, 2024

Solution:
Claim 1. Let $a$ and $b$ be positive integers. Then, $\operatorname{gcd}\left(a^{3}-b^{3},(a-b)^{3}\right)$ is squarefree if and only if $\operatorname{gcd}(a, b)=1$, $a-b$ is squarefree, and $a-b$ is not divisible by 3.

Proof. If $\operatorname{gcd}(a, b)=d>1$, then $g$ is divisible by $d^{3}$, hence not squarefree. Thus, we now restrict our attention to the case $\operatorname{gcd}(a, b)=1$. In that case, we factor out $a-b$ from the gcd and simplify it as follows:
$$\begin{aligned} \operatorname{gcd}\left(a^{3}-b^{3},(a-b)^{3}\right) & =(a-b) \operatorname{gcd}\left(a^{2}+a b+b^{2},(a-b)^{2}\right) \\ & =(a-b) \operatorname{gcd}\left((a-b)^{2}+3 a b,(a-b)^{2}\right) \\ & =(a-b) \operatorname{gcd}\left((a-b)^{2}, 3 a b\right) \end{aligned}$$
Moreover, since $\operatorname{gcd}(a, b)=1$, we have that $\operatorname{gcd}(a-b, a)=\operatorname{gcd}(a-b, b)=1$, and so $\operatorname{gcd}\left((a-b)^{2}, a b\right)=1$, which implies that $\operatorname{gcd}\left((a-b)^{2}, 3 a b\right)$ is either 1 or 3.

Thus, each of the following is equivalent to the next.
- $\operatorname{gcd}\left(a^{3}-b^{3},(a-b)^{3}\right)$ is squarefree.
- $(a-b) \operatorname{gcd}\left((a-b)^{2}, 3 a b\right)$ is squarefree.
- $a-b$ is squarefree and at least one of $a-b$ and $\operatorname{gcd}\left((a-b)^{2}, 3 a b\right)$ is not divisible by 3.
- $a-b$ is squarefree and $a-b$ is not divisible by 3.

The claim implies that $c=a-b$ is possible only if $c$ is squarefree and not divisible by 3. For any such $c$, we may construct $(a, b)$ that works by taking $a=c+1$ and $b=1$. Hence, the answer is simply the number of squarefree integers up to 50 that are not divisible by 3.

There are 16 multiples of 3 from 1 to 50. Among the remaining numbers, $4,8,16,20,28,32,40,44,25,50$, and $49$ are not squarefree. This leaves $50-16-11=23$ possible values of $a-b$.