A note on Jeśmanowicz’ conjecture for non-primitive Pythagorean triples

1. Introduction

Let \((a, b, c)\) be a primitive Pythagorean triple. Clearly for such a triple with \(2\mid b\) one has the following parameterization \[a=u^2-v^2,\ b=2uv,\ c=u^2+v^2\] with

In 1956 L. Jesmanowicz ([1]) made the following conjecture:

Conjecture 1. For any positive integer \(n\), the Diophantine equation

has only the positive integer solution \((x,y,z)=(2,2,2)\).

The primitive case of the conjecture (\(n=1\)) was investigated thoroughly. Although the conjecture is still open, many special cases are shown to be true. We refer to a recent survey [2] for a detailed account.

Much less known about the non-primitive case (\(n>1\)). A positive integer solution \((x, y, z, n)\) of (2) is called exceptional if \((x, y, z)\ne (2, 2, 2)\) and \(n > 1\). For a positive integer \(t\), let \(P(t)\) denote the set of distinct prime factors of \(t\) and \(P(t) -\) their product. The first known result in this direction was obtained in 1998 by M.-J. Deng and G.L. Cohen ([3]), namely if \(u=v+1\) \(a\) is a prime power, and either \(P(b)\mid n\), or \(P(n)\nmid b\), then (2) has only positive integer solution \((x,y,z)=(2,2,2)\). In 1999, M.-H. Le gave necessary conditions for (2) to have exceptional solutions.

Theorem 1. [4] If \((x,y,z,n)\) is an exceptional solution of (2), then one of the following three conditions is satisfied:

• (i)   \(\max\{ x,y\}>\min\{ x,y\}>z\), \(P(n)\subsetneqq P(c);\)
• (ii)   \(x>z>y\), \(P(n)\subset P(b);\)
• (iii)   \(y>z>x\), \(P(n)\subset P(a).\)

However, as noted in [5] by H. Yang and R.-Q. Fu, the case \(x=y>z\) is not completely handled by the arguments used in [4]. Furthermore they completed the unhandled case ([5], Theorem 1) based on a powerful result of Zsigmondy ([6], cf. [7,8]). In fact one can give a unified simple proof of Theorem of Le-Yang-Fu (Theorem 1) by using a weaker version of the Zsigmondy theorem as stated in Lemma 3 of [3].

Since many works [3,4] intensively investigated the first interesting family of primitive triples:

Most recently, X.-W. Zhang and W.-P. Zhang [9], and T. Miyazaki [10] independently proved Conjecture 1 for the (infinite) family (3).

It is natural to treat the next interesting case: \(v=2,\ u\) is an odd prime which was known recently for few values \(u\): \(u=3\) ([3]) \(u=5\) - 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. [2] for references). Let's formulate our main results. We rewrite (2) as

An arithmetical argument (given in Lemma 7 below) shows that \(u^2-4\) admits a proper decomposition \(u^2-4=u_1 u_2,\ \gcd(u_1,u_2)=1\), so that there are three possibilities to consider: \(u_1\equiv \pm 1, 5\ (\mathrm{mod}\ 8)\).

Theorem 2. If \((x,y,z,n)\) is an exceptional solution of (4) and \(u_1\equiv \pm 1\ (\mathrm{mod}\ 8)\), then \(y\) is even.

In view of Theorem 2 the possibility \(u_1\equiv -1\ (\mathrm{mod}\ 8)\) is eliminated, because in this case \(x, y, z\) are even, which is in general impossible by an auxiliary argument (Lemma 8 below).

Let \(\nu_q(t)\), for a prime \(q\), denote the exponent of \(q\) in the prime factorization of \(t\), and let \(\Big( \dfrac{}{m}\Big)\) denote the Jacobi quadratic residue symbol.

Theorem 3. If \((x,y,z,n)\) is an exceptional solution of (4), then one of the following cases is satisfied

• (1)   \(\nu_2(u_1-1)=3:\) \((\nu_2(x), \nu_2(y), \nu_2(z))=(0, \geq 2, 1);\) \(u_1\) admits a proper decomposition \(u_1=t_1 t_2,\ \gcd(t_1,t_2)=1\) and \(t_1, t_2\equiv 5\ (\mathrm{mod}\ 8)\) satisfying certain special Diophantine equations;
• (2)   \(u_1\equiv 5\ (\mathrm{mod}\ 8)\), \(u_2=w^{2^s}\), where \(s=\nu_2(z-x)-\nu_2(x)\) and either of the following
• (2.1)   \(w\equiv \pm 3\ (\mathrm{mod}\ 8):\) \((\nu_2(x), \nu_2(y), \nu_2(z))=(0, \geq 1, 0);\) \(u\equiv 1\ (\mathrm{mod}\ 4);\) \(\Big( \dfrac{u_1}{p}\Big)=\Big( \dfrac{w}{p}\Big),\ \forall\ p\mid (u^2+4)\) and \(\Big( \dfrac{w}{p}\Big)=\Big( \dfrac{u^2+4}{p}\Big),\ \forall\ p\mid u_1;\)
• (2.2)   \(w\equiv \pm 1\ (\mathrm{mod}\ 8):\) \((\nu_2(x), \nu_2(y), \nu_2(z))=(\beta, 0, \beta),\ \beta\geq 1;\) \(u\equiv\pm 3\ (\mathrm{mod}\ 8);\) \(\Big( \dfrac{w}{p}\Big)=1,\ \forall\ p\mid (u^2+4)\) and \(\Big( \dfrac{w}{p}\Big)=\Big( \dfrac{u}{p}\Big),\ \forall\ p\mid u_1\). Moreover, if \(u\equiv 3\ (\mathrm{mod}\ 8)\), then \(w\) can not be a square.

