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For a positive integer $n$ with $0 \leq n<2^{12}-1$, let $f(n)$ be equal to $n$ if there is a perfect square $m$ which leaves a remainder of $n$ when divided by $2^{12}$, and let $f(n)$ be $0$ otherwise. Determine the remainder when $$\sum_{n = 0}^{2^{12}-1}f(n)$$ is divided by $2^{12}$.

For $m = k^2$ we have $k^2 \equiv n \pmod{2^{12}}$. We are then trying to determine the sum of the quadratic residues modulo $2^{12}$. How can we do this?

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    Start with this: an _odd_ residue is a $2$-adics square, thus a quadratic residue mod $2^{12}$, if and only if it is one greater than a multiple of $8$. – 2017-01-08
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    @OsarLanzi How do we do the even residues? – 2017-01-08
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    Multilpy any odd square succeasivy by $4$ ... . – 2017-01-08
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    I'd rather you work it out from my ideas. :-) – 2017-01-08

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