dealing with sum of squares (1) I need to be able to conclude that there are $a, b \in \Bbb Z$, not 0, such that $|a| < √p,\ |b| < √p$
and $$a^2 + 2b^2 ≡ 0\ (mod\ p)$$
I'm not sure how to go about this at all.  But apparently it is supposed to help me show (2) that 
there are $a, b \in \Bbb Z$, such that either $$a^2 + 2b^2 = p$$ or $$a^2 + 2b^2 = 2p$$
Any idea how to go about 1 or 2?

Hint $\ $ Apply the following result of Aubry-Thue

