I'm wanting to work out how many solutions there are to an equation of the form $x^2 = 1 \pmod {1155}$ in $\mathbb Z/1155\mathbb Z$, and I'm a little unsure of what the correct way to work this out is. I've attempted it several times now and know that I'm not answering it correctly, so I'm hoping someone on here will be able to help me. A step by step method would be really really really helpful!!
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1Welcome to Mathematics Stack Exchange. Do you know the Chinese remainder theorem? Have you factored $1155$? – J. W. Tanner Jun 01 '20 at 17:32
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$1155$ is factored into primes as $3\times5\times7\times11$.
If $x^2\equiv1\bmod1155$, then $x^2\equiv1\bmod3, 5, 7, $ and $11$, so $x\equiv\pm1\bmod3, 5, 7, $ and $11$.
By the Chinese remainder theorem,
each solution mod $1155$ corresponds to one of the solutions modulo $3, 5, 7, $ and $11$;
for an easy example, $x\equiv-1\bmod3, 5, 7, $ and $11$ corresponds to $x\equiv-1\bmod1155$.
There are two possibilities for $x\bmod3$, two for $x\bmod5$, two for $x\bmod7$, and two for $x\bmod 11$,
so, altogether, there are $2^4$ possibilities for $x\bmod1155$.
J. W. Tanner
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There are no solutions to $x^2\equiv-1\pmod{1155}$, because there are no solutions to $x^2\equiv-1\pmod3$ – J. W. Tanner Jun 01 '20 at 20:05