# Has the diophantine equation $z^2+zx=2x^2y$ infinitely many solutions over integers $x,y,z\geq 2$?

After I did a reasoning involving even perfect numbers (see it if you want below my question) I am interested to know next

Question. Has $$z^2+zx=2x^2y\tag{D}$$ infinitelty many solutions $(x,y,z)$ over integers $x\geq 2$, $y\geq 2$ and $z\geq 2$? Many thanks.

To solve the question I did few reasonings like to try factorizations $z(z+x)$, or $x(2xy-z)$, and the code solve z^2+zx=2x^2 y, over integers using the Wolfram Alpha online calculator.

I don't know if our problem (D) was in the literature. I would like to know a parametric solution having infinitely many solutions, if you know it from the literature refer the literature, and I try to find it. If solve the Question is very hard, feel free to add your reasonings with this purpose to study this problem, or a remarkable heuristic if your experiments with a computer can provide us it.

I hope that these calculations are rights.

Claim. If there are infinitely many even perfect numbers, then (D) has infinitely many solutions*.

*As those that were stated, I say integer tuples $(x,y,z)$ with $x,y,z\geq 2$.

Proof. Let $M_p=2^p-1$ a Mersenne prime. The celebrated Lucas-Lehmer test, see [1] (is a free access journal) implies that there exist an integer $x\geq 2$ such that $$S_{p-2}=x\cdot M_p.\tag{1}$$ Thus denoting $z=S_{p-2}$ (see in [1] the definition, that is the sequence in the Lucas-Lehmer test), and since we know that an even perfect number can be written as $$y:=y_p=\frac{(M_p+1)}{2}M_p,\tag{2}$$ one gets solving $(2)$ for $M_p$ and doing the substitution in $(1)$, this
$$z=x\left(\frac{-1+\sqrt{1+8y}}{2}\right).\tag{3}$$ Write $(3)$ as $2z+x=x\sqrt{1+8y}$, and taking the square as $4z^2+4zx=8x^2y$.$\square$

Example. The tuple $(x,y,z)=(2,28,14)$ solves (D) since $$14^2+14\cdot 2-2\cdot 2^2\cdot 28=0,$$ and $x,y$ and $z$ are integers greater or equal than $2$.

## References:

[1] Berrizbeitia and Luca and Melham, On a Compositeness Test for $(2^p+1)/3$, Journal of Integer Sequences, Vol. 13 (2010), Article 10.1.7.

• Feel free also to do feedback in comments about my Question. I am asking it because I didn't a lot of calculations with a computer. On the other hand I don't know if this diophantine equation is easy to solve (in the past when I was a student I known different methods to solve nice diophantine equations), there are examples of diophantine equations, for which one can find parametric solutions having infinitely many terms, but there are other examples where it is very hard. – user243301 Oct 13 '17 at 10:19

One family of infinite solutions of the diophantine equation $z^2+zx=2x^2y$ is $$(x,y,z)=\left(n,\frac{mn(mn-1)}{2},n(mn-1)\right)$$ with $n,m\in\mathbb{Z}$. If $n,m\geq 2$, then $x,y,z\geq 2$. Your solution $(4,28,14)$ can be obtained by letting $n=2$ and $m=4$.

• Many thanks, I am going to check it. – user243301 Oct 13 '17 at 10:26
• I found a more general family of infinite solutions with TWO parameters. Your solution (4,28,14) can be obtained by letting $n=2$ and $m=4$. – Robert Z Oct 13 '17 at 11:00
• Many thanks for your answer and effort. Now I see then that my question (my thoughts realated to perfect nubmers) was miscellaneous. – user243301 Oct 13 '17 at 11:37

Take any $x \ge 2$ and let $z=2nx^2$ for some integer $n$. Then

$z^2 + zx = 2nx^2(2nx^2+x)$

so if you let $y=n(2nx^2+x)$ then $z^2 + zx = 2x^2y$.

• Many thanks for your details, I am going to study your answer. – user243301 Oct 13 '17 at 10:27