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Prove that $a$, where $0 \leq a < 2^n$, is a quadratic residue modulo $2^n$ is odd if and only if $a = 4^m(1+8k)$ for some nonnegative integers $m,k$.

If $a$ is odd we have $r^2 \equiv a \pmod{2^n}$ where $r = 1+2k$. Then $$r^2 = (1+2k)^2 = 1+4k+4k^2 = 1+4k(1+k).$$ If $k$ is even then $1+4k(1+k) = 1+8\left(\frac{k}{2}\right)(1+k)$. How do we continue? We also have the case where $r^2 \geq 2^n$. How do we continue?

user19405892
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  • If $k$ is odd, then $1+4k(k+1) = 1+8k\frac{k+1}2$, so $a$ is still of that form. Now to take care of the case where $r$ is even. – Arthur Jan 08 '17 at 13:19
  • Very closely related. There may be a perfect match. Did you search? – Jyrki Lahtonen Jan 08 '17 at 13:24
  • @JyrkiLahtonen Your linked question is only one direction, this one is an iff statement. so very related, but not a perfect match. – Arthur Jan 08 '17 at 13:25
  • @Arthur I didn't find a better one right away (the search engine is anything but ideal). My answer to that question gives the "iff" for odd $a$ but I agree that it is not a perfect match. – Jyrki Lahtonen Jan 08 '17 at 13:43

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