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Taken from here (last page).
Let $t\in\mathbb Z$ and let $0\le j\le g-1$ such that $j\equiv t \pmod g$.

Then,$$t\cdot \frac {m}{g}\equiv j\cdot \frac{m}{g}\pmod m$$

($g = \gcd (m,a)$ if it matters)

Why is that?

Also, if I'm trying to divide by $m$ I get:

$$\frac{t}{g} \equiv \frac{j}{g} \pmod {1} \implies \frac{t}{g} = \frac{j}{g} \implies j=t$$

but it doesn't have to be the case at all that $j=t$.

I'd be glad if you could help me arranging my thoughts around this.

Thanks!

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By your assumption, $(t-j)/g$ is an integer $c$. Therefore $$t\frac{m}{g}-j\frac{m}{g}=cm.$$ Is this divisible by $m$?

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  • $\begingroup$ okay, I get it. From here $t \frac{m}{g} = j\frac{m}{g} + cm \implies t \frac{m}{g} = j \frac{m}{g} \pmod m$ $\endgroup$ – Elimination Dec 10 '17 at 10:56
  • $\begingroup$ but why is dividing by $m$ leading to the might-be-false-conclusion that $j=t$? $\endgroup$ – Elimination Dec 10 '17 at 10:57

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