Corollary 1. Conjecture 1 is true for \(v=2\), \(u\) - an odd prime \(< 100\).

Let's explain the ideas in proving our main results. As for Theorem 2 and Theorem 3 we exploit a total analysis of Jacobi quadratic and quartic residues. In the case \(u_1\equiv 1\ (\mathrm{mod}\ 8)\) we have a further proper decomposition \(u_1=t_1 t_2\), which leads to certain special Diophantine equations. Theorem 3 helps us substantially in reducing the verification process, as the possibility \(u_1\equiv 5\ (\mathrm{mod}\ 8)\) occurs quite sparsely. We demonstrate this for \(u < 100\) in proving Corollary 1.

The paper is organized as follows. In Section 2 we give a unified simple proof of Theorem 1. Section 3 provides some reduction of the problem and preliminary results. Theorem 2 will be proved in Section 4. The case \(u_1\equiv 5\ (\mathrm{mod}\ 8)\) and Theorem 3 will be treated in Section 5. The verification for \(u < 100\) in Corollary 1 will be given in the last Section 6.

2. A Simple Proof of Theorem 1

We shall use the following weaker version of Zsigmondy's theorem.

Lemma 1. ( cf. [3], Lemma 3) For \(X>Y>0\) co-prime integers,

• (1)   if \(q\) is a prime, then \[\gcd\big( X-Y,\dfrac{X^q-Y^q}{X-Y}\big)=1,\ {\text{ or}}\ q; \]
• (2)   if \(q\) is an odd prime, then \[\gcd\big( X+Y,\dfrac{X^q+Y^q}{X+Y}\big)=1,\ {\text{ or}}\ q. \]

Proof. Part \((2)\) is Lemma 3 of [3]. As for part \((1)\) one argues similarly: if \(\ell^r\) is a common prime power divisor of \(X-Y\) and \((X^q-Y^q)/(X-Y)\). Clearly

On the other hand from the fact that \(X\equiv Y\ (\mathrm{mod}\ \ell^r)\) it follows Since \(\ell\nmid Y\), (5)-(6) imply that \(\ell=q\), and \(r=1\).

Remark 1. Part \((1)\) of Lemma 1 is a special case of Theorem IV in [7].

Lemma 2. For a prime divisor \(q\) of \((X-Y)\) and positive integer \(\beta\)

Proof. Applying part \((1)\) of Lemma 1 \(\beta\) times one has \[\gcd\big( X^{q^{\beta-1}}-Y^{q^{\beta-1}},\dfrac{X^{q^\beta}-Y^{q^\beta}}{X^{q^{\beta-1}}-Y^{q^{\beta-1}}}\big)=q; \] \[\cdots\] \[\gcd\big( X-Y,\dfrac{X^q-Y^q}{X-Y}\big)=q.\] Hence the formula 7.

In view of Lemma 2 of [3] there are no exceptional solutions with \(z\geq\max\{x,y\}\), so as in [4] 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 (2) by \(n^y\) one gets
  \begin{equation} a^x n^{x-y}=c^y-b^y.\label{(2.4)}\end{equation}

  (8)
  By considering \(\mathrm{mod}\ c+b\), and taking into account \((c+b) (c-b)=a^2\), one sees that \(y\) must be even, say \(y=2 y_1\). Now put \(X=c^2,\ Y=b^2\), so \(X\equiv Y (\mathrm{mod}\ a^2),\ \gcd(Y,a)=1\). Taking \(\mathrm{mod}\ a\) and in view of (8) \[0\equiv \dfrac{X^{y_1}-Y^{y_1}}{X-Y}=X^{y_1-1}+X^{y_1-2} Y+\cdots +X Y^{y_1-2}+Y^{y_1-1}\equiv y_1 Y^{y_1-1}\ (\mathrm{mod}\ a),\] one concludes that \(a\mid y_1\).

  For any \(q\in P(a)\) let \(\beta=\nu_q(y_1)\), so that \(y_1=q^\beta y_2\) with \(q\nmid y_2\). Putting \(U=X^{q^\beta}, V=Y^{q^\beta}\) for short, we have

  \begin{equation} X^{y_1}-Y^{y_1}=(U-V) (U^{y_2-1}+U^{y_2-2} V+\cdots +U V^{y_2-2}+V^{y_2-1}),\end{equation}

  (9)
  and
  \begin{equation} U^{y_2-1}+U^{y_2-2} V+\cdots +U V^{y_2-2}+V^{y_2-1}\equiv y_2 V^{y_2-1}\not\equiv 0\ (\mathrm{mod}\ q). \label{(2.6)}\end{equation}

  (10)
  Lemma 2 and (9), (10) imply that
  \begin{equation} \nu_q(X^{y_1}-Y^{y_1})=\nu_q(U-V)=\beta+2\nu_q(a). \label{(2.7)}\end{equation}

  (11)
  In view of (8) the equality (11) means that \(a^{x-2}\mid y_1\) in contradiction with \(y_1=y/2< a^{x-2}\) as \(x>y,\ a>1\).
• (II) y>x=z : Similarly dividing both sides of (2) by \(n^z\) one gets
  \begin{equation} b^y n^{y-x}=c^x-a^x.\label{(2.8)}\end{equation}

  (12)
  Arguing as above with \(\mathrm{mod}\ c+a\), one sees that \(x\) must be even, say \(x=2 x_1\). Put \(X=c^2,\ Y=a^2\). Considering \(\mathrm{mod}\ b\) and from (12) it follows that \(b\mid x_1\). So \(\nu_q(X^{x_1}-Y^{x_1})=\nu_q(x_1)+2\nu_q(b)\) for any \(q\in P(b)\), therefore \(b^{y-2}\mid x_1\) in contradiction with \(x_1=x/2< b^{y-2}\) as \(y>x,\ b>1\).
• (III) x=y>z: Dividing both sides of (2) by \(n^z\) one gets
  \begin{equation} (a^x+b^x) n^{x-z}=c^z.\label{(2.9)}\end{equation}

  (13)
  First we claim that \(x\) must be even. Indeed, if \(x\) is odd, then from (13) it follows that there is an odd prime \(q\in P(a+b)\cap P(c)\), so \(q\in P(ab)\), as \(c^2=a^2+b^2\). A contradiction with \(\gcd(a,b)=1\).

