Show that no number of the form 8k + 3 or 8k + 7 can be written in the form $a^2 +5b^2$ I'm studying for a number theory exam, and have got stuck on this question. 

Show that no number of the form $8k + 3$ or $8k + 7$ can be written in the form $a^2 +5b^2$

I know that there is a theorem which tells us that $p$ is expressible as a sum of $2$ squares if $p\equiv$  $1\pmod 4$.
This is really all I have found to work with so far, and I'm not really sure how/if it relates.
Many thanks!
 A: Hint If $x$ is even $x^2 \equiv 0,4 \pmod 8$ while if $x$ is odd we have $x^2 \equiv 1 \pmod 8$.
Thus 
$$a^2+5b^2 \in \{0,1,4\} + 5 \cdot \{0,1,4 \} \pmod 8 \,.$$
Actually you can prove that $a^2+5b^2$ is $0,1$ or $2 \pmod 4$.
A: This is a very basic way I am suggesting you. Assume that, for example $8k+3=a^2+5b^2$ for some integers $k,a,b$. And assume that $a$ and $b$ are both even. What will you get? $$a^2+5b^2=4q^2+5(4q'^2)\\\\\ a^2+5b^2=8k+3$$ So $4|3$ which is a contradiction. Another assumption. Let $a=(2q+1),b=(2q'+1)$ for some integers $q$ and $q'$. So $$a^2+5b^2=(2q+1)^2+5(2q'+1)^2\\\\\ a^2+5b^2=8k+3$$ So $4\mid3$ which is a contradiction again. Continue to have two contradiction for other two cases.
A: $8k+3,8k+7$ can be merged into $4c+3$ where $k,c$ are integers
Now, $a^2+5b^2=4c+3\implies a^2+b^2=4c+3-4b^2=4(c-b^2)+3\equiv3\pmod 4,$ 
But as $(2c)^2\equiv0\pmod 4,(2d+1)^2\equiv1\pmod 4,$
$a^2+b^2\equiv0,1,2\pmod 4\not\equiv3$
Clearly, $a^2+5b^2$ in the question can be generalized $(4m+1)x^2+(4n+1)y^2$ where $m,n$ are  any integers
