Problem 1613

To find all prime numbers $p$ which satisfy the given condition, we need to show that for any prime $q < p$, if $p = kq + r$ with $0 \leq r < q$, there does not exist an integer $a > 1$ such that $a^2 \mid r$.

We will prove that if $p > 13$ is prime, there is a prime $q < p$ such that the least residue of $p$ modulo $q$ is not square-free. In fact, we will prove a stronger statement: if $p > 13$ is odd (not necessarily prime), there is a prime $q$ such that the least residue of $p$ modulo $q$ is either $4$, $8$, or $9$.

1. **Assume $p > 13$ is a counterexample**:
Suppose that some odd $p > 13$ is a counterexample: there is no prime $q < p$ for which the remainder upon dividing $p$ by $q$ belongs to $\{4, 8, 9\}$.

2. **Consider $p - 4$**:
If there is a prime $q > 3$ dividing $p - 4$, then $4$ is the least residue of $p$ modulo $q$, which contradicts our assumption on $p$. Since $p$ is odd, $p - 4$ is also not divisible by $2$, so the only prime divisor of $p - 4$ must be $3$. In other words, $p - 4 = 3^k$ for some natural number $k$.

3. **Consider $p - 8$**:
If $p - 8$ has any prime factor greater than $8$, we once again contradict our assumption on $p$. Thus, we may assume that all of its prime factors belong to $\{2, 3, 5, 7\}$. But $2$ is impossible since $p - 8$ is odd, and $3$ is impossible since $p - 4$ is divisible by $3$. Thus, $p - 8$ is divisible only by $5$, $7$, or both.

4. **Consider $p - 9$**:
For the same reason as above, it too can only have prime factors among $\{2, 3, 5, 7\}$. Since $p - 4$ is divisible by $3$ and not by $5$, $p - 9$ is divisible by neither $3$ nor $5$. It may or may not be divisible by $7$, but we claim that in either case, $p - 8$ cannot be divisible by $7$. Here's why: If $p - 9$ is divisible by $7$, then $p - 8$ certainly isn't. If $p - 9$ is not divisible by $7$, then it must be a power of $2$ (since no other prime factors are available). But powers of $2$ can only leave remainders of $1, 2, 4$ upon division by $7$, so $p - 8$, which is one greater than a power of $2$, cannot be divisible by $7$.

5. **Conclude $p - 8$ is a power of $5$**:
Recall that the only possible prime factors of $p - 8$ were $5$ and $7$, and we just ruled out $7$. We may therefore conclude that $p - 8$ is a power of $5$: $p - 8 = 5^m$. Comparing this with $p - 4 = 3^k$, we obtain:
$$5^m + 4 = 3^k$$

6. **Solve the equation $5^m + 4 = 3^k$**:
Reducing modulo $4$, we see that $k$ must be even, so $k = 2t$ for some $t$. Now,
$$5^m = 3^{2t} - 4 = (3^t - 2)(3^t + 2)$$
But these factors cannot both be divisible by $5$ (they differ by $4$), so the only way their product can be a power of $5$ is if one of them is $1$ and the other is a power of $5$. This forces $t = 1$, $k = 2$ as required. Thus, the only solution is:
$$5 + 4 = 3^2 \implies p = 13$$

Therefore, the only prime number $p$ that satisfies the given condition is $p = 13$.

The final answer is $\boxed{13}$.