  Writing now \(x=2 x_1\) one sees that \(x_1\) must be odd. Since otherwise for an odd prime \(q\in P(a^x+b^x)\cap P(c)\) taking \(\mathrm{mod}\ q\) and by (13)

  \[0\equiv a^x+b^x=a^{2 x_1}+(c^2-a^2)^{x_1}\equiv 2 a^{2 x_1} \ (\mathrm{mod}\ q),\] one gets a contradiction with \(\gcd(a,c)=1\).

  Now from (13) we see that

  \begin{equation} \dfrac{(a^2)^{ x_1}+(b^2)^{x_1}}{a^2+b^2}=\dfrac{c^{z-2}}{n^{x-z}}>1. \label{(2.10)}\end{equation}

  (14)
  as \(x>z\geq 2\). So there is an odd prime \(q\in P(c)\) dividing \(((a^2)^{ x_1}+(b^2)^{x_1})/(a^2+b^2)\). Considering \(\mathrm{mod}\ q\) and taking into account \(a^2\equiv -b^2\ \mathrm{mod}\ q,\ q\nmid a\) one has \[0\equiv \dfrac{(a^2)^{ x_1}+(b^2)^{x_1}}{a^2+b^2}=(a^2)^{x_1-1}-(a^2)^{x_1-2} b^2+\cdots-a^2 (b^2)^{x_1-2}+(b^2)^{x_1-1}\equiv x_1 a^{2 x_1-2}\ (\mathrm{mod}\ q).\] Hence \(q\mid x_1\), and so \(((a^2)^q+(b^2)^q)\mid ((a^2)^{ x_1}+(b^2)^{x_1})\). Applying part \((1)\) of Lemma 1 we get
  \begin{equation} \gcd\big( a^2+b^2,\dfrac{(a^2)^q+(b^2)^q}{a^2+b^2}\big)=q.\label{(2.11)}\end{equation}

  (15)
  On the other hand from (14) one knows that \(((a^2)^q+(b^2)^q)/(a^2+b^2)\) is a product of primes in \(P(c)\). It is easy to see that \(((a^2)^q+(b^2)^q)/(a^2+b^2)>q\). So either \(\nu_q\big(((a^2)^q+(b^2)^q)/(a^2+b^2)\big)\geq 2\) and \(\nu_q(a^2+b^2)\geq 2\), or both of them must have another common prime factor in \(P(c)\), a contradiction with (15).

3. Preliminary reduction

We need some reduction of the problem. The following result is due to N. Terai [11].

Lemma 3. Conjecture 1 is true for \(n=1,\ v=2\).

Because of Lemma 3 we will assume henceforth \(n>1\).

M.-J. Deng ([12], from the proof of Lemma 2), and H. Yang, R.-Q. Fu ([5]) showed that we can remove the condition \((i)\) in Theorem 1.

Lemma 4. If \((x, y, z,n)\) is an exceptional solution, then either \(x>z>y\), or \(y>z>x.\) Note that the proof of Lemma 4 relies essentially on the condition \(n>1\). It could be interesting 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 [13] succeeded to eliminate the possibility \((ii)\) in Theorem 1.

Lemma 5. Suppose that \(u\) is an odd prime and \(v=2\). Then equation (2) has no exceptional solutions \((x, y, z,n)\) with \(x > z > y\).

Lemma 6. For a positive integer \(w\)

• (1)   if \(\nu_2(w)\geq 2,\) then \(\nu_2[(1+w)^x-1]=\nu_2(w)+\nu_2(x);\)
• (2)   if \(\nu_2(w)=1\) and \(x\) is odd, then \(\nu_2[(1+w)^x-1]=1;\)
• (3)   if \(\nu_2(w)=1\) and \(x\) is even, then \(\nu_2[(1+w)^x-1]=\nu_2(2+w)+\nu_2(x).\)

In particular \(\nu_2[(1+w)^x-1]=2+\nu_2(x),\) if \(w\equiv 4\ (\mathrm{mod}\ 8);\) or if \(w\equiv 2\ (\mathrm{mod}\ 8)\) and \(x\) is even.

Proof.

• (1)   The conclusions of Lemma 6 are true trivially for \(x=1\). Assuming now \(x\geq 2\) we have
  \begin{equation} (1+w)^x-1=w (C^1_x+C^2_x w+\cdots +C^{x-1}_x w^{x-2}+C^x_x w^{x-1}).\label{(3.1)}\end{equation}

  (16)
  Clearly \(\nu_2(j)\leq j-1\) for \(j=2, \cdots, x,\) and so \[\nu_2(C^j_x w^{j-1})=\nu_2\Big( \dfrac{x}{j} C^{j-1}_{x-1} w^{j-1}\Big) \geq \nu_2(x)+j-1>\nu_2(x),\] as \(\nu_2(w)\geq 2\). Hence the conclusion follows from taking \(\nu_2(\cdot)\) on both sides of (16).
• (2)   Obvious from (16), since \(C^1_x+C^2_x w+\cdots +C^{x-1}_x w^{x-2}+C^x_x w^{x-1}\) is odd in this case.
• (3)   Writing \(x=2 x_1\) we have
  \begin{equation} (1+w)^x-1=[(1+w)^{x_1}-1] [(1+w)^{x_1}+1] .\label{(3.2)}\end{equation}

  (17)
  If \(x_1\) is odd, i.e., \(\nu_2(x)=1\), then \(\nu_2[(1+w)^{x_1}-1]=1\) by the part \((2)\) above, and \(\nu_2[(1+w)^{x_1}+1]=\nu_2(2+w)\), as \[(1+w)^{x_1}+1=(2+w) [(1+w)^{x_1-1}-(1+w)^{x_1-2}+\cdots -(1+w)+1] \] and \((1+w)^{x_1-1}-(1+w)^{x_1-2}+\cdots -(1+w)+1\) is odd.

