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Find all the solutions of $x^2 \equiv 1 \mbox{ mod }365$.

We know that $365=5\cdot 73$. So if I could find the solutions of $x^2 \equiv 1 \mbox{ mod }5$ and $x^2 \equiv 1 \mbox{ mod }73$, using CRT I could find the solutions of the given equation.
I can solve $x^2 \equiv 1 \mbox{ mod }5$ by hand, but I'm sure that there is an easier way to solve $x^2 \equiv 1 \mbox{ mod }73$.
Would appreciate your help:)

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For any prime $p$, you can show that $x^2\equiv1\bmod p\iff x\equiv \pm1\bmod p$. To this end, notice that $$x^2\equiv 1\bmod p\iff \exists k\in\mathbb Z: x^2=k\cdot p+1\iff (x+1)(x-1)=k\cdot p$$ Euclid's Lemma implies now that either $p\mid x+1\iff x\equiv -1\bmod p$ or $p\mid x-1\iff x\equiv 1\bmod p$.

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I'm sure that there is an easier way to solve $x^2 \equiv 1 \mbox{ mod }73$.
Would appreciate your help:)

$x^2\equiv1\bmod73\iff x\equiv\pm1\bmod73$

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  • $\begingroup$ Yes, now to solve $x^2 \equiv 1 \mod 365$, apply the Chinese Remainder theorem with $x \equiv ±1 \mod 5$. $\endgroup$ – J. Linne Dec 10 '20 at 0:34

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