Let $p$ and $q$ be two primes such that $p≠q$ and $6|(p + q)$. Then $(p²+q²)$ is not divisible by 6

Question

Let p and q be two primes such that p≠q and 6 divides (p + q). Prove that 6 does not divide (p² + q²).

Actually if p+q is divisible by 6 then also divisible by 2 and 3 then p+q=2k and p+q=3r. Then how to proceed from here please help...

Answer 1

$\textbf{Hint:}$

Write $p^2+q^2$ as $(p+q)^2-2pq$. Check for divisibility by $3$. Argue that none of the two primes can be $3$ because $6\lvert p+q$ and $p\neq q$ to complete the proof.

Answer 2

Any prime except $2$ and $3$ should have a form of $6 k + 1$ or $6 k - 1$.

I.e., $p \equiv \pm 1 \pmod{6}$ for prime $p \ne 2, 3$.

Because $p \ne q$, then $p, q$ cannot be $3$. Hence $p, q \ge 5$ and satisfy the property above.

Then $p^2 + q^2 \equiv (\pm 1)^2 + (\pm 1)^2 \equiv 2 \pmod{6}$. Not divisible by $6$.

Answer 3

Just prove that $p^2+q^2$ is not divisible by $3$. This follows from the fact that $p^2+q^2 = (p+q)^2 - 2pq$. But $p+q$ is coprime to $pq$ (why?), so $2pq$, and hence $p^2+q^2$ is not divisible by 3.