## Task 4.

Determine all pairs of distinct natural numbers $(a, b)$ for which

$$
a b(b-a) \mid a^{4}-b^{2}
$$
## First Solution.

Let $d = M(a, b)$. Then $a = d x$ and $b = d y$ for relatively prime $x, y$, and the expression becomes

$$
d^{3} x y(y-x) \mid d^{4} x^{4} - d^{2} y^{2}
$$

which simplifies to $d x y(y-x) \mid d^{2} x^{4} - y^{2}$. Now we see that $x \mid y^{2}$, and since $x$ and $y$ are relatively prime, it follows that $x = 1$, and the expression becomes

$$
d y(y-1) \mid d^{2} - y^{2}
$$

Now let $k = M(d, y)$. Then $d = k u$ and $y = k v$ for relatively prime $u, v$, and the expression becomes

$$
k^{2} u v(k v - 1) \mid k^{2}(u^{2} - v^{2})
$$

which simplifies to $u v(k v - 1) \mid u^{2} - v^{2}$. From this, it follows that $u \mid v^{2}$ and $v \mid u^{2}$, and since $u$ and $v$ are relatively prime, it follows that $u = v = 1$, so $d = y$ and we conclude that $(a, b) = (d, d^{2})$.

We easily see that all pairs $(a, b) = (d, d^{2}), d \neq 1$, are solutions to the problem.