Problem source

The integers $a,b$ and $c$ satisfy 
\[
2a^{2}+b^{2}=5c^{2}.
\]
By considering the possible values of $a\pmod5$ and $b\pmod5$, show
that $a$ and $b$ must both be divisible by $5$. 

By considering how many times $a,b$ and $c$ can be divided by $5$,
show that the only solution is $a=b=c=0.$

Solution source

\begin{array}{c|ccccc}
a & 0 & 1 & 2 & 3 & 4 \\
a^2 & 0 & 1 & 4 & 4 & 1
\end{array}

Therefore $a^2 \in \{0,1,4\}$ and so we can have

\begin{array}
$2a^2+b^2 & 0 & 1 & 4 \\ \hline
0 & 0 & 1 & 4 \\
1 & 2 & 3 & 1 \\
4 & 3 & 4 & 2
\end{array}

Therefore the only solution must have $5 \mid a,b$, but then we can write them has $5a'$ and $5b'$ so the equation becomes

$2\cdot25 a'^2 + 25b'^2 = 5c^2$ ie $5 \mid c^2 \Rightarrow 5 \mid c$. But that means we can always divide $(a,b,c)$ by $5$, which is clearly a contradiction if we consider the lowest power of $5$ dividing $a,b,c$ for any solution.