4.2(4 . (cid:16) u p (cid:17) x/ = (cid:16) − p (cid:17) (cid:16) np (cid:17) ( y − z ) / = (cid:16) u p (cid:17) y/ , p ≡ − (cid:16) u p (cid:17) y/ , p ≡ z/ r denote the number of prime divisors p | ( u + 4), p ≡ r is odd , as u + 4 ≡ . − (4 . .
7) over all (not necessarily distinct) prime divisors p | ( u + 4) we get1 = (cid:16) u u + 4 (cid:17) x/ = ( − r (cid:16) u u + 4 (cid:17) y/ = − ( − y/ . Hence y/ y − z ) / p | ( u + 4) taking (cid:16) p (cid:17) onequation A = t x from (3 .
14) now gives us(4 . (cid:16) np (cid:17) = 1 (cid:18) = (cid:16) u p (cid:17) by Lemma 4 . (cid:19) On the other hand from (4 .
4) it follows that there exists a prime p | ( u + 4) such that (cid:16) u p (cid:17) = −
1, a contradiction with (4 . .
14) (and hence (3 .
5. The case u ≡ ν ( z ) = ν ( x ), hence from (3 .
4) it follows that u = w s , where s = ν ( z − x ) − ν ( x ). The following lemma can be proved similarly asLemma 4.2. Lemma 5.1. If (3 . is consistent and u ≡ , then (cid:16) np (cid:17) = (cid:16) wp (cid:17) for anyprime p .Proof . Indeed, in this case n m = w k with k, m odd by the above argument, and there-fore the conclusion of Lemma 5.1. (cid:3) Lemma 5.2. If x, z are even and (3 . is consistent, then y is odd and u ≡ .Moreover n admits a decomposition n = n n such that gcd( n , n ) = 1 and (5 . ( u x/ = u y n y − z − y − n y − z ( u + 4) z/ = u y n y − z + 2 y − n y − z Proof . By Lemma 3.6 y is odd. In view of Lemma 3.5 and Theorem 1.3 we are in thesituation (3) of Lemma 3.5. Now one rewrites (3 .
5) as C · D = 2 y u y n y − z with gcd( C , D ) = 2 , k D , where C = ( u + 4) z/ − u x/ , D = ( u + 4) z/ + u x/ . As 2 k D we obtain either(5 . C = 2 y − n y − z , D = 2 u y n y − z . C = 2 y − u y n y − z , D = 2 n y − z where n = n n , gcd( n , n ) = 1 and(5 . w = w w , n m = w k , n m = w k with k, m odd from Lemma 5.1. Note that this is not used in the proof here, we label it forconvenience in proving Proposition 5.5 below.Clearly (5 .
2) is equivalent to (5 . .
3) can’t happen byrewriting it as(5 . ( u x/ = n y − z − y − u y n y − z ( u + 4) z/ = n y − z + 2 y − u y n y − z which is impossible, since ( u + 4) z/ < y − u y . (cid:3) Lemma 5.3. If (cid:16) u u (cid:17) = 1 and u is a square, then u ≡ .Proof . We have obviously1 = (cid:16) u u (cid:17) = (cid:16) u u u (cid:17) = (cid:16) u − u (cid:17) = (cid:16) − u (cid:17) , so the conclusion of the lemma. (cid:3) Lemma 5.4.
In the notations of Lemma 5.1 we have (1) if w ≡ ± , then x, z are odd, y is even ;(2) if w ≡ ± , then x, z are even, y is odd.Proof . For a prime p | ( u + 4) by taking (cid:16) p (cid:17) on (3 .
5) and using Lemmas 4.1, 5.1 onesees that(5 . (cid:16) u p (cid:17) x = (cid:16) np (cid:17) y − z = (cid:16) wp (cid:17) y − z By taking the product of both sides of (5 .
6) over all (not necessarily distinct) primedivisors p | ( u + 4) and using the reciprocity law we have(5 . Y p | ( u +4) (cid:16) u p (cid:17) x = (cid:16) u u + 4 (cid:17) x = (cid:16) u + 4 u (cid:17) x = (cid:16) u (cid:17) x = ( − x (5 . Y p | ( u +4) (cid:16) wp (cid:17) y − z = (cid:16) wu + 4 (cid:17) y − z = (cid:16) u + 4 w (cid:17) y − z == (cid:16) w (cid:17) y − z = ( ( − y − z , w ≡ ± , w ≡ ± w ≡ ± . , (5 . − x = ( − y − z . Thus y must be even, as ν ( z ) = ν ( x ). In view of Lemma 3.6 x, z are odd.In the case w ≡ ± . , (5 .
8) we see that ( − x = 1,therefore x is even, and so is z . By Lemma 3.6 y must be odd. (cid:3) Proposition 5.5.
