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Let $a$ be an odd integer. Show that the congruence $x^2 \equiv a \pmod {2^e}$, where $e$ is an integer, $e \geq 3$, has either no solutions or exactly 4 incongruent solutions.

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closed as off-topic by TMM, Old John, Thomas Andrews, Branimir Ćaćić, Dan Rust Jan 5 '14 at 16:19

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If $x^2\equiv a\pmod{2^e}$ the also $x^2\equiv a\pmod{8}$. A square is odd iff the number itself is odd, and the square of an odd number is always $\equiv1\pmod 8$. Therefore, there is no solution if $a\not\equiv 1\pmod 8$. On the other hand, if $a\equiv 1\pmod 8$ there are exactly four solutions. This can be shown by induction on $e$. Indeed, if $e=3$, then $1,3,5,7\bmod 8$ are the four soulutions. Assume we have exactly four soulutions $x_1, x_2, x_3, x_4$ modulo $2^e$. Then there are at most eight solutions modulo $2^{e+1}$, namely $x_1, x_2, x_3, x_4, x_1+2^e, x_2+2^e, x_3+2^e, x_4+2^e$. But $$(x_i+2^e)^2=x_i^2+2^{e+1}x_i+2^{2e}\equiv x_i^2+2^{e+1}\pmod{2^{e+1}}$$ because $x_i$ is odd, so that of each pair $x_i, x_i+2^e$ there is at most one solution modulo $2^{e+1}$. On the other hand, $x_i^2\equiv a\pmod {2^e}$ implies that $x_i^2\equiv a\pmod {2^{e+1}}$ or $x_i^2\equiv a+2^e\pmod {2^{e+1}}$ so that eihter $x_i$ or $x_i+e^2$ gives us a solution modulo $2^{e+1}$.

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    $\begingroup$ Why are there at most $8$ solutions? How does your argument show that only one of each pair is a solution? $\endgroup$ – chubakueno Jan 5 '14 at 19:50
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    $\begingroup$ It looks to me that it shows that if $x$ is a solution, then so is $x+2^e$ $\endgroup$ – chubakueno Jan 5 '14 at 20:04

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