  If \(x_1\) is even, then \(\nu_2[(1+w)^{x_1}+1]=1\), since

  \[(1+w)^{x_1}+1=2+C^1_{x_1} w+C^2_{x_1} w^2+\cdots +C^{x_1-1}_{x_1} w^{x_1-1}+w^{x_1}. \] Therefore \(\nu_2[(1+w)^x-1]=\nu_2[(1+w)^{x_1}-1]+1\) by (17). Now the descending argument yields the conclusion.

The following claims play a central role in the next sections.

Lemma 7. If \((x,y,z,n)\) is an exceptional solution of (4), then \(u^2-4\) admits a proper decomposition \(u^2-4=u_1 u_2,\ \gcd(u_1,u_2)=1\) and with one of the following conditions satisfied:

• (1)   \(u_1\equiv 1\ (\mathrm{mod}\ 8)\) and \(\nu_2(z)=\nu_2(u_1-1)+\nu_2(x)-2;\)
• (2)   \(u_1\equiv 7\ (\mathrm{mod}\ 8)\), \(\nu_2(z)=\nu_2(u_1+1)+\nu_2(x)-2\), and \(\nu_2(x)\geq 1;\)
• (3)   \(u_1\equiv 5\ (\mathrm{mod}\ 8)\), \(u_2\) is a square and and \(\nu_2(z)=\nu_2(x).\)

Proof. In view of Lemmas 4, 5 we may assume the existence of an exceptional solution with \(y>z>x\) (the case \((iii)\) of Theorem 1). Dividing both sides of (4) by \(n^x\) one gets

It is easy to see that \(\gcd(u^2+4, n)=1\). So (18) is equivalent to the following system with \(u^2-4=u_1 u_2,\ \gcd(u_1,u_2)=1.\) The system (19) can be rewritten as or equivalently with \(k(z-x)=mx\), and \(n^m=u_2^k\).

Clearly \(u_2>1\). Assume now \(u_1=1\). As \(u^2\equiv 1\ (\mathrm{mod}\ 8)\), by comparing \(\nu_2(\cdot)\) both sides of (20) and by \((1)\) of Lemma 6 we have \(\nu_2[(u^2+4)^z-1]=2+\nu_2(z)< 2y\). So (21) is inconsistent. So \(u_1>1\) and

If \(u_1\equiv 1\ \mathrm{mod}\ 8\), then by \((1)\) of Lemma 6 we get \(\nu_2(z)=\nu_2(u_1-1)+\nu_2(x)-2\).

If \(u_1 \equiv 7\ \mathrm{mod}\ 8\) and \(x\) is odd, then by \((2)\) of Lemma 6: \(\nu_2(u_1^x-1)=1\), impossible by (22). Thus (21) is inconsistent.

If \(u_1\equiv 7\ \mathrm{mod}\ 8\) and \(x\) is even, then by \((3)\) of Lemma 6: \(\nu_2(u_1^x-1)=\nu_2(u_1+1)+\nu_2(x)\). Hence by (22) one gets \(\nu_2(z)=\nu_2(u_1+1)+\nu_2(x)-2\).

For \(u_1\equiv 3\ \mathrm{mod}\ 8\), we have \(\nu_2(u_1^x-1)=1\), if \(x\) is odd (by \((2)\) of Lemma 6), and \(\nu_2(u_1^x-1)=2+\nu_2(x)\), if \(x\) is even (by \((3)\) of Lemma 6). Hence for (20) to be consistent one has necessarily \(\nu_2(z)=\nu_2(x)\), which implies \(\nu_2(z-x)\geq \nu_2(x)+1\). So from the second equation of (19): \(n^{z-x}=u_2^x\) it follows that \(u_2\) must be a square, hence \(u_2\equiv 1\ \mathrm{mod}\ 8\). Thus \(u_1 u_2\equiv 3\mod 8\), a contradiction with \(u_1 u_2=u^2-4\equiv 5\ \mathrm{mod}\ 8\).

Similarly, for \(u_1\equiv 5\ \mathrm{mod}\ 8\), by using \((1)\) of Lemma 6 we have \(\nu_2(u_1^x-1)=2+\nu_2(x)\), and by the same reason \(\nu_2(z)=\nu_2(x)\). Hence the system (19) is inconsistent, if \(u_2\) is not a square.

Lemma 8. In the notations above if \(x, y, z\) are even, then (20) is inconsistent.

Proof. In this case we can rewrite (20) in the form of Pythagorian equation \[\big(u_1^{x/2}\big)^2+\big[2^y u^{y/2} n^{(y-z)/2}\big]^2=\big[(u^2+4)^{z/2}\big]^2.\] Hence ( cf. (1)) there are integers \(X, Y\), say with \(2\mid Y\) such that

In view of Lemma 2.2 of [9], Equation (23) has solutions Since \(u^2+4\equiv 5\ \mathrm{mod}\ 8\) it follows from (25) that \(\nu_2(B)=1\). From (24) we have \(\nu_2(Y)=y-1\) which together with (26) implies \[y=\nu_2(z)+1,\] a contradiction with \(y>z\).

