On a Diophantine Equation

Abstract

This paper proves that for all positive integers n, the equation

$$ \frac{x}{y}+p\frac{y}{z}+\frac{z}{w}+p\frac{w}{x}=8np, $$

where p = 1 or p is a prime congruent to 1 (mod8), does not have solutions in positive integers.

1 Introduction

The problem concerning the sum of rationals whose product is 1 has been studied by many authors. Cassels [5] showed that the equation

$$ \frac{x}{y}+\frac{y}{z}+\frac{z}{x}=1 $$

does not have solutions in integers. Bremner and Guy [3] found integer solutions to the equation

$$ \frac{x}{y}+\frac{y}{z}+\frac{z}{x}=n $$

for many values of n in the range |n|≤ 1000. Sierpinski [6] asked if the equation

$$ \frac{x}{y}+\frac{y}{z}+\frac{z}{x}=4 $$

has solutions in positive integers? Bondarenko [1] showed that the equation

$$ \frac{x}{y}+\frac{y}{z}+\frac{z}{x}=4k^{2} $$

does not have solutions in positive integers if $3\nmid k$. Using the technique developed by Bremner and Tho [4], which is based on Stoll’s idea [8], we will prove the following results:

Theorem 1

Let n be a positive integer. Then the equation

$$ \frac{x}{y}+\frac{y}{z}+\frac{z}{w}+\frac{w}{x}=8n $$

does not have solutions in positive integers.

Theorem 2

Let n be a positive integer, p - a prime congruent to 1 (mod8). Then the equation

$$ \frac{x}{y}+p\frac{y}{z}+\frac{z}{w}+p\frac{w}{x}=8pn $$

does not have solutions in positive integers.

An equivalent form of Theorem 1 is that there are no four positive rationals whose product is 1 and sum is an integer divisible by 8. In the next section, we give a proof for Theorem 2. Theorem 1 can be proven in a similar (and simpler) way. All computations in the paper are done in Magma [2].

2 Proof of Theorem 2

2.1 Notation

For a prime q and a nonzero q-adic number a, denote v_q(a) the highest power of q dividing a. By definition, $\mathbb {Q}_{\infty }=\mathbb {R}$. Let $k=\mathbb {Q}_{q}$ or $k=\mathbb {R}$. For a,b in k^∗, the Hilbert symbol (a,b)_q is defined by

$$ (a,b)_{q}=\left\{\begin{array}{ll} 1&\quad \text{ if } ax^{2}+by^{2}=z^{2} \text{ has a solution $(x,y,z)\neq (0,0,0)$ in $k^{3}$,}\\ -1&\quad \text{ otherwise.} \end{array}\right. $$

When $k=\mathbb {Q}_{\infty }$, the symbol $(a,b)_{\infty }$ is defined similarly. The following properties of Hilbert symbol are true, see Serre [7, Chap. III]:

• (i) For $a,b,c\in \mathbb {Q}_{q}^{\ast }$,

  $$ \begin{array}{@{}rcl@{}} (a,bc)_{q}&=&(a,b)_{q}(a,c)_{q},\\ (a,b^{2})_{q}&=&1. \end{array} $$

• (ii) For $a,b\in \mathbb {Q}^{\ast }$,

  $$ (a,b)_{\infty}{\prod}_{q\text{ prime}}(a,b)_{q}=1. $$

• (iii) For $a,b\in \mathbb {Q}_{q}^{\ast }$, let a = q^αu, b = q^βv, where α = v_q(a) and β = v_q(b). Then

  $$ \begin{array}{@{}rcl@{}} (a,b)_{q}&=&(-1)^{\alpha\beta(q-1)/2}\left( \frac{u}{q}\right)^{\beta}\left( \frac{v}{q}\right)^{\alpha},\quad\text{ if } q\neq 2,\\ (a,b)_{q}&=&(-1)^{\frac{(u-1)(v-1)}{4}+\frac{\alpha(v^{2}-1)}{8}+\frac{\beta(u^{2}-1)}{8}},\quad\text{ if } q=2, \end{array} $$

  where $\left (\frac {u}{q}\right )$ denotes the Legendre symbol.

2.2 Proof

Assume that (x,y,z,w) is a positive integer solution to

$$ \frac{x}{y}+p\frac{y}{z}+\frac{z}{w}+p\frac{w}{x}=8pn $$

(1)

with $\gcd (x,y,z,w)=1$.

