Let p, q and r be prime other than 3. Show that 3 divides p²+q²+r² The question:Let $p, q$ and $r$ be prime other than $3.$ Show that $3$ divides $p^2+q^2+r^2.$
I am not sure how to start. 
Should I use" Fundamental Theorem of Arithmetic"? I know every integer greater than $1$ is a prime or a product of primes, but I don't know how to apply on this question
 A: Hint: Work modulo $3$.  All primes except $3$ are congruent to either $1 \pmod 3$ or $-1 \pmod 3$.
A: Given a prime number $x_{i}$ other than $3$, it can be written as $x_{i} = 3n_{i} + r_{i}$ where $n_{i}$ and $r_{i}$ are integers and $1 \leq r_{i} \leq 2$. Then
$$
\sum_{i = 1}^{3}x_{i}^{2}
=
3\left(3\sum_{i = 1}^{3}n_{i}^{2} + 2\sum_{i = 1}^{3}n_{i}r_{i}\right)
+
\sum_{i = 1}^{3}\overbrace{\quad r_{i}^{2}\quad}^{\displaystyle{3r_{i} - 2}}
=
3\left\lbrack%
3\sum_{i = 1}^{3}n_{i}^{2} + 2\sum_{i = 1}^{3}n_{i}r_{i}
+
\sum_{i = 1}^{3}r_{i} - 2
\right\rbrack
$$ 
A: Brief lemma:
If $s\neq 3$ is prime, then $s^{2}=3t+1$, where $t \in \mathbb{N}$.
pf.: 'Note the case where $s=2$ is obvious. Now suppose $s\neq 3$ is an odd prime. Consider the expression $s^{2}-1=(s+1)(s-1)$. Recall that for any three consecutive integers, exactly one is divisible by the integer three. By our givens, $3 \nmid s$, and so either $$3|s-1 \text{ or } 3|s+1.$$In any case, we have $3|s^{2}-1$. Hence there exists $t \in \mathbb{N}$ such that $s^{2}-1=3t$. Therefore $s \neq 3$ is prime implies $s^{2}=3t+1$, where $t \in \mathbb{N}$. QED.'

OP's question:
pf.: 'Suppose $p,q,r$ are primes other than three. We complete the proof in several cases.
Case 1 ($p=q=r=2$): Trivial.
Case 2 ($p = 2 \land q,r \neq 2$): WLOG suppose $p=2$ and $q,r \neq 2$. By our above lemma, $p=3s+1$ and $r=3t+1$, where $s,t \in \mathbb{N}$. Hence 
\begin{equation}
\begin{split}
p^{2}+q^{2}+r^{2} & =4+(3s+1)^{2}+(3t+1)^{2}\\
       & = ...\\
       & = 9s^{2}+9t^{2}+6s+6t+6\\
       & = 3(3s^{2}+3t^{2}+2s+2t+2).\\
\end{split}
\end{equation}
Since $3s^{2}+3t^{2}+2s+2t+2 \in \mathbb{Z}$, we have $3|p^{2}+q^{2}+r^{2}.$
Case 3 ($p,q=2 \land r \neq 2$): Suppose without loss of generality that $p=q=2$ and $r\neq 2$. Hence by the above lemma, $r=3s+1 (\text{for }s \in \mathbb{N})$. It follows that
\begin{equation}
\begin{split}
p^{2}+q^{2}+r^{2} & =4+4+(3s+1)^{2}\\
       & = ...\\
       & = 9s^{2}+6s+9\\
       & = 3(3s^{2}+2s+3).\\
\end{split}
\end{equation}
Since $3s^{2}+2s+3 \in \mathbb{Z}$, we have $3|p^{2}+q^{2}+r^{2}.$
Case 4 ($p,q,r \neq 2$): If $p,q,r \neq 2,$ then by the above lemma, $$p=3s+1, \;\; q=3t+1, \;\; \text{ and } \;\; r=3u+1$$for $s,t,u \in \mathbb{N}.$ It follows that 
\begin{equation}
\begin{split}
p^{2}+q^{2}+r^{2} & =(3s+1)^{2}+(3t+1)^{2}+(3u+1)^{2}\\
       & = ...\\
       & = 9s^{2}+9t^{2}+9u^{2}+6s+6t+6u+3\\
       & = 3(3s^{2}+3t^{2}+3u^{2}+2s+2t+2u+1).\\
\end{split}
\end{equation}
Since $3s^{2}+3t^{2}+3u^{2}+2s+2t+2u+1 \in \mathbb{Z}$, we have $3|p^{2}+q^{2}+r^{2}.$
This proves that if $p,q,r \neq 3$ are primes, then $3|p^{2}+q^{2}+r^{2}.$ QED.'
A: We know that (any prime n other than 3)mod 3 is 1 or 2 implies that $n^2$ mod 3 is 1. And so $(p^2 + q^2 + r^2)$ mod 3 = $(p^2mod3 + q^2mod3 + r^2mod3)mod3 = 3mod3 = 0$.
Hence, 3 divides $(p^2 + q^2 + r^2)$.