Corollary 2. In the notations above if \(y, z\) are even and (20) is consistent, then \(x\) is odd and \(u_1\equiv 1\ (\mathrm{mod}\ 8)\). Moreover \(u_1\) admits a proper decomposition \(u_1=t_1 t_2\) such that \(\gcd(t_1,t_2)=1\) and

Proof. By Lemma 8 \(x\) is odd. In fact one can rewrite (20) as \[A\cdot B=u_1^x\ \mathrm{with}\ \gcd(A,B)=1, \] where \[A=(u^2+4)^{z/2}-2^y u^{y/2} n^{(y-z)/2},\ \ \ \ B=(u^2+4)^{z/2}+2^y u^{y/2} n^{(y-z)/2}.\] Hence

If \(t_1=1\), then by \((1)\) of Lemma 6: \(\nu_2[(u^2+4)^{z/2}-1]=2+\nu_2(z/2)< y=\nu_2(2^y u^{y/2} n^{(y-z)/2})\). So \(A=1\) is impossible.

Now from (30) we have two possibilities:

• (1)   \(z/2\) is odd: \(t_1\equiv t_2\equiv 5\ (\mathrm{mod}\ 8);\)
• (2)   \(z/2\) is even: \(t_1\equiv t_2\equiv 1\ (\mathrm{mod}\ 8);\)

both of them imply \(u_1\equiv 1\ (\mathrm{mod}\ 8)\).

Also (27)-(29) follow immediately from (30).

Corollary 3. In the situation of Corollary 2 we have \(t_1, t_2\equiv 5\ (\mathrm{mod}\ 8)\) and \(\nu_2(u_1-1)=3.\)

Proof. We will show that \(z/2\) must be odd, from which the conclusion immediately follows by the proof above, noting that \(\nu_2(u_1-1)=\nu_2(u_1^x-1)=\nu_2(A-1)+1=3\).

Assume on the contrary that \(\nu_2(z)\geq 2\). In view of (30) one has \(x\geq 3\), as \(t_1< t_2< u^2-4\). We claim that \(x>3\). Indeed, if \(x=3\), then \(n=u_2^3\) by (19), noting that \(z=4\) by \(B=t_2^x\) of (30), so \(y=6\) as \(A=t_1^x>0\). Now from the equation \(t_1^x=A\) in (30) we see that \((t_1,4 u u_2,u^2+4)\) is a primitive solution of

Euler ([14], pp. 578-579) indicated a primitive parameterization for the Diophantine Equation (31) with \(3\nmid Z,\ 2\mid Y\) as follows \[X=(s-t) (3s-t) (3s^2+t^2),\ \ \ \ \ Y=4 s t (3s^2-3st+t^2), \] with \(s, t\) co-prime, \(3\nmid t\) and \(s\not\equiv t\ (\mathrm{mod}\ 2)\). Hence \(8\mid Y\) which shows that \(t_1^x=A\) in (30) is impossible.

Furthermore, if \(x\geq 4\), then by Theorem 1.1 of [15], (27) is again impossible.

4. Proof of Theorem 2

The aim of this section is to show that the case \(u_1\equiv 7\ (\mathrm{mod}\ 8)\) in Lemma 7 is not realized. We refer the reader to [16] for basic properties of Jacobi quadratic and quartic residue symbols \(\Big( \dfrac{}{m}\Big),\ \Big( \dfrac{}{m}\Big)_4\) we shall use in the following lemmas.

Lemma 9. For a prime \(p\mid (u^2+4)\) one has \(p\equiv 1\ (\mathrm{mod}\ 4)\) and \(\Big( \dfrac{u}{p}\Big)=1\).

Proof. Since \(u^2\equiv -4\ (\mathrm{mod}\ p)\), so \(\Big( \dfrac{-1}{p}\Big)=1\), i.e., \(p\equiv 1\ (\mathrm{mod}\ 4)\). Furthermore we include the following simple argument due to the referee instead of ours in the original version: \[\Big( \dfrac{u}{p}\Big)=\Big( \dfrac{4u}{p}\Big)=\Big( \dfrac{4u+u^2+4}{p}\Big)=\Big( \dfrac{(u+2)^2}{p}\Big)=1.\]

Lemma 10. If (20) is consistent and \(u_1\equiv \pm 1\ (\mathrm{mod}\ 8)\), then \(\Big( \dfrac{n}{p}\Big)=\Big( \dfrac{u_2}{p}\Big)\) for any prime \(p\).

Proof. Indeed, in this case by Lemma 7 \(\nu_2(z)>\nu_2(x)\). Hence \(\nu_2(z-x)=\nu_2(x)\), so we have in (21) \(n^m=u_2^k\) with \(k, m\) odd, and therefore the conclusion of Lemma 10.

We are ready now to prove Theorem 2. Let \(p\mid (u^2+4)\). By taking \(\Big( \dfrac{}{p}\Big)\) on (20) and using Lemmas 9, 10 one sees that

(as \(z\) is even). Now taking the product of (32) over all (not necessarily distinct) prime divisors \(p\mid (u^2+4)\) we have By the quadratic reciprocity law as \(u_1\equiv \pm 1\ (\mathrm{mod}\ 8)\), \(u_2\equiv \pm 5\ (\mathrm{mod}\ 8)\). Altogether (33)-(35) imply that \(\Big( \dfrac{u_2}{p}\Big)^y=(-1)^y=1\), i.e., \(y\) must be even.

Corollary 4. The possibility \(u_1\equiv 7\ (\mathrm{mod}\ 8)\) in Lemma 7 is not realized.

Proof. Indeed, in this case \(\nu_2(z)>\nu_2(x)\geq 1\), so (20) is inconsistent by Lemma 8.

