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 $u^2\equiv 1(\mathrm{mod}3)$, so $w^2=2u^2\equiv 2(\mathrm{mod}3)$, 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. $\square$