Given two positive integers $a$ and $k>3$ : From experimental data, it appears the diophantine equation


has infinitely many solutions in positive integers $x,y, z$.

To motivate the question, it can easily be shown that if $k <3$, the given diophantine equation has no solutions in positive integers $x, y ,z$ with $z>a$.

Proof: $(xz+1)(yz+1)=az^{k}+1$ may be simplified to $xyz^{2}+(x+y)z=az^{k}$. If $k=1$, this reduces to $xyz+x+y=a$. Its clear that $a>z$ therefore there are no positive integral solutions in $x$ and $y$ when $z>a$. if $k=2$, we have the reduced equation $xyz+x+y=az$. We have $z$ | $x+y$, $z \le(x+y) \le xy$. Therefore $LHS=xyz+x+y>z^{2}$. Because $RHS=az$, we must have $a>z$ thus there are no solutions in positive integers $x ,y$ when $z>a$.

I would like to prove that given two positive integers $a$ and $k>3 $, the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many positive integer solutions $x, y, z$. I do not know how to start the proof.

  • $\begingroup$ For $z=1$ the diophantine equation becomes $$(x+1)(z+1)=a+1,$$ which boils down to factoring $a+1$. For $k>3$, are you also looking only for solutions with $z>a^2+2a$? $\endgroup$
    – Servaes
    Sep 26, 2020 at 7:18
  • $\begingroup$ For the case $k=2$, note that $x+y\leq xy$ fails if $x=y=1$. $\endgroup$
    – Servaes
    Sep 26, 2020 at 10:23
  • $\begingroup$ The case $x=1$, $y=1$ can be proved separately. We may assume $x>1$, $y>1$. $\endgroup$
    – ASP
    Sep 27, 2020 at 9:21
  • $\begingroup$ I've got finiteness for $z^3.$ The bound is $2a^2 + 2a-1,$ some sort of additional work required to get it down to $a^2 + 2a.$ Take me a while to typeset, it is just inequalities. $\endgroup$
    – Will Jagy
    Sep 27, 2020 at 15:20
  • 2
    $\begingroup$ David, it is time, long past time, to say exactly where you got this question and what it means to you. $\endgroup$
    – Will Jagy
    Sep 28, 2020 at 15:07

2 Answers 2


This answer is based on the excellent work of Will Jagy. This solves all cases of $k>3.$

Let $p<k$ be an odd prime such that $p\not\mid k.$

Solve $kd\equiv -1\pmod{p}.$ Let $n=(kd+1)/p.$ Note that since $p<k,$ $n>d.$

Then for any integer $t,$ we can take $z=a^{d}t^p$ so that $$\begin{align}az^k+1&=a^{kd+1}t^{kp}+1\\&=\left(a^nt^k\right)^p+1\\ &=(a^nt^k+1)\left(1+a^nt^k\sum_{j=1}^{p-1} (-1)^j\left(a^nt^k\right)^{j-1}\right) \end{align}$$

Where the last equation is because when $p$ is odd, $$ \begin{align}u^p+1&=(u+1) \sum_{j=0}^{p-1} (-1)^ju^j \\&=(u+1)\left(1+u\sum_{j=1}^{p-1}(-1)^ju^{j-1}\right)\end{align}$$

Now, since $n>d,$ we can set $$ \begin{align}x&=a^{n-d}t^{k-p}\\ y&=a^{n-d}t^{k-p}\sum_{j=1}^{p-1} (-1)^j\left(a^nt^k\right)^{j-1} \end{align}$$

For $k\geq 4$ we can always find such a $p$ by taking a prime factor of $n-1$ or $n-2$ if $n$ is even or odd, respectively.

So this solves all cases $k>3.$

You don't need $p$ prime, just that $1<p<k$ is odd and $\gcd(p,k)=1.$

k even

So when $k$ is even, we can take $p=k-1.$ Then $d=p-1$ and $n=p.$

Then for any integer $t,$ $$\begin{align}z&=a^{k-2}t^{k-1}\\x&=at\\y&=at\sum_{j=1}^{k-2}(-1)^j\left(a^{k-1}t^k\right)^{j-1}.\end{align}$$

k odd

Likewise, if $k=2m+1$ is odd, then you can take $p=2m-1,$ $d=m-1$ and $n=m.$ Then for any integer $t$:

$$\begin{align}z&=a^{m-1}t^{2m-1}\\ x&=at^2\\ y&=at^2\sum_{j=1}^{2m-2}(-1)^j\left(a^mt^{2m+1}\right)^{j-1} \end{align}$$

is a solution.

In particular, for $k>3$ there are infinitely many solutions $(x,y,z)$ with $a\mid x$ and $x\mid y$ and $x\mid z.$

  • 2
    $\begingroup$ Good. I found something that works, but $x,y,z$ are all going to be powers of $a.$ Nicer if you can work in $a^v t^u$ for each. $\endgroup$
    – Will Jagy
    Sep 29, 2020 at 2:40
  • 1
    $\begingroup$ Thomas, please read this short transcript, it is the reason i was willing to continue work on this: chat.stackexchange.com/transcript/113484 $\endgroup$
    – Will Jagy
    Sep 29, 2020 at 23:26

Getting there. Here is $k=4.$ a family of solutions to $$ a z^4 + 1 = (xz+1)(yz+1) $$

is parametrized by integer $t$ with

$$ y=at $$

$$ x = a^4 t^5 - at $$

$$ z = a^2 t^3 $$

Both sides of the equation are $$ a^9 t^{12} + 1 $$


For that matter, we can take care of all $k \neq 0 \pmod 3$ this way.

When $k > 3$ and $k \equiv 1 \pmod 3,$ we may take $$ y = a^{\frac{2k-5}{3}} \; t^{k-3} $$ $$ z = a^2 t^3 $$ followed by $$ x = y \left( y^2 z^2 - 1 \right) $$

When $k > 3$ and $k \equiv 2 \pmod 3,$ we may take $$ y = a^{\frac{k+1}{3}} \; t^{k-3} $$ $$ z = a t^3 $$ followed by $$ x = y \left( y^2 z^2 - 1 \right) $$

  • 1
    $\begingroup$ +1 Nice work. I came up with a complete answer based on your work here. Basically, you need an odd prime $p<k$ which does not divide $k.$ $\endgroup$ Sep 29, 2020 at 0:53

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