Corollary 5. In the case \(u_1\equiv 1\ (\mathrm{mod}\ 8)\) of Lemma 7 we have \[(\nu_2(x), \nu_2(y), \nu_2(z))=(0, \geq 2, 1).\]

Proof. By Lemma 7 and Theorem 2: \(y, z\) are even, hence \(x\) is odd by Lemma 8. From the proof of Corollary 3 it follows that \(\nu_2(z)=1\). For a prime \(p\mid (u^2+4)\) by taking \(\Big( \dfrac{}{p}\Big)\) on \(A=t_1^x\) of (30) and using Lemma 9 one gets

By the same reason of (35) we have \(\Big( \dfrac{t_1}{u^2+4}\Big)=-1\), as \(t_1\equiv 5\ (\mathrm{mod}\ 8)\) by Corollary 3. Hence there exists a prime \(p_0\mid (u^2+4)\) such that From (36), (37) one concludes that \((y-z)/2\) must be odd (and \(\Big( \dfrac{n}{p_0}\Big)=-1\)), so the conclusion of Corollary 5 follows.

Remark 2. One can have another proof of Lemma 8 as shown in several steps below. Assuming \(y, z\) even, and arguing as in the proof of Corollary 2 one gets Equation (30) together with (27)-(29).

• (1)   If \(u_1\equiv 5\ (\mathrm{mod}\ 8)\) we have four possibilities for \((t_1,t_2)\):
  • (i)   \( t_1\equiv 1\ (\mathrm{mod}\ 8),\ t_2\equiv 5\ (\mathrm{mod}\ 8);\)
  • (ii)   \( t_1\equiv 5\ (\mathrm{mod}\ 8),\ t_2\equiv 1\ (\mathrm{mod}\ 8);\)
  • (iii)   \(t_1\equiv 3\ (\mathrm{mod}\ 8),\ t_2\equiv 7\ (\mathrm{mod}\ 8);\)
  • (iv)   \( t_1\equiv 7\ (\mathrm{mod}\ 8),\ t_2\equiv 3\ (\mathrm{mod}\ 8);\)
  all of them violate (29).
• (2)   Assume now \(u_1\equiv \pm 1\ (\mathrm{mod}\ 8)\) and \(x\) even, hence \(\nu_2(z)\geq 2\) by Lemma 7. We will shows that \(\nu_2(y)=1\). Indeed, considering \(p\mid (u^2+4)\) and taking \(\Big( \dfrac{}{p}\Big)_4\) on (20) one has by using Lemmas 9, 10
  \begin{equation} \Big( \dfrac{u_1}{p}\Big)^{x/2}=\Big( \dfrac{-1}{p}\Big)_4 \Big( \dfrac{n}{p}\Big)^{(y-z)/2}=\begin{cases}\quad \Big( \dfrac{u_2}{p}\Big)^{y/2},\ p\equiv 1\ (\mathrm{mod}\ 8)\\ -\Big( \dfrac{u_2}{p}\Big)^{y/2},\ p\equiv 5\ (\mathrm{mod}\ 8)\end{cases}\label{(4.7)}\end{equation}

  (38)
  as \(z/2\) is even. Let \(r\) denote the number of prime divisors \(p\mid (u^2+4)\), \(p\equiv 5\ (\mathrm{mod}\ 8)\). Clearly \(r\) is odd, as \(u^2+4\equiv 5\ (\mathrm{mod}\ 8)\). In a similar way as in (33)-(35), taking the product of (38) over all (not necessarily distinct) prime divisors \(p\mid (u^2+4)\) we get \[1=\Big( \dfrac{u_1}{u^2+4}\Big)^{x/2}=(-1)^r \Big( \dfrac{u_2}{u^2+4}\Big)^{y/2}=-(-1)^{y/2}.\] Hence \(y/2\) must be odd, so \((y-z)/2\) is odd. For any prime \(p\mid (u^2+4)\) taking \(\Big( \dfrac{}{p}\Big)\) on equation \(A=t_1^x\) from (30) now gives us
  \begin{equation} \Big( \dfrac{n}{p}\Big)=1\ \bigg(=\Big( \dfrac{u_2}{p}\Big)\ \ \mathrm{by\ Lemma\ 10}\bigg)\label{(4.8)}\end{equation}

  (39)
  On the other hand from (35) it follows that there exists a prime \(p_0\mid (u^2+4)\) such that \(\Big( \dfrac{u_2}{p_0}\Big)=-1\), a contradiction with (39). Thus (30) (and hence (20)) is inconsistent.

5. The case \(u_1\equiv 5\ (\mathrm{mod}\ 8)\)

In this case by \((3)\) of Lemma 7 we have \(\nu_2(z)=\nu_2(x)\), hence from (19) it follows that \(u_2=w^{2^s}\), where \(s=\nu_2(z-x)-\nu_2(x)\). The following lemma can be proved similarly as Lemma 10.

Lemma 11. If (20) is consistent and \(u_1\equiv 5\ (\mathrm{mod}\ 8)\), then \(\Big( \dfrac{n}{p}\Big)=\Big( \dfrac{w}{p}\Big)\) for any prime \(p\).

Proof. Indeed, in this case \(n^m=w^k\) with \(k, m\) odd by the above argument, and therefore the conclusion of Lemma 11.

Lemma 12. If \(x, z\) are even and (20) is consistent, then \(y\) is odd and \(u_1\equiv 5\ (\mathrm{mod}\ 8)\). Moreover \(n\) admits a decomposition \(n=n_{1} n_{2}\) such that \(\gcd(n_{1},n_{2})=1\) and

Proof. By Lemma 8 \(y\) is odd. In view of Lemma 7 and Theorem 2 we are in the situation \((3)\) of Lemma 7. Now one rewrites (20) as \[C_1\cdot D_1=2^{2y} u^y n^{y-z}\ \mathrm{with}\ \gcd(C_1,D_1)=2,\ 2\| D_1,\] where \[C_1=(u^2+4)^{z/2}-u_1^{x/2},\ D_1=(u^2+4)^{z/2}+u_1^{x/2}.\] As \(2\| D_1\) we obtain either

or where \(n=n_{1} n_{2}\), \(\gcd(n_{1},n_{2})=1\) and with \(k, m\) odd from Lemma 11. Note that this is not used in the proof here, we label it for convenience in proving Proposition 1 below.