Consider two quadratic forms:

$$ \begin{array}{@{}rcl@{}} D(X,Z) &=& X^{2}+Z^{2}-2XZ(8pn^{2}-1),\\ H(X,Z) &=& X^{2}+Z^{2}-2XZ(8pn^{2}+1). \end{array} $$

Lemma 1

$$ D(x,z) <0,\quad H(y,w) < 0. $$

Proof

From (1) and the AM-GM inequality, we have

$$ \begin{array}{@{}rcl@{}} 8pn=\left( \frac{x}{y}+\frac{pw}{x}\right)+\left( \frac{py}{z}+\frac{z}{w}\right)&\geq& 2\sqrt{\frac{x}{y}\frac{pw}{x}}+2\sqrt{\frac{py}{z}\frac{z}{w}}\\ &=& 2\sqrt{p}\frac{y+w}{\sqrt{yw}}. \end{array} $$

Thus,

$$ 4n\sqrt{pyw}\geq y+w. $$

Hence,

$$ y^{2}-2(8pn^{2}-1)yw+w^{2}\leq 0. $$

Similarly,

$$ \begin{array}{@{}rcl@{}} 8np=\left( \frac{x}{y}+\frac{py}{z}\right)+\left( \frac{z}{w}+\frac{pw}{x}\right)&\geq& 2\left( \sqrt{\frac{x}{y}\frac{py}{z}}+\sqrt{\frac{z}{w}\frac{pw}{x}}\right)\\ &=&2\sqrt{p}\frac{x+z}{\sqrt{xz}}. \end{array} $$

Thus,

$$ 4n\sqrt{pxz}\geq x+z. $$

Hence,

$$ x^{2}-2(8pn^{2}-1)xz+z^{2}\leq 0. $$

Since (8pn² − 1)² − 1 is not a perfect square, we have y² − 2(8pn² − 1)yw + w² < 0 and x² − 2(8pn² − 1)xz + z² < 0. Hence D(x,z) < 0 and H(y,w) < D(y,w) < 0. □

From (1):

$$ x^{2}zw+py^{2}wx+z^{2}xy+pw^{2}yz-8npxyzw=0. $$

(2)

Fix x, z and consider the projective curve F_x,z(Y,W,d) = 0, where

$$ F_{x,z}(Y,W,d)=pxWY^{2}+pW^{2}Yz+(xz^{2}Y+x^{2}zW)d^{2}-8npxzYWd. $$

Then, F_x,z(y,w,1) = F_x,z(0,1,0) = 0. So, F_x,z(Y,W,d) = 0 is isomorphic to the elliptic curve

$$ C_{x,z}\colon \omega^{2}=u(u^{2}+pxz(16n^{2}pxz-x^{2}-z^{2})u+p^{2}x^{4}z^{4}) $$

(3)

via the rational maps ϕ: F_x,z → C_x,z,

$$ \phi(Y:W:d)=\left( \frac{-x^{2}z^{2}Wp}{Y},\frac{x^{2}z^{2}W(4nxzd-xY-zW)}{Yd}\right), $$

and $\psi \colon C_{x,z}\rightarrow F_{x,z}$,

$$ \psi(u,\omega)=\big(px^{2}z^{2}(4nxzu+p\omega):-u(4nxznu+p\omega):zu(u-px^{3}z)\big). $$

Let D = D(x,z), H = H(x,z). Let

$$ \begin{array}{@{}rcl@{}} A&=&pxz(16n^{2}pxz-x^{2}-z^{2}),\\ B&=&p^{2}x^{4}z^{4}. \end{array} $$

Lemma 2

If q is an odd prime, then

$$ (u,D)_{q}=1. $$

Proof

Let $d=\gcd (x,z)$, x = dx₁, y = dy₁, where $x_{1},z_{1}\in \mathbb {Z}^{+}$ with $\gcd (x_{1},z_{1})=1$, $u_{1}=\frac {u}{d^{4}}$ and $\omega _{1}=\frac {\omega }{d^{6}}$. Then

$$ \begin{array}{@{}rcl@{}} (D(x,z),u)_{q}&=&(d^{2}({x_{1}^{2}}+{z_{1}^{2}}-2(8pn^{2}-1)x_{1}z_{1}),d^{4}u_{1})_{q}\\ &=&(D(x_{1},z_{1}),u_{1})_{q}. \end{array} $$

From (3), we also have

$$ {\omega_{1}^{2}}=u_{1}\left( {u_{1}^{2}}+px_{1}z_{1}(16pn^{2}x_{1}z_{1}-{x_{1}^{2}}-{z_{1}^{2}})u_{1}+p^{2}{x_{1}^{4}}{z_{1}^{4}}\right). $$

Therefore, it is enough to prove Lemma 2 in the case $\gcd (x,z)=1$.

