I tried using quadratic residues, but I couldn't get much further in determining what $m$ those are.
Additionally, Fermat's Little Theorem seems to be a special case rather than a general solution.
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$a^2\equiv1\pmod m$ implies $\gcd(a,m)=1,$ so those,conditions are redundant. – Thomas Andrews Sep 25 '24 at 22:01
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Weird that neither "duplicate" is a duplicate. Is this supposed t9 be true for all $a?$ – Thomas Andrews Sep 25 '24 at 22:03
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@ThomasAndrews yes, I improved the question. It should be true for all $a$ – Diego Cândido Sep 25 '24 at 22:21
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Honestly, I don't know how the "duplicated" questions could help me. – Diego Cândido Sep 25 '24 at 22:22
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@OP if you seek the moduli $,m,$ such that $,a^2\equiv 1,$ for all $a$ coprime to $m$ then see the linked dupes for proof that it holds $\iff m\mid 24.,$ If that is not your question then please elaborate more precsely. – Bill Dubuque Sep 25 '24 at 22:24
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Direction $(\Rightarrow)$ is proved in the first dupe and direction $(\Leftarrow)$ is proved in the 2nd dupe. See also the many linked posts there for more on this FAQ. – Bill Dubuque Sep 25 '24 at 22:33
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The title is incorrect. It's essentially nonsense. – suckling pig Sep 26 '24 at 05:39