In the situation of Lemma 5.4 we have (1) if w ≡ ± , then u ≡ if w ≡ ± , then u ≡ ± . Moreover, if u ≡ , then w cannot be a square.Proof . (1) If w ≡ ± x, z are odd in view of Lemma 5.4. So by taking (cid:16) u (cid:17) on (3 .
5) one gets (cid:16) u u (cid:17) = 1, hence u ≡ w ≡ ± x, z are even, y is odd by Lemma 5.4. There are twosubcases to consider. I. x/ , z/ are odd . For a prime p | ( u + 4) by taking (cid:16) p (cid:17) on D = 2 u y n y − z from(5 . , (5 .
4) and using Lemmas 4.1, 5.1 one sees that(5 . (cid:16) u p (cid:17) = (cid:16) p (cid:17)(cid:16) n p (cid:17) = (cid:16) p (cid:17)(cid:16) w p (cid:17) = (cid:16) w p (cid:17) , p ≡ − (cid:16) w p (cid:17) , p ≡ p | ( u + 4), p ≡ Q p | ( u +4) (cid:16) p (cid:17) = −
1. Now taking the product of both sides of (5 .
9) overall (not necessarily distinct) prime divisors p | ( u + 4) and using the reciprocity law one has(5 . Y p | ( u +4) (cid:16) u p (cid:17) = (cid:16) u u + 4 (cid:17) = (cid:16) u + 4 u (cid:17) = (cid:16) u (cid:17) = − . Y p | ( u +4) (cid:16) p (cid:17)(cid:16) w p (cid:17) = − Y p | ( u +4) (cid:16) w p (cid:17) = − (cid:16) w u + 4 (cid:17) = − (cid:16) u + 4 w (cid:17) = − (cid:16) w (cid:17) Equalizing (5 . .
11) we get w ≡ ± . n ≡ ± .
1) it follows that u ≡ ± u ≡ w ≡ − . w can not be a square. II. x/ , z/ are even . If one takes (cid:16) u (cid:17) on the second equation of (5 . (cid:16) n u (cid:17) = 1.Now taking (cid:16) u (cid:17) on the first equation of (5 .
1) we get 1 = (cid:16) − u (cid:17)(cid:16) n u (cid:17) . Thus u ≡ (cid:3)
14s for Theorem 1.4 notice that the case u ≡ ± i.e. the case u ≡ .
5) and Lemma 5.1. (cid:3)
6. Proof of Corollary 1.5
In this section we shall apply results of previous parts for establishing the truth ofJe´smanowicz’ conjecture for u <
100 and v = 2. In view of Theorem 1.4 one has to consideronly two cases: u ≡ u ≡ Observation 6.1. If u ≡ and (3 . is consistent, then u > .Proof . Indeed, it was noted that x ≥ . ν ( z ) = 1, so z ≥
6, hence y ≥
8. From (3 .
12) it follows that2 y +1 | t − t , as x is odd. Since t , u are co-prime and ≡ t u ≥ · u > √ t t u ≥ p (2 + 5) · > (cid:3) Observation 6.2. If u ≡ and (3 . is consistent, then in fact u > .Proof . By Corollary 4.4 one knows 4 | y . We claim that y ≥
12. Assuming on thecontrary y = 8, then by the above z = 6. In view of (3 .
11) and [14] we must have x > x = 5, which gives us a non-trivial solution of X + Y = 2 Z . This is impossible by [2](Theorem 1.5).Therefore y ≥
12, and by the argument above u > p (2 + 5) · > (cid:3) It remains to consider the case u ≡ < u − u u and u is a square, namely u = 7 , , , , , , , , . In view of Proposition 5.5 we shall exclude the possibilities u =7 , , , For ( u, u , u ) = (11 , , ), (43 , · , ), (83 , · , ) we have w ≡ ± u ≡ , · , ) and w = 9 we used implicitly the fact that if u ≡ w can not be a square, which we include a proof in the revised version( cf. Proposition 5.5 above). The referee provides another argument by choosing p = 5 | u which leads also to a contradiction as follows1 = (cid:16) (cid:17) = (cid:16) wp (cid:17) = (cid:16) up (cid:17) = (cid:16) (cid:17) = − . For ( u, u , u ) = (61 , · , ) one has w = 3, so − (cid:16) w (cid:17) = (cid:16) u + 47 (cid:17) = 1a contradiction with (2 .
1) of Theorem 1.4. 15 .5.
For ( u, u , u ) = (73 , · , ) we have w = 5, hence x, z are odd and y is even by(2 .
1) of Theorem 1.4. Taking modulo 73 on (3 .
5) one gets(6 .
1) 4 z ≡ ( − x (mod 73)Working in F ∗ we have(6 .
2) ord(4) = 9 , ord( −
6) = 36Therefore from (6 . , (6 .
2) it follows that 36 | x , so 4 | x , a contradiction. For ( u, u , u ) = (97 , · · , ) one has w = 3, so1 = (cid:16) w (cid:17) = (cid:16) u + 411 (cid:17) = − .
1) of Theorem 1.4.
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