Clearly (41) is equivalent to (40). It remains to show that (42) can't happen by rewriting it as

which is impossible, since \((u^2+4)^{z/2} < 2^{2y-2} u^y\).

Lemma 13. If \(\Big( \dfrac{u_1}{u}\Big)=1\) and \(u_2\) is a square, then \(u\equiv 1\ (\mathrm{mod}\ 4)\).

Proof. We have obviously \[1=\Big( \dfrac{u_1}{u}\Big)=\Big( \dfrac{u_1 u_2}{u}\Big)=\Big( \dfrac{u^2-4}{u}\Big)=\Big( \dfrac{-1}{u}\Big),\] so the conclusion of the lemma.

Lemma 14. In the notations of Lemma 11 we have

• (1)   if \(w\equiv \pm 3\ (\mathrm{mod}\ 8)\), then \(x, z\) are odd, \(y\) is even;
• (2)   if \(w\equiv \pm 1\ (\mathrm{mod}\ 8)\), then \(x, z\) are even, \(y\) is odd.

Proof. For a prime \(p\mid (u^2+4)\) by taking \(\Big( \dfrac{}{p}\Big)\) on (20) and using Lemmas 9, 11 one sees that

By taking the product of both sides of (45) over all (not necessarily distinct) prime divisors \(p\mid (u^2+4)\) and using the reciprocity law we have Hence if \(w\equiv \pm 3\ (\mathrm{mod}\ 8)\), then by equalizing (46), (47): \((-1)^x=(-1)^{y-z}\). Thus \(y\) must be even, as \(\nu_2(z)=\nu_2(x)\). In view of Lemma 8 \(x, z\) are odd.

In the case \(w\equiv \pm 1\ (\mathrm{mod}\ 8)\), again equalizing (46), (47) we see that \((-1)^x=1\), therefore \(x\) is even, and so is \(z\). By Lemma 8 \(y\) must be odd.

Proposition 1. In the situation of Lemma 14 we have

• (1)   if \(w\equiv \pm 3\ (\mathrm{mod}\ 8)\), then \(u\equiv 1\ (\mathrm{mod}\ 4);\)
• (2)   if \(w\equiv \pm 1\ (\mathrm{mod}\ 8)\), then \(u\equiv\pm 3\ (\mathrm{mod}\ 8)\). Moreover, if \(u\equiv 3\ (\mathrm{mod}\ 8)\), then \(w\) can not be a square.

Proof.

• (1)   If \(w\equiv \pm 3\ (\mathrm{mod}\ 8)\), then \(x, z\) are odd in view of Lemma 14. So by taking \(\Big( \dfrac{}{u}\Big)\) on (20) one gets \(\Big( \dfrac{u_1}{u}\Big)=1\), hence \(u\equiv 1\ (\mathrm{mod}\ 4)\) by Lemma 13.
• (2) In the case \(w\equiv \pm 1\ (\mathrm{mod}\ 8)\): \(x, z\) are even, \(y\) is odd by Lemma 14. There are two subcases to consider.
  • I. \(x/2, z/2\) are odd. For a prime \(p\mid (u^2+4)\) by taking \(\Big( \dfrac{}{p}\Big)\) on \(D_1=2 u^y n_{2}^{y-z}\) from (41), (43) and using Lemmas 9, 11 one sees that
    \begin{equation} \Big( \dfrac{u_1}{p}\Big)=\Big( \dfrac{2}{p}\Big) \Big( \dfrac{n_{2}}{p}\Big)=\Big( \dfrac{2}{p}\Big) \Big( \dfrac{w_{2}}{p}\Big)=\begin{cases} \Big( \dfrac{w_{2}}{p}\Big),& p\equiv 1 \ (\mathrm{mod}\ 8),\\ -\Big( \dfrac{w_{2}}{p}\Big),& p\equiv 5\ (\mathrm{mod}\ 8).\end{cases}\label{(5.9)}\end{equation}

    (48)
    Recall that the number of (not necessarily distinct) prime divisors \(p\mid (u^2+4)\), \(p\equiv 5\ (\mathrm{mod}\ 8)\) is odd, so \( \prod_{p\mid (u^2+4)}\ \Big( \dfrac{2}{p}\Big)=-1\). Now taking the product of both sides of (48) over all (not necessarily distinct) prime divisors \(p\mid (u^2+4)\) and using the reciprocity law one has
    \begin{equation} \prod_{p\mid (u^2+4)}\ \Big( \dfrac{u_1}{p}\Big)=\Big( \dfrac{u_1}{u^2+4}\Big)=\Big( \dfrac{u^2+4}{u_1}\Big)=\Big( \dfrac{2}{u_1}\Big)=-1,\label{(5.10)}\end{equation}

    (49)
    and
    \begin{equation} \prod_{p\mid (u^2+4)}\ \Big( \dfrac{2}{p}\Big) \Big( \dfrac{w_{2}}{p}\Big)=-\prod_{p\mid (u^2+4)}\ \Big( \dfrac{w_{2}}{p}\Big)=-\Big( \dfrac{w_{2}}{u^2+4}\Big)=-\Big( \dfrac{u^2+4}{w_{2}}\Big)=-\Big( \dfrac{2}{w_{2}}\Big).\label{(5.11)}\end{equation}

