IMO Team Selection Test 3, June, 2020

For $i \ge 0$, let $a = 2 \cdot 5^i$ and $b = 3 \cdot 5^i$. Then both $a$ and $b$ have $2(i+1)$ divisors. Moreover, we have $2a+3b = 4 \cdot 5^i + 9 \cdot 5^i = 13 \cdot 5^i$, which also has $2(i+1)$ divisors. Therefore all even values of $k$ satisfy the condition in the problem.

Now suppose that $k$ is odd. Then $a$ has an odd number of divisors and therefore is a square, say $a = x^2$. The same reasoning shows that $b$ is also a square, say $b = y^2$, and $2a+3b$ is also a square, say $2a+3b = z^2$. Therefore we have
$$2x^2 + 3y^2 = z^2.$$
We show that this equation has no positive integer solutions.

Suppose for a contradiction that this equation does have a positive integer solution. Let $(x, y, z) = (u, v, w)$ be the solution with minimal $x+y+z$. So $2u^2 + 3v^2 = w^2$. Modulo $3$ this equation is $2u^2 \equiv w^2$. If $u$ is not divisible by $3$, then $u^2 \equiv 1 \pmod{3}$, so $w^2 = 2u^2 \equiv 2 \pmod{3}$, however, that isn't possible. Therefore $u$ is divisible by $3$, from which it follows that $w$ is divisible by $3$ as well. Now $2u^2$ and $w^2$ are both divisible by $9$, so $3v^2$ is divisible by $9$. It follows that $v$ is divisible by $3$ as well. However, now $(x, y, z) = (\frac{u}{3}, \frac{v}{3}, \frac{w}{3})$ also satisfies the equation, and it is a solution with smaller $x+y+z$ than the supposed minimal one. This is a contradiction. Therefore the equation $2x^2 + 3y^2 = z^2$ has no positive integer solutions.

It follows that no odd $k$ satisfies the condition in the problem. The positive integers $k$ that satisfy that condition are therefore the even integers. $\square$