Problem 2. Find all primes $p \geq 3$ such that $p-\left[\frac{p}{q}\right] q$ is a square-free integer for any prime $q<p$.
2. It is easy to see that the remainder $p$ modulo $q$ is equal to $p-\left[\frac{p}{q}\right] q$. A direct verification shows that $p=3,5,7$ and 13 are solutions of the problem. Assume that $p \geq 11$ is a solution and let $q$ be a prime divisor of $p-4$. If $q>3$, the remainder under consideration is 4 - not square-free. So $q=3$ and $p=3^{k}+4$ for some $k \in \mathbb{N}$. Analogously, the prime divisors of $p-8$ are 5 or 7 , and these of $p-9$ are 2 or 7 . Since 7 cannot be a divisor in both cases, we get that $p=5^{m}+8$ or $p=2^{n}+9$. Hence $3^{k}=5^{m}+4$ or $3^{k}=2^{n}+5$.

In the first case, $3^{k} \equiv 1(\bmod 4)$, i.e. $k=2 k_{1}$, and so $\left(3^{k_{1}}-2\right)\left(3^{k_{1}}+2\right)=5^{m}$ which gives $k_{1}=1, m=1$, i.e. $k=2$. In the second case one has that $n \geq 2$ and we conclude as above (using modulo 4 and 3) that $k=2 k_{1}$ and $n=2 n_{1}$ are even integers. Then $\left(3^{k_{1}}-2^{n_{1}}\right)\left(3^{k_{1}}+2^{n_{1}}\right)=5$ which implies $k_{1}=n_{1}=1$, i.e. $k=n=2$. Thus, in both cases, new solutions do not appear.