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?

Some ideas: for $\,a,b,\in\Bbb F_p^*\,$ :
$$a^2+2b^2=0\pmod p\iff -2=x^2\pmod p\iff 1=\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)\iff$$
$$p=1,3\pmod 8\;\;(\text{why?})$$
For example, with $\,p=7\,\;$ we get
$$\left(\Bbb F_7^*\right)^2=\{1,2,4\}\;,\;\;2\cdot\left(\Bbb F_7^*\right)^2=\{1,2,4\}\implies a^2+2b^2\neq0\pmod 7\;\;\forall\;a,b\in\Bbb F_7^*$$
but with $\,p=11\,$ we get
$$\left(\Bbb F_{11}^*\right)^2=\{1,4,9,5,3\}\;,\;\;2\cdot\left(\Bbb F_{11}^*\right)^2=\{2,8,7,10,6\}$$
and we certainly have several (how many?) solutions: 
$$1^2+2\cdot4^2=0\pmod{11}\;,\;\;2^2+2\cdot 3^2=0\pmod{11}\;,\;\;etc.$$
