= Solution
For an odd <prime number> $p$, the <Legendre symbol> is zero when $p\mid a$, one when $a$ is a nonzero <quadratic residue> modulo $p$, and minus one otherwise. The number-theoretic <Gauss lemma (number theory)> says that, for $p\nmid a$, $(a/p)=(-1)^r$, where $r$ counts those least positive residues of $a,2a,\ldots,(p-1)a/2$ exceeding $p/2$.
For $a=2$, the residues are simply $2,4,\ldots,p-1$, so $r=(p-1)/2-\lfloor p/4\rfloor$. Examining the four odd residue classes modulo eight gives
$$
\boxed{\left(\frac2p\right)=(-1)^{(p^2-1)/8}
=\begin{cases}1&p\equiv1,7\pmod8,\\-1&p\equiv3,5\pmod8.\end{cases}}
$$
Now $2^m\equiv-1\pmod p$ and $p$ is odd. The <multiplicative order> $d$ of $2$ divides $2m$ but not $m$. Since $m$ is a power of two, every proper divisor of $2m$ divides $m$, whence $d=2m$. <Lagrange's theorem for finite groups> gives $2m\mid p-1$. Because $m\ge4$, this implies $p\equiv1\pmod8$, so the supplementary law just proved gives $(2/p)=1$. By <Euler's criterion>, $2^{(p-1)/2}\equiv1\pmod p$, and consequently $2m\mid(p-1)/2$. Therefore
$$
\boxed{p\equiv1\pmod{4m}}.
$$
Back to article page
