2. Find all primes $p$ such that $a^{3} b-a b^{3}$ is divisible by $p$ for any integers $a$ and $b$.
Answer: 2 and 3.

Solution. Let \( a = 2 \) and \( b = 1 \), then \( a^3 b - a b^3 = 8 - 2 = 6 = 2 \cdot 3 \).

Therefore, there can be no other prime divisors besides 2 and 3.

We will prove that 2 and 3 are suitable.
Factorize this expression:

\( a^3 b - a b^3 = a b (a^2 - b^2) = a b (a - b)(a + b) \).

We will prove that it is divisible by 2. If one of the numbers \( a \) and \( b \) is divisible by 2, then one of the first two factors is divisible by 2, and thus the entire product. If neither is divisible by 2, then both are odd, and their difference is divisible by 2, meaning the entire product is divisible by 2.

We will prove that it is divisible by 3. If one of the numbers \( a \) and \( b \) is divisible by 3, then one of the first two factors is divisible by 3, and thus the entire product. If neither is divisible by 3, then they can give the same remainder when divided by 3, and in this case, their difference is divisible by 3, or if they give different remainders, which are 1 and 2, then their sum is divisible by 3, and thus the entire product.

## Grading Criteria.

If it is not proven that there are no other prime divisors besides 2 and 3: no more than 5 points.

Proving divisibility by 2: 2 points.

Proving divisibility by 3: 3 points.