Circumcircle wrote: Find all prime numbers $p$ and $q$ such that $p^q + 5q - 2$ is also a prime number. By parity $p=2$ or $q=2$. If $p=2$, then by FLT $q|2^q+5q-2 \Rightarrow q=2^q+5q-2$ which is impossible. If $q=2$ then $p^2+8$ is prime. But if $p \not = 3$, then $3|p^2+8>3$. So, $p=3$ and it works. Answer: $p=3,q=2$.

I don't know how I missed FLT during the contest

a jan zgjell ekipa ce kan me represent ks

NO_SQUARES wrote: Circumcircle wrote: Find all prime numbers $p$ and $q$ such that $p^q + 5q - 2$ is also a prime number. By parity $p=2$ or $q=2$. If $p=2$, then by FLT $q|2^q+5q-2 \Rightarrow q=2^q+5q-2$ which is impossible. If $q=2$ then $p^2+8$ is prime. But if $p \not = 3$, then $3|p^2+8>3$. So, $p=3$ and it works. Answer: $p=3,q=2$. waoo, that is an amazing sol. So basic and beautiful!! Good day.

Atlasi wrote: a jan zgjell ekipa ce kan me represent ks Pak vone

Note that (p, q) both odd doesn't work. If q=2, mod 3 guarantees p=3, and if p=2, mod 1 guarantees no solution. So the only solution (which can be checked to work) is (3, 2).

We claim that the only solution is $(p,q)=(3,2)$. If $q$ is odd then $p$ cannot be odd so must $p$ be 2. But note that $2^q +5q-2 \equiv 2^q-2\equiv 0 (\mod q)$, so this will not give any result. If $q$ is even then it is equal to $2$. If $p>3$, then $p^2 +8 \equiv 0 (\mod 3)$, so it cannot be prime. If $p=3$ then $3^2+8=17$ is indeed a prime. $\blacksquare$.

Huh only to cheak parity