Let u = q^ru₀, where r = v_q(u). We consider the following cases:

Case 1: r < 0. From (3), we have

$$ \omega^{2}=q^{3r}u_{0}({u_{0}^{2}}+q^{-r}Au_{0}+q^{-2r}B). $$

Thus

$$ 3r=v_{q}(\omega^{2})=2v_{q}(\omega). $$

Therefore, 2|r. Now

$$ (q^{-3r/2}\omega)^{2}=u_{0}({u_{0}^{2}}+q^{-r}Au_{0}+q^{-2r}B). $$

Taking reduction (mod p) shows u₀ is a square (mod q). Thus $u_{0} \in \mathbb {Z}_{q}^{2}$. Because r is even, we have $u=q^{r}u_{0}\in \mathbb {Q}_{q}^{2}$. Hence (D,u)_q = 1.

Case 2: r = 0. Let D = q^kD₁, where k = v_q(D). Because $D\in \mathbb {Z}$, we have k ≥ 0. Suppose 2|k. Because u and D₁ are units in $\mathbb {Z}_{q}$, we have (u,D₁)_q = 1. Thus

$$ (u,D)_{q}=(u,q^{k}D_{1})_{q}=(u,D_{1})_{q}=1. $$

We consider the case $2\nmid k$. Then,

$$ x^{2}+z^{2}\equiv 2(8pn^{2}-1)xz \pmod q. $$

Hence,

$$ (x+z)^{2}\equiv 16pn^{2}xz\pmod q. $$

We have

$$ \begin{array}{@{}rcl@{}} u^{2}+Au+B&=&\left( u+\frac{A}{2}\right)^{2}-\frac{p^{2}x^{2}z^{2}HD}{4}\\ &\equiv& \left( u+\frac{A}{2}\right)^{2} \pmod q. \end{array} $$

• (i) Suppose $q\nmid u+\frac {A}{2}$. Then $u^{2}+Au+B\in \mathbb {Z}_{q}^{2}$. Because ω² = u(u² + Au + B), we have $u\in \mathbb {Z}_{q}^{2}$. Thus (u,D)_q = 1.

• (ii) Suppose $q | u+\frac {A}{2}$. Thus $u\equiv -\frac {A}{2}~ (\text {mod} q)$. Therefore, $q\nmid A$. Because q|D, we have

  $$ A=pxz(2xz-D)\equiv 2px^{2}z^{2}\pmod q. $$

  Because $q\nmid u$ and $q\nmid A$, we have $q\nmid 2px^{2}z^{2}$. Now $q\nmid 2pxz$, so $\gcd (D,H)=1$. Let $S=u+\frac {A}{2}$ and $T=\frac {HD}{4}$. Because $\gcd (H,D)=1$ and q|D, we have v_q(T) = v_q(D) = k. Let S = q^lS₁, T = q^kT₁, where l = v_q(S). From ω² = u(u² + Au + B) and $q\nmid u$, we have

  $$ v_{q}(u^{2}+Au+B)=2v_{q}(\omega). $$

  Thus 2|v_q(u² + Au + B). On the other hand,

  $$ v_{q}(u^{2}+Au+B)=v_{q}(S^{2}+T)=v_{q}(q^{2l}{S_{1}^{2}}+q^{k}T_{1}). $$

  Since $2\nmid k$, we must have 2l < k. Thus,

  $$ u^{2}+Au+B=q^{2l}({S_{1}^{2}}+q^{k-2l}T_{1})\in \mathbb{Q}_{q}^{2}. $$

  Hence, $u=\frac {\omega ^{2}}{u^{2}+Au+B}\in \mathbb {Q}_{q}^{2}$. So (u,D)_q = 1.

Case 3: r > 0.

• (a) Suppose $q\nmid pxz$. Since (u² + Au + B)α² + Dβ² = γ² has a solution (1,0,px²z²) (mod q), it has a nontrivial solution in $\mathbb {Q}_{q}$. Therefore,

  $$ (u^{2}+Au+B,D)_{q}=1. $$

  Because u(u² + Au + B) = ω²≠ 0, we have (u,D)_q = 1.

• (b) Suppose q|xz. Then q|x or q|z. If q|x, then

  $$ D=x^{2}+z^{2}-2p(8n^{2}-1)xz\equiv z^{2}\pmod q. $$

  Note that $\gcd (x,z)=1$, thus $D\in \mathbb {Z}_{q}^{2}$. Therefore, (u,D)_q = 1. Similarly, we also have (u,D)_q = 1 if q|z.

