Prove that if $a \equiv b \mod n$, then for all positives integers $e$ that divide both $a$ and $b$: $$ \frac{a}{e} \equiv \frac{b}{e} \mod \frac{n}{\gcd(n,e)}$$
my try
we know that $$ a\equiv b \mod n\Leftrightarrow ad \equiv bd \mod nd $$ let $$ d:= \frac{1}{\gcd(n,e)} $$ then we have $$ \frac{a}{\gcd(n,e)} \equiv \frac{b}{\gcd(n,e)} \mod \frac{n}{\gcd(n,e)}$$ but also we know that $$ \frac{\gcd(n,e)}{e} \equiv \frac{\gcd(n,e)}{e} \mod \frac{n}{\gcd(n,e)}$$ so $$ \frac{a}{\gcd(n,e)} \cdot \frac{\gcd(n,e)}{e} \equiv \frac{b}{\gcd(n,e)} \cdot \frac{\gcd(n,e)}{e} \mod \frac{n}{\gcd(n,e)}$$ and we have thesis: $$ \frac{a}{e} \equiv \frac{b}{e} \mod \frac{n}{\gcd(n,e)}$$ but probably it is wrong because I didn't use all assumptions... can somebody give me hint, how can I solve that and tell me where I failed?
