Let p, q be prime; if p | q + 1 then also p | q^(q + 1) + q, proof? Let $p$, $q$ be prime. if $p | q + 1$ then $p | q^{q + 1} + q$
Any elementary proof will be appreciated!
 A: Let us add the condition that $q$ is odd. We are told that $q\equiv -1\pmod{p}$. Since $q+1$ is even, it follows that $q^{q+1}\equiv 1\pmod{p}$, and the result follows.
Remark: We could write that $q+1=kp$ for some integer $k$, and therefore $q=kp-1$. Then we could imagine expanding
$$(kp-1)^{q+1} +(kp-1).$$
Since $q+1$ is even, the term of $(-1)^{q+1}$ in the expansion is $1$, which cancels the $-1$ from the $kp-1$ term. All the rest of the terms are divisible by $p$. 
We mention this way of doing it to point out how much more efficiently the congruence machinery does the job.
A: *

*if neither p nor q is not 2 then $oddnumber∤evennumber$ so the first term will be false.

*if only p is 2 and q is an odd prime number , then q+1 is even too and q^(q+1) is odd so q^(q+1)+q is even and then p | q^(q + 1) + q

*if only q is 2 and p is an odd prime number , then only if p=3 the first term will be true. but in this case we will get to 3|10 which is false! 

*if both q and p are two then the first term will be false.
so based on number 3 it can't be prooved.