• (c) Suppose $q\nmid xz$ and q = p. Then u = p^ru₀. So

  $$ \omega^{2}=p^{r}u_{0}(p^{2r}{u_{0}^{2}}+Ap^{r}u_{0}+p^{2}x^{4}z^{4}). $$

  We have two subcases:

  • (i) r ≥ 2. Then 2v_p(ω) = r + 2. Thus 2|r. We now have

    $$ (\omega p^{-r/2})^{2}=u_{0}(p^{2r-2}{u_{0}^{2}}+Ap^{r-2}u_{0}+x^{4}z^{4}). $$

    Because p|A, a reduction (mod p) gives u₀x⁴z⁴ is a square (mod p). Therefore, $u_{0}\in \mathbb {Z}_{q}^{2}$. Thus,

    $$ (u,D)_{p}=(2^{s}u_{0},D)_{p}=1. $$

  • (ii) r = 1. Then

    $$ \begin{array}{@{}rcl@{}} \omega^{2}&=&pu_{0}(p^{2}{u_{0}^{2}}+pAu_{0}+p^{2}x^{4}z^{4})\\ &=&p^{3}u_{0}({u_{0}^{2}}+xz(16pn^{2}xz-x^{2}-z^{2})u_{0}+x^{4}z^{4}). \end{array} $$

    Thus p divides

    $$ {u_{0}^{2}}+xz(-x^{2}-z^{2})u_{0}+x^{4}z^{4}=(u_{0}-x^{3}z)(u_{0}-xz^{3}). $$

    Therefore,

    $$ u_{0}\equiv x^{3}z \pmod p\qquad \text{ or }\qquad u_{0}\equiv xz^{3} \pmod p. $$

    If u₀ ≡ x³z (modp), then

    $$ \begin{array}{@{}rcl@{}} (pu_{0},p)_{p}&=&(-1)^{(p-1)/2}\left( \frac{u_{0}}{p}\right)=\left( \frac{x^{3}z}{p}\right)\\ &=&\left( \frac{-x^{4}}{p}\right)=\left( \frac{-1}{p}\right)\\ &=&(-1)^{\frac{p-1}{2}}\\ &=&1. \end{array} $$

    Similarly, if xz³ ≡ 0 (modp), then (pu₀,p)_p = 1. So it is always true that

    $$ (pu_{0},p)_{p}=1. $$

    (4)

    Now

    $$ D=(x+z)^{2}-16pn^{2}xz\equiv (x+z)^{2}\pmod p. $$

    Suppose $p\nmid x+z$. Then $D\in \mathbb {Z}_{p}^{2}$. Hence (u,D)_p = 1. We consider the case p|x + z. Let x + z = p^sf, where s = v_p(x + y) > 0. Then

    $$ D=p(p^{2s-1}f^{2}-16n^{2}xz). $$

    • Suppose that $p\nmid n$, then

      $$ p^{2s-1}f^{2}-16n^{2}xz \equiv (4nx)^{2} \pmod p. $$

      Hence, $p^{2s-1}f^{2}-16n^{2}xz \in \mathbb {Z}_{p}^{2}$. Let $D=p{D_{1}^{2}}$, where $D_{1}\in \mathbb {Z}_{p}$. Then from (4),

      $$ (u,D)_{p}=(pu_{0},p{D_{1}^{2}})_{p}=(pu_{0},p)_{p}=1. $$

    • Suppose that p|n. Let n = p^tn₁, where t = v_p(n) > 0. Then

      $$ D=p^{2s}f^{2}-16p^{2t+1}xz. $$

      If s ≤ t, then

      $$ D=p^{2s}(f^{2}-16p^{2t+1-2s}{n_{1}^{2}}xz). $$

      Thus $D\in \mathbb {Z}_{p}^{2}$. Hence (u,D)_p = 1. We consider the case s > t. Then

      $$ D=p^{2t+1}(p^{2s-2t-1}f^{2}-16{n_{1}^{2}}xz). $$

      Because

      $$ p^{2s-2t-1}f^{2}-16{n_{1}^{2}}xz\equiv 16{n_{1}^{2}}x^{2}\pmod p, $$

      we have $D=p^{2s+1}{D_{2}^{2}}$, where $D_{2}\in \mathbb {Z}_{p}$. From (4), we have

      $$ (u,D)_{p}=(pu_{0},p^{2r+1}{D_{2}^{2}})_{p}=(pu_{0},p)_{p}=1. $$

□

Lemma 3

If $4\nmid x+z$, then

$$ (D,u)_{2}=1. $$

Proof

Let $d=\gcd (x,z)$, let x = dx₁, y = dy₁, where $x_{1},z_{1}\in \mathbb {Z}^{+}$ with $\gcd (x_{1},z_{1})=1$, let $u_{1}=\frac {u}{d^{4}}$ and $\omega _{1}=\frac {\omega }{d^{6}}$. Then

$$ \begin{array}{@{}rcl@{}} (D(x,z),u)_{q}&=&\big(d^{2}({x_{1}^{2}}+{z_{1}^{2}}-2(8pn^{2}-1)x_{1}z_{1}),d^{4}u_{1}\big)_{q}\\ &=&\big(D(x_{1},z_{1}),u_{1}\big)_{q}. \end{array} $$

