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We have $r_2(n)$ is the number of ordered pairs of integers $(x,y)$ where $x^2 + y^2= n$

Using the identity $(a + b)^2 + (a - b)^2= 2(a^2 + b^2)$, prove $r_2 (2n) = r_2 (n)$

This means that the calculation of $r_2(n) $ can be reduced to the case when $n$ is odd.

I can see how the identity fits in to the question, but is there anything else to it?

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You have to demonstrate that the solution of the equations $x^2+y^2=n$ are in one-to-one correspondence with the solutions of $x^2+y^2=2n$. This is done by the following bijective map: $$ \begin{matrix} f: & \mathbb R^2 & \longrightarrow & \mathbb R^2 \\ & (a,b) & \longrightarrow & (a+b,\ a-b) \end{matrix} $$

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