<s>[INST] $p \equiv 5 \mod8\Rightarrow p=(2x+y)^{2}+4y^{2}$ If $p \equiv 5 \mod8$ , then $p=(2x+y)^{2}+4y^{2}$,for some x and y integers.
Thanks
Here is my approach:
I know $p \equiv 5 \mod8\Rightarrow $
$p \equiv 1 \mod4\Rightarrow $
$n^{2}+m^{2}=p\equiv 5 \mod8 \Rightarrow$ 
from $t^{2}\equiv 0,1~or ~4 \mod8$ we get $n^{2} \equiv 1 \mod8$ and $m^{2} \equiv 4 \mod8$
$\Rightarrow n=4x-1$ and $m=2y$.
So I need $n=2x+y$.
 [/INST] $p \equiv 5 \mod8\Rightarrow $
$p \equiv 1 \mod4\Rightarrow $
$n^{2}+m^{2}=p\equiv 5 \mod8 \Rightarrow$ 
from $t^{2}\equiv 0,1~or ~4 \mod8$ we get $n^{2} \equiv 1 \mod8$ and $m^{2} \equiv 4 \mod8 \Rightarrow$
n is odd and $ m=2y$ where y is odd $\Rightarrow$
n-y is even $\Rightarrow$
n-y=2x and m=2y $\Rightarrow$
$p=(2x+y)^{2}+4y^{2}\square$.
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