From (3), we also have

$$ {\omega_{1}^{2}}=u_{1}\left( {u_{1}^{2}}+px_{1}z_{1}(16pn^{2}x_{1}z_{1}-{x_{1}^{2}}-{z_{1}^{2}})+p^{2}{x_{1}^{4}}{z_{1}^{4}}\right). $$

Of course, $4\nmid x_{1}+z_{1}$ if $4\nmid x+z$. Therefore, it is enough to prove Lemma 3 in the case $\gcd (x,z)=1$.

If $2\nmid x+z$, then

$$ D=(x+z)^{2}-16pn^{2}xz\equiv 1\pmod 8. $$

So $D\in \mathbb {Z}_{2}^{2}$, hence (D,u)₂ = 1. Let us consider the case 2|x + z. Because $4\nmid x+z$, we can write x + z = 2h with $2\nmid h$. Then

$$ D=4(h^{2}-4pn^{2}xz). $$

We consider two cases:

• (a) 2|n. Then h² − 4pn²xz ≡ 1 (mod8). Thus $D\in \mathbb {Z}_{2}^{2}$. Hence (D,u)₂ = 1.

• (b) $2\nmid n$. Then pn²xz ≡ 1 (mod4), and so h² − 4pn²xz ≡ 5 (mod8). Thus D = 4D₁, where D₁ ≡ 5 (mod8).

  Let u = 2^ru₁, where r = v₂(u). Then

  $$\omega^{2}=2^{r}u_{1}(2^{2r}{u_{1}^{2}}+2^{r}Au_{1}+B).$$

  • (i) Suppose r ≥ 3. Then r = 2v₂(ω). We have

    $$ (2^{-r/2}\omega)^{2}=u_{1}(2^{2r}{u_{1}^{2}}+2^{r}Au_{1}+B). $$

    Because

    $$ \begin{array}{@{}rcl@{}} 2^{2r}{u_{1}^{2}}+2^{r}Au_{1}+B&\equiv& B \pmod 8\\ &\equiv& p^{2}x^{4}z^{4} \pmod 8\\ &\equiv& 1 \pmod 8, \end{array} $$

    we have u₁ ≡ 1 (mod8). Thus $u_{1} \in \mathbb {Z}_{2}^{2}$, so $u=2^{r}u_{1}\in \mathbb {Z}_{2}^{2}$. Hence (u,D)₂ = 1.

  • (ii) Suppose r = 2. Then

    $$ \left( \frac{\omega}{2}\right)^{2}=u_{1}(2^{4}{u_{1}^{2}}+2^{2}Au_{1}+B). $$

    Taking a reduction (mod 8) gives u₁ ≡ 1 (mod8). Therefore, $u=2^{2}u_{1}\in \mathbb {Z}_{2}^{2}$. Hence (u,D)₂ = 1.

  • (iii) Suppose r = 1. Then

    $$ \omega^{2}=2u_{1}(4{u_{1}^{2}}+2Au_{1}+B), $$

    what is impossible (mod 2).

  • (iv) Suppose r = 0. Then u = u₁ and D = 2²D₁, where D₁ ≡ 5 (mod8). Therefore,

    $$ \begin{array}{@{}rcl@{}} (u,D)_{2}&=&(u_{1},2^{2}D_{1})_{2}\\ &=&(u_{1},D_{1})_{2}\\ &=&(-1)^{(u_{1}-1)(D_{1}-1)/4}\\ &=&1. \end{array} $$

  • (v) Suppose r < 0. Then

    $$ \omega^{2}=2^{3r}u_{1}\left( {u_{1}^{2}}+2^{-r}Au_{1}+2^{-2r}B\right). $$

    Therefore, 3r = 2v₂(ω). Thus r ≤− 2. Then

    $$ \left( 2^{-3r/2}\omega\right)^{2}=u_{1}\left( {u_{1}^{2}}+2^{-r}Au_{1}+2^{-2r}B\right). $$

    Note that 2|A, so taking a reduction (mod 8) gives u₁ ≡ 1 (mod8). Thus $u_{1}\in \mathbb {Z}_{2}^{2}$, so $u\in \mathbb {Z}_{2}^{2}$. Hence (u,D)₂ = 1.

