If for some positive integer $x,y,m$ we get $x^m \equiv -1 \pmod{y}$, then can I surely say that $2m$ is the order of $x$ w.r.t. $y$? If so, then how?
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1No. $2^3\equiv -1\pmod 3$ but the order of $2\pmod 3$ is not $6$. – lulu Feb 07 '23 at 16:15
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Restricting to the nontrivial case where the modulus $y>2,,$ notice that by invoking the order test we infer that $,x$ has order $2m\iff x^{2m/p}\not \equiv 1,$ for all odd primes $p\mid m.,$ But your hypotheses don't imply this need hold, e.g. as here this is true in the case $,m = p,$ an odd prime $\iff x\not\equiv -1,,$ and numerous other counterexamples are easy to construct using the above criterion. – Bill Dubuque Feb 07 '23 at 17:51
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If by order you mean that given the natural projection $\phi_y:\mathbb{Z}\to \mathbb{Z}/y\mathbb{Z}$ the image of $x$ has order $2m$, then it is true that the order of $x$ divides $2m$ , because $(-1)^2$, so $((\phi_y(x))^m)^2 = 1$.
But it doesn't have to be true, lulu gave a counterexample, in general just take any element $x$ such that $(\phi_y(x))^m = -1$ then $(\phi_y(x))^{3m} = -1$ but not both $2m$ and $6m$ can be the order of $x$. It is true, if $m$ is the smallest number such that this is the case (by definition of order as the smallest number such that the element to this power is 1).
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