Prove that if both p and p^2 + 8 are prime numbers, then p^3 + 10 is also prime
thanks.
If $\displaystyle p=3$, then $\displaystyle p^2+8=17$ which is again prime, and surely enough $\displaystyle p^3+10=37$ is prime.
Now if $\displaystyle p\neq 3$ - and p prime-, then $\displaystyle p^2\equiv{1}(\bmod.3)$ -since p must be coprime to 3- thus $\displaystyle p^2+8\equiv{0}(\bmod.3)$ so $\displaystyle p^2+8$ cannot be prime now.
Then, since $\displaystyle p$ and $\displaystyle p^2+8$ are primes simultaneously only for $\displaystyle p=3$, and in this case the assertion is true, we are done.