<s>[INST] Sums of squares 
Let $R(n)$ be the number of ordered pairs of integers $(x,y)$ with $x^2+y^2 = n$. Using the identity $$(a+b)^2+(a-b)^2 =2(a^2 +b^2),$$
  prove that for all integers $n$, $R(2n) = R(n)$.

I'm not really sure how to start this question. Any help would be appreciated.
 [/INST] If $n$ is even and $$n = x^2 + y^2$$ for integers $x$ and $y$, then $$\frac{n}{2} = (\frac{x+y}{2})^2 + (\frac{x-y}{2})^2,$$ (using the mentioned identity) and both these terms are integers, as $x$ and $y$ must have the same parity.
On the other hand, if $n$ is an integer with $$n = a^2 +b^2$$ for integers $a$ and $b,$ then $$2n = (a+b)^2 + (a-b)^2.$$ Hence if $n$ is even, we can replace $n$ by $\frac{n}{2},$ and $R(n)$ doesn't change.
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