Prove that the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many solutions in positive integers.

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

$$(xz+1)(yz+1)=az^{k}+1$$

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.

• 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 $z>a^2+2a$? Sep 26, 2020 at 7:18
• For the case $k=2$, note that $x+y\leq xy$ fails if $x=y=1$. Sep 26, 2020 at 10:23
• The case $x=1$, $y=1$ can be proved separately. We may assume $x>1$, $y>1$.
– ASP
Sep 27, 2020 at 9:21
• 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. Sep 27, 2020 at 15:20
• David, it is time, long past time, to say exactly where you got this question and what it means to you. Sep 28, 2020 at 15:07

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

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

Solve $$kd\equiv -1\pmod{p}.$$ Let $$n=(kd+1)/p.$$ Note that since $$p $$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 $$1 is odd and $$\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 $$x\mid z.$$

• 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. Sep 29, 2020 at 2:40
• Thomas, please read this short transcript, it is the reason i was willing to continue work on this: chat.stackexchange.com/transcript/113484 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 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 $k.$ Sep 29, 2020 at 0:53