□

Lemma 4

If $4\nmid x+z$, then

$$ (u,D)_{\infty}=1. $$

Proof

Since (D,u)_q = 1 for all prime q, and

$$ (D,u)_{\infty}{\prod}_{q\text{ prime}, q<\infty}=1, $$

we have

$$ (u,D)_{\infty}=1. $$

□

Lemma 5

If q is an odd prime, then

$$ (H,u)_{q}=1. $$

Proof

Because A² − 4B = p²x²z²DH, we have

$$ \omega^{2}=u\left( \left( u+\frac{A}{2}\right)^{2} - DH\left( \frac{pxz}{2}\right)^{2}\right). $$

So

$$ u\alpha^{2}-uDH\beta^{2}=\omega^{2}, $$

where $\alpha =u+\frac {A}{2}$ and $\beta =\frac {pxz}{2}$. Thus

$$ (u,-uHD)_{q}=1. $$

On the other hand, (u,−u)_q = 1 and (u,D)_q = 1; therefore,

$$ (u,H)_{q}=1. $$

□

Lemma 6

If $4\nmid x-z$, then

$$ (u,H)_{2}=1. $$

Proof

Similar to the proof of Lemma 3, it is enough to consider the case $\gcd (x,z)=1$. Suppose $2\nmid x-z$, then

$$ H=(x-z)^{2}-16pn^{2}xz\equiv 1\pmod 8. $$

Thus $H\in \mathbb {Z}_{2}^{2}$, and hence (u,H)₂ = 1. We consider the case 2|x − z. Because $4\nmid x-z$, let x − z = 2k, where $2\nmid k$. Then

$$ H=4(k^{2}-4pn^{2}xz). $$

We consider two cases:

• (a) 2|n. Then k² − 4pn²xz ≡ 1 (mod8). Thus $H\in \mathbb {Z}_{2}^{2}$ and (u,H)₂ = 1.

• (b) $2\nmid n$. Then 4pn²xz ≡ 4 (mod8). Thus k² − 4pn²xz ≡ 5 (mod8), so H = 4H₁, where H₁ ≡ 5 (mod8). Let u = 2^ru₁, where r = v₂(u). Then

  $$ \omega^{2}=2^{r}u_{1}(2^{2r}{u_{1}^{2}}+2^{r}Au_{1}+B). $$

  • (i) Suppose r ≥ 3. Then r = 2v₂(ω). Thus 2|r. We have

    $$ (2^{-r/2}\omega)^{2}=u_{1}(2^{2r}{u_{1}^{2}}+2^{r}Au_{1}+B). $$

    Taking reduction (mod 8) gives $2^{2r}{u_{1}^{2}}+2^{r}Au_{1}+B\equiv 1~(\text {mod} 8)$. Hence $u_{1} \in \mathbb {Z}_{2}^{2}$, thus $u=2^{r}u_{1}\in \mathbb {Z}_{2}^{2}$. So (u,H)₂ = 1.

  • (ii) Suppose r = 2. Then

    $$ \left( \frac{\omega}{2}\right)^{2}=u_{1}(2^{4}{u_{1}^{2}}+2^{2}Au_{1}+B). $$

    Taking reduction (mod 8) gives u₁ ≡ 1 (mod8). Therefore, $u\in \mathbb {Z}_{2}^{2}$. Hence (u,H)₂ = 1.

  • (iii) Suppose r = 1. Then

    $$ \omega^{2}=2u_{1}(4{u_{1}^{2}}+2Au_{1}+B), $$

    what is impossible (mod 2).

  • (iv) Suppose r = 0. Then u = u₁ and H = 2²H₁, where H₁ ≡ 5 (mod8). Therefore,

    $$ \begin{array}{@{}rcl@{}} (u,H)_{2}&=&(u_{1},2^{2}H_{1})_{2}\\ &=&(u_{1},H_{1})_{2}\\ &=&(-1)^{(u_{1}-1)(H_{1}-1)/4}\\ &=&1. \end{array} $$

  • (v) Suppose r < 0. Then

    $$ \omega^{2}=\frac{u_{1}({u_{1}^{2}}+2^{-r}Au_{1}+2^{-2r}B)}{2^{-3r}}. $$

    Therefore, 2|r. Thus r ≤− 2. Now

    $$ (2^{3r/2}\omega)^{2}=u_{1}({u_{1}^{2}}+2^{-r}Au_{1}+2^{-2r}B). $$

    Note that 2|A, so taking (mod 8) gives u₁ ≡ 1 (mod8). Therefore, $u_{1}\in \mathbb {Z}_{2}^{2}$, so $u\in \mathbb {Z}_{2}^{2}$. Thus (u,H)₂ = 1.