    (50)
    Equalizing (49), (50) we get \(w_{2}\equiv\pm 1\ (\mathrm{mod}\ 8)\), so in view of (43): \(n_{2}\equiv\pm 1\ (\mathrm{mod}\ 8)\). From this and (40) it follows that \(u\equiv\pm 3\ (\mathrm{mod}\ 8)\). Moreover, if \(u\equiv 3\ (\mathrm{mod}\ 8)\), then \(w_{2}\equiv -1\ (\mathrm{mod}\ 8)\), hence by (43) \(w\) can not be a square.
  • II. \(x/2, z/2\) are even. If one takes \(\Big( \dfrac{}{u}\Big)\) on the second equation of (40), then \(\Big( \dfrac{n_{1}}{u}\Big)=1\). Now taking \(\Big( \dfrac{}{u}\Big)\) on the first equation of (40) we get \(1=\Big( \dfrac{-1}{u}\Big) \Big( \dfrac{n_{1}}{u}\Big)\). Thus \(u\equiv 1\ (\mathrm{mod}\ 4)\).

The proof of Proposition 1 is completed.

As for Theorem 3 notice that the case \(u_1\equiv\pm 1\ (\mathrm{mod}\ 8)\) follows from Corollaries 2, 3, 4 and 5. The rest of Theorem 3, i.e., the case \(u_1\equiv 5\ (\mathrm{mod}\ 8)\), follows from Lemma 14 and Proposition 1.

The equalities for Jacobi symbols are immediate from (20) and Lemma 11.

6. Proof of Corollary 1

In this section we shall apply results of previous parts for establishing the truth of Jesmanowicz' conjecture for \(u< 100\) and \(v = 2\). In view of Theorem 3 one has to consider only two cases: \(u_1\equiv 1\ (\mathrm{mod}\ 8)\) and \(u_1\equiv 5\ (\mathrm{mod}\ 8)\).

Observation 1. If \(u_1\equiv 1\ (\mathrm{mod}\ 8)\) and (20) is consistent, then \(u>183\).

Proof. Indeed, it was noted that \(x\geq 3\) by (30). On the other hand from the proof of Corollary 3 we have \(\nu_2(z)=1\), so \(z\geq 6\), hence \(y\geq 8\). From (28) it follows that \(2^{y+1}\mid t_2-t_1\), as \(x\) is odd. Since \(t_1, u_2\) are co-prime and \(\equiv 5\ (\mathrm{mod}\ 8)\), so \(t_1 u_2\geq 5\cdot 13\). Therefore \(u>\sqrt{t_1 t_2 u_2}\geq \sqrt{(2^9+5)\cdot 65}>183\).

Observation 2. If \(u_1\equiv 1\ (\mathrm{mod}\ 8)\) and (20) is consistent, then in fact \(u>729\).

Proof. By Corollary 5 one knows \(4\mid y\). We claim that \(y\geq 12\). Assuming on the contrary \(y=8\), then by the above \(z=6\). In view of (27) and [17] we must have \(x>3\), so \(x=5\), which gives us a non-trivial solution of \(X^5+Y^5=2 Z^3\). This is impossible by [18] (Theorem 1.5).

Therefore \(y\geq 12\), and by the argument above \(u>\sqrt{(2^{13}+5)\cdot 65}>729\).

It remains to consider the case \(u_1\equiv 5\ (\mathrm{mod}\ 8)\). In the range of odd primes \(< 100\) there are ten possibilities with \(u^2-4=u_1 u_2\) and \(u_2\) is a square, namely \(u=7, 11, 23\), \(43, 47, 61, 73, 79, 83, 97.\) In view of Proposition 1 we shall exclude the possibilities \(u=7, 23, 47, 79\).

Observation 3. For \((u,u_1,u_2)=(11,13,3^2)\), \((43,5\cdot 41,3^2)\), \((83,5\cdot 17,3^4)\) we have \(w\equiv \pm 3\ (\mathrm{mod}\ 8)\), hence \(u\equiv 1\ (\mathrm{mod}\ 4)\) by Proposition 1, a contradiction. Note that in the original version to eliminate the possibility \((83,5\cdot 17,3^4)\) and \(w=9\) we used implicitly the fact that if \(u\equiv 3\ (\mathrm{mod}\ 8)\), then \(w\) can not be a square, which we include a proof in the revised version ( cf. Proposition 1 above). The referee provides another argument by choosing \(p=5\mid u_1\) which leads also to a contradiction as follows \[1=\Big( \dfrac{9}{5}\Big)=\Big( \dfrac{w}{p}\Big)\ne \Big( \dfrac{u}{p}\Big)=\Big( \dfrac{83}{5}\Big)=-1.\]

Observation 4. For \((u,u_1,u_2)=(61,7\cdot 59,3^2)\) one has \(w=3\), so \[-1=\Big( \dfrac{w}{7}\Big)\ne \Big( \dfrac{u^2+4}{7}\Big)=1, \] a contradiction with \((2.1)\) of Theorem 3.

Observation 5. For \((u,u_1,u_2)=(73,3\cdot 71,5^2)\) we have \(w=5\), hence \(x, z\) are odd and \(y\) is even by \((2.1)\) of Theorem 3. Taking modulo \(73\) on (20) one gets

Working in \(F_{73}^*\) we have Therefore from (51), (52) it follows that \(36\mid 9 x\), so \(4\mid x\), a contradiction.

Observation 6. For \((u,u_1,u_2)=(97,5\cdot 11\cdot 19,3^2)\) one has \(w=3\), so \[1=\Big( \dfrac{w}{11}\Big)\ne \Big( \dfrac{u^2+4}{11}\Big)=-1, \] again a contradiction with \((2.1)\) of Theorem 3.

Acknowledgments:

The authors would like to thank the referee for many valuable comments and suggestions greatly improving the content of the paper.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

The authors declare no conflict of interest.