65th Czech and Slovak Mathematical Olympiad

Firstly, let us note that $5^3-4^3=61$, $2^3-1^3=7$ and $3^3-2^3=19$ are nice primes, so 1, 7 and 9 belong to desired digits. We show that they are all desired digits. Let $p=m^3-n^3$ be a nice prime, where $m>n$ are positive integers. Second factor in rewriting $$p=m^3-n^3=(m-n)(m^2+mn+n^2),$$ is greater than 1, thus the first one is 1 and therefore $m=n+1$. After substitution we obtain $$p=3n^2+3n+1.(1)$$ An estimate $3n^2+3n+1>6$ gives that the prime $p$ is odd and greater than 5. This excludes 0, 2, 4, 5, 6 and 8 as the last digits and 3 stays the only remaining digit to exclude. It is sufficient to find remainders of the numbers $3n^2+3n+1$ after division by 5. For remainders 0, 1, 2, 3 and 4 of $n$ we obtain remainders 1, 2, 4, 2, 1 of (1) which ones really exclude the last digit 3.

Answer. The last digits of the nice primes are 1, 7 and 9.