<s>[INST] if $5\nmid a$ or $5\nmid b$, then $5\nmid a^2-2b^2$. I have  a homework as follow:
if $5\nmid a$ or $5\nmid b$, then $5\nmid a^2-2b^2$.
Please help to prove it.
EDIT: MY ATTEMPT
Suppose that $5\mid a^2-2b^2$, then $a^2-2b^2=5n$,where $n\in Z$,
then $a^2-2b^2=(a+\sqrt2b)(a-\sqrt2b)=5n$, 
Since 5 is a prime number, we get that $5\mid (a+\sqrt2b)$or $5\mid (a-\sqrt2b)$,
If $5\mid (a+\sqrt2b)$, then $5\mid a$ and $5\mid b$, contradiction.
If $5\mid (a-\sqrt2b)$, then $5\mid a$ and $5\mid b$, contradiction.
Thus $5\nmid a^2-2b^2$.
 [/INST] You just have to make a table for the function $a^2-5b^2$ in $\mathbf Z/5\mathbf Z$ above the horizontal line, the possible values for $a^2$; to the left of the vertical line, those for $b^2$):
$$\begin{array}{r|rrr}
&0&1&-1\\
\hline
0&0&1&-1\\1&-2&-1&2\\-1&2&-2&1
\end{array}$$
which shows the only case with $a^2-5b^2\equiv 0\mod5$ is when $a\equiv b\equiv 0\mod5$.
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