Originally Posted by

**JaneBennet** I don’t think so. What the question is saying is that if *a* and *m* are coprime, then *a* can be made congruent to any integer modulo *m* by multiplying by a suitable *x*.

I think you can do it this way. If $\displaystyle b\mid m$, then $\displaystyle x=0$ is a solution to the congruence equation. So assume $\displaystyle b\nmid m$ and write $\displaystyle b=qm+r,\ 0<r<m$.

Now consider the numbers $\displaystyle a,2a,\ldots,(m-1)a$. If $\displaystyle ai\equiv aj\!\!\!\pmod{m}$, where $\displaystyle 1\leq i,j\leq m$, then $\displaystyle m\mid a(i-j)$. But $\displaystyle \gcd{(a,n)}=1$. Hence $\displaystyle m\mid i-j\ \Rightarrow\ i-j=0$ (since $\displaystyle 0<i,j<m)$. So $\displaystyle i\ne j\ \Rightarrow\ ai\not\equiv aj\!\!\!\pmod{m}$. It follows that $\displaystyle a,2a,\ldots,(m-1)a$ are all distinct modulo *m*; thus $\displaystyle a,2a,\ldots,(m-1)a\!\!\!\pmod{m}$ is a permutation of $\displaystyle 1,2,\ldots,m-1\!\!\!\pmod{m}$.

Now *r* is one of the numbers $\displaystyle 1,2,\ldots,m-1$. Thus there is an integer *x* such that $\displaystyle ax\equiv r\!\!\!\pmod{m}$. This concludes the proof as $\displaystyle r\equiv b\!\!\!\pmod{m}$.