□

Lemma 7

If $4\nmid x-z$, then

$$ (u,H)_{\infty}=1. $$

Proof

Since (u,H)_q = 1 for all primes q and

$$ (u,H)_{\infty}{\prod}_{q\text{ prime}}(u,H)_{q}=1, $$

we have

$$ (u,H)_{\infty}=1. $$

□

We summarize Lemmas 2, 3, 4, 5, 6, and 7 into the following proposition:

Proposition 1

• (D(x,z),u)_q = (H(x,z),u)_q = 1 if q is an odd prime or $q=\infty$.

• $(D(x,z),u)_{2}=(D(x,z),u)_{\infty }=1$ if $4\nmid x+z$.

• $(H(x,z),u)_{2}=(H(x,z),u)_{\infty }=1$ if $4\nmid x-z$.

In order for (2) to have a positive integer solutions, we seek for points (u,ω) on C_x,z such that ψ(u,ω) = (Y : W : d) satisfies d≠ 0, $\frac {Y}{d}>0$ and $\frac {W}{d}>0$. If u = 0, then ω = 0. Because ψ(0,0) = (1 : 0 : 0), we have u≠ 0. Therefore,

$$ \left\{\begin{array}{l} \frac{px^{2}z(4nxzu+p\omega)}{u(u-px^{3}z)}>0,\\ -\frac{4xznu+p\omega}{u-px^{3}z}>0. \end{array}\right. $$

Multiplying two inequalities gives u < 0. Let

$$ (u,\omega)=\phi(y:w:1)=\left( \frac{-x^{2}z^{2}wp}{y},\frac{x^{2}z^{2}w(4nxz-xy-zw)}{y}\right). $$

(5)

If ω≠ 0, then we consider the following cases:

• $4\nmid x+z$. From Proposition 1, we have $(D(x,z),u)_{\infty }=1$. Because D(x,z) < 0, by Lemma 1 we have u > 0, contradicting u < 0.

• $4\nmid x-z$. From Proposition 1, we have $(H(x,z),u)_{\infty }=1$. Because H(x,z) < 0, by Lemma 1 we have u > 0, contradicting u₀ < 0.

• 4|x + z and 4|x − z. Let x = 2x₁ and z = 2z₁ with $x_{1},z_{1}\in \mathbb {Z}$ and $2\nmid x_{1},z_{1}$. Then $4\nmid x_{1}+z_{1}$ or $4\nmid x_{1}-z_{1}$. From ω² = u(u² + Au + B), we have

  $$ \left( \frac{\omega}{2^{6}}\right)^{2}=\frac{u}{2^{4}}\left( \left( \frac{u}{2^{4}}\right)^{2}+A_{1}\left( \frac{u}{2^{4}}\right)+B_{1}\right), $$

  where $A_{1}=px_{1}z_{1}(16pn^{2}x_{1}z_{1}-{x_{1}^{2}}-{z_{1}^{2}})$ and $B_{1}=p^{2}{x_{1}^{4}}{z_{1}^{4}}$. Now $4\nmid x_{1}-z_{1}$ or $4\nmid x_{1}+z_{1}$, so we have $\big (D(x_{1},z_{1}),\frac {u}{2^{4}}\big )_{2}=1$ or $\big (H(x_{1},z_{1}),\frac {u}{2^{4}}\big )_{2}=1$. In addition,

  $$ (D(x,z),u)_{2}=\left( 2^{2}D(x_{1},z_{1}),2^{4}\frac{u}{2^{4}}\right)_{2}=\left( D(x_{1},z_{1}),\frac{u}{2^{4}}\right)_{2}. $$

  Similarly

  $$ (H(x,z),u)_{2}=\left( H(x_{1},z_{1}),\frac{u}{2^{4}}\right)_{2}. $$

  So (D(x,z),u)₂ = 1 or (H(x,z),u)₂ = 1. Hence u > 0, contradicting u < 0.

