when p is sum of two square integer, prove (a/p) which is legendre symbol = 1
Let a,b be integers and p be an odd prime. if $p$=$a^2+b^2$ and a is odd, prove $(a/p)$ which is legendre symbol = $1$
what i have done is that :
because p and a are odd, b must be even and p is the form of $4k+1$ ($k$ is integer)
and after this, how to prove it ?
It is enough to show that $(q/p)=1$ for any (necessarily odd) prime divisor $q$ of $a$.
By Quadratic Reciprocity, we have $(q/p)=(p/q)$. But since $a^2+b^2\equiv b^2\pmod{q}$, we have $((a^2+b^2)/q)=(b^2/q)=1$.