
5. Let $n$ be a positive integer and let $d(n)$ denote the number of ordered pairs of positive integers $(x, y)$ such that

$$
(x+1)^{2}-x y(2 x-x y+2 y)+(y+1)^{2}=n .
$$


Solution. First, we modify the examined expression:

$$
\begin{array}{r}
(x+1)^{2}-x y(2 x-x y+2 y)+(y+1)^{2}= \\
\left(x^{2}+2 x+1\right)-2 x^{2} y+x^{2} y^{2}-2 x y^{2}+(y+1)^{2}= \\
x^{2}\left(y^{2}-2 y+1\right)-2 x\left(y^{2}-1\right)+(y+1)^{2}+1= \\
(x(y-1)-(y+1))^{2}+1
\end{array}
$$

We see that we want to find the smallest integer $n$ of the form $t^{2}+1$ such that the equation

$$
x(y-1)-(y+1)= \pm t
$$

has exactly 61 solutions over positive integers.

We may assume $t \geq 0$. For $y=1$ we have $x(y-1)-(y+1)=-2$, for all $x$. So for $t=2$ we have infinitely many solutions.

Assume that $y=1$ yields no solution of (1). Then

$$
x=\frac{ \pm t+y+1}{y-1}=1+\frac{ \pm t+2}{y-1} .
$$

Clearly for $t=0$ the solutions $(x, y)$ are $(1,3),(1,3)$ and for $t=1$ the solutions are $(2,4),(4,2),(2,2)$. Assume that $t \geq 3$. Then the negative sign before $t$ cannot yield a solution, since

$$
1+\frac{-t+2}{y-1} \leq 1+\frac{-3+2}{y-1}=1-\frac{1}{y-1}<1 \text {. }
$$

Therefore it must hold

$$
x=1+\frac{t+2}{y-1} .
$$

The number of solutions of this equation is clearly equal to the number of divisors of $t+2$. So we want to find $t$ such that the number of divisors of $t+2$ is exactly 61 . Since 61 is a prime, $t+2$ must be of form $p^{60}$, where $p$ is a prime. Since we want to minimize $n$, we want to minimize $t$, therefore take $p=2$. The minimal $t$ is therefore $2^{60}-2$, which gives us $n=\left(2^{60}-2\right)^{2}+1$.
