
NT1 Find all the positive integers $x$ and $y$ that satisfy the equation

$$
x(x-y)=8 y-7
$$


## Solution 1:

The given equation can be written as:

$$
\begin{aligned}
& x(x-y)=8 y-7 \\
& x^{2}+7=y(x+8)
\end{aligned}
$$

Let $x+8=m, m \in \mathbb{N}$. Then we have: $x^{2}+7 \equiv 0(\bmod m)$, and $x^{2}+8 x \equiv 0(\bmod m)$. So we obtain that $8 x-7 \equiv 0(\bmod m) \quad(1)$.

Also we obtain $8 x+8^{2}=8(x+8) \equiv 0(\bmod m) \quad(2)$.

From (1) and $(2)$ we obtain $(8 x+64)-(8 x-7)=71 \equiv 0(\bmod m)$, therefore $m \mid 71$, since 71 is a prime number, we have:

$x+8=1$ or $x+8=71$. The only accepted solution is $x=63$, and from the initial equation we obtain $y=56$.

Therefore the equation has a unique solution, namely $(x, y)=(63,56)$.
