I know some elementary proofs of this fact. I was wondering if there's some short slick proof of this fact using the structure of the $2$-adic integers? I'm looking for a proof of this fact that's easy to remember.
Proof that $a\equiv 1\,(\textrm{mod }8)$ implies $a$ is a square modulo $2^n$ for all $n$
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number-theory
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0$a=17, n=5.\ a\equiv17\mod2^5,$ but 17 is not a square... or didn't I understand your statement? – 2011-09-22
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0@FUZxxl: Is 17 a square *modulo* $2^5$? That is, does there exist $x$ such that $x^2\equiv 17\pmod{2^5}$? – 2011-09-22
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0Ah! I missunderstood. Thank you. – 2011-09-22
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2@asdfg: the shortest proof I can think of uses Hensel's Lemma. Do you count that as "using the structure of the $2$-adic integers"? – 2011-09-22
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2@Pete: Now that you mention it, letting $f(X)=X^2-a$, we have $|f(1)|_2=1/8$ while $|f'(1)^2|_2=1/4$, so $|f(1)|<|f'(1)|_2$ i.e. by Hensel, we find a root in $\mathbb{Z}_2$ and hence modulo every $2^n$. I guess this is as short as it gets. – 2011-09-22