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The answer is no. Take $\alpha=\sqrt{2}$ and note that if $|\sqrt{2}-m/n|\le 1/n^2$ then we have $0<|2n^2-m^2| \le (\sqrt{2}n+m)/n \le 3$. Now suppose we want $n\equiv 4\pmod p$ say. Then we must have that $32-m^2 \equiv b \pmod p$ for some $|b|\le 3$. But we can find a prime $p$ for which the numbers $29$ to $35$ are all quadratic non-residues $\pmod p$ (for example, choose $p$ so that $2, 7, 11, 29, 31$ are all non-residues $\pmod p$, and $3, 5, 17$ are residues). Thus there are no good approximations to $\sqrt{2}$ with $n\equiv 4\pmod p$ for such a prime $p$. One can clearly vary this argument a fair bit.