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> Suppose $a,b,m$ are integers and $(a,m) = 1$. Does $a \equiv b \pmod m \Rightarrow (b,m) = 1$ ? 
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I know the least residues of $a$ and $b$ are the same. And I know $$(m,a)=(a, a \mod m)$$, so the least residue of $a$ is relatively prime to $m$. But $$b = qm + a \mod m$$ and $$(b,m) = (m, b \mod m) = (m, a \mod m) = 1$$

Is this correct ? Is there an easier way to show this ?