  Therefore, ω = 0. From (5), we have

  $$ 4nxz-xy-zw=0. $$

  (6)

  Thus

  $$ \frac{y}{z}+\frac{w}{x}=4n. $$

  Hence

  $$ \frac{x}{y}+\frac{z}{w}=4np. $$

  (7)

  Now fix y, w and consider the equation F_y,w(X,Z,d) = 0, where

  $$ F_{y,w}(X,Z,d)=X^{2}yZ+py^{2}Zwd^{2}+Z^{2}wX+pw^{2}Xyd^{2}-8npXZdyw. $$

  Then F_y,w(0,1,0) = F_y,w(x,z,1) = 0. So F_y,w(X,Z,d) = 0 is isomorphic to the elliptic curve

  $$ C_{y,w}\colon \omega^{\prime2}=u^{\prime}\left( u^{\prime2}+pyw\left( 16n^{2}pyw-y^{2}-w^{2}\right)u^{\prime}+p^{2}y^{4}w^{4}\right), $$

  via the rational maps $\alpha \colon F_{y,w}\rightarrow C_{y,w}$,

  $$ \alpha(X:Z:d)=\left( \frac{-w^{2}pZy^{2}}{X},\frac{-py^{2}w^{2}Z(Xy+Zw-4npywd)}{Xd}\right), $$

  and $\beta \colon E_{y,w}\rightarrow F_{y,w}$,

  $$ \beta(u^{\prime},\omega^{\prime})=\left( pw^{2}(4npywu^{\prime}+\omega^{\prime}):-u^{\prime}(4npywu^{\prime}+\omega^{\prime}):wu^{\prime}(u^{\prime}-py^{3}w)\right). $$

We have the following result:

Proposition 2

• $(D(y,w),u^{\prime })_{q}=(H(y,w),u^{\prime })_{q}=1$ if q is an odd prime.

• $(D(y,w),u^{\prime })_{2}=(D(y,w),u^{\prime })_{\infty }=1$ if $4\nmid y+w$.

• $(H(y,w),u^{\prime })_{2}=(H(y,w),u^{\prime })_{\infty }=1$ if $4\nmid y-w$.

Proof

The same as for Proposition 1. □

In order for (2) to have positive integer solutions, we seek for points $(u^{\prime },\omega ^{\prime })$ on C_y,w such that $\beta (u^{\prime },\omega ^{\prime })=(X:Z:d)$ satisfies d≠ 0, $\frac {X}{d}>0$ and $\frac {Z}{d}>0$. If u = 0, then ω = 0. Because β(0,0) = (1 : 0 : 0). We must have u≠ 0. So

$$ \left\{\begin{array}{l} \frac{pw(4npywu^{\prime}+\omega^{\prime})}{u^{\prime}(u^{\prime}-py^{3}w)}>0,\\ -\frac{4npywu^{\prime}+\omega^{\prime}}{w(u^{\prime}-py^{3}w)}>0. \end{array}\right. $$

Multiplying together the two inequalities gives $u^{\prime }<0$. Assume

$$ (u^{\prime},\omega^{\prime})=\alpha(x:z:1)=\left( \frac{-py^{2}w^{2}z}{x},\frac{-py^{2}w^{2}z(yx+wz-4npyw)}{x}\right). $$

(8)

If $\omega ^{\prime }\neq 0$, we then consider the following cases:

• $4\nmid y+w$. From Proposition 2, we have $(D(y,w),u^{\prime })_{\infty }=1$, thus $u^{\prime }>0$ because D(y,w) < 0 by Lemma 1. This contradicts $u^{\prime }<0$.

• $4\nmid y-w$. From Proposition 2, we have $(H(y,w),u^{\prime })_{\infty }=1$. Because H(y,w) < 0, we have $u^{\prime }>0$, contradicting $u^{\prime }<0$.

• 4|y + w and 4|y − w. Then y = 2y₁ and w = 2w₁, where $2\nmid y_{1},w_{1}$. Then $4\nmid y_{1}+w_{1}$ or $4\nmid y_{1}-w_{1}$. Then similar to the case $4\nmid x+z$ and $4\nmid x-z$, we have $(D(y,w),u^{\prime })_{\infty }=1$ or $(H(y,w),u^{\prime })_{\infty }=1$; the either case implies $u^{\prime }>0$, which contradicts $u^{\prime }<0$.

  Therefore, $\omega ^{\prime }=0$. From (8), we have

  $$ xy+zw-4npyw=0. $$

  (9)

  From (6) and (9), we have

  $$ 4nxz=4npyw. $$

  Thus,

  $$ \frac{x}{y}\frac{z}{w}=p. $$

  (10)

  From (7) and (10), we have

  $$ (4np)^{2}-4p=\left( \frac{x}{y}-\frac{z}{w}\right)^{2}. $$

  Thus $4n^{2}p^{2}-p\in \mathbb {Q}^{2}$, hence $4n^{2}p^{2}-p \in \mathbb {Z}^{2}$, impossible because $p^{2}\nmid 4n^{2}p^{2}-p$. Therefore, there are no positive integer solutions to (1).