# Why does $\genfrac(){.4pt}{}{a+b}p=\genfrac(){.4pt}{}2{a+b}$ if $p=a^2+b^2$?

$\def\legendre{\genfrac(){.4pt}{}}$Part of Dirichlet's proof of the quartic character of 2 is confusing me.

Suppose $p\equiv 1\mod 4$, so $p$ is the sum of two squares $p=a^2+b^2$. The claim is that $\legendre{a+b}p=\legendre2{a+b}$.

Using the Jacobi symbol, I can fill in that $\legendre{a+b}p=\legendre p{a+b}=\legendre{(a+b)^2-2ab}{a+b}=\legendre{-2ab}{a+b}=\legendre2{a+b}$, although I can't explain the last equality.

How does the last equality work? I can't figure out why $\legendre{-ab}{a+b}=1$. Can somebody please elucidate?

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Since $a \equiv -b$ (mod $a+b$), we have ($\frac{-ab}{a+b}$) $=$ ($\frac{b^2}{a+b}$) $= 1$.