
N6. Find all ordered triples $(x, y, z)$ of integers satisfying the following system of equations:

$$
\begin{aligned}
x^{2}-y^{2} & =z \\
3 x y+(x-y) z & =z^{2}
\end{aligned}
$$


Solution. If $z=0$, then $x=0$ and $y=0$, and $(x, y, z)=(0,0,0)$.

Let us assume that $z \neq 0$, and $x+y=a$ and $x-y=b$ where $a$ and $b$ are nonzero integers such that $z=a b$. Then $x=(a+b) / 2$ and $y=(a-b) / 2$, and the second equations gives $3 a^{2}-3 b^{2}+4 a b^{2}=4 a^{2} b^{2}$.

Hence

$$
b^{2}=\frac{3 a^{2}}{4 a^{2}-4 a+3}
$$

and

$$
3 a^{2} \geq 4 a^{2}-4 a+3
$$

which is satisfied only if $a=1,2$ or 3 .

- If $a=1$, then $b^{2}=1 .(x, y, z)=(1,0,1)$ and $(0,1,-1)$ are the only solutions in this case.
- If $a=2$, then $b^{2}=12 / 11$. There are no solutions in this case
- If $a=3$, then $b^{2}=1 .(x, y, z)=(1,2,-3)$ and $(2,1,3)$ are the only solutions in this case.

To summarize, $(x, y, z)=(0,0,0),(1,0,1),(0,1,-1),(1,2,-3)$ and $(2,1,3)$ are the only solutions.

Comments. 1. The original problem proposal asks for the solutions when $z=p$ is a prime number.

2. The problem can be asked with a single equation in the form:

$$
3 x y+(x-y)^{2}(x+y)=\left(x^{2}-y^{2}\right)^{2}
$$

