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I’ve found this problem in a number theory book, it says:

If $a \equiv b \pmod n$ with $c>0$ and $\gcd(c,n)=1$, prove that $$\frac{a}{c} \equiv \frac{b}{c} \pmod n$$ It seems trivial but unfortunately i don’t know how to prove it .

I don’t know is it necessary to $c\mid a,b$ Because if $c\nmid a,b \implies \frac{a}{c} \notin \mathbb Z$ But modular arithmetic deals only with integers (I’m new to congruences so maybe I’m wrong.)

metamorphy
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PNT
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    In your header you are assuming $a\equiv b\pmod n$. In your post you assume $\frac bc\equiv a \pmod n$. I assume you meant the former? But even the former is false unless you add something like $\gcd (n,c)=1$ as an assumption. – lulu Feb 16 '21 at 21:49
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    For instance: letting $(a,b,n)=(10,20,5)$ we certainly have $a\equiv b \pmod n$. But now taking $c=10$ we have $\frac {10}{10}\not \equiv \frac {20}{10}\pmod 5$. – lulu Feb 16 '21 at 21:51
  • See the edit @lulu – PNT Feb 16 '21 at 22:03
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    Well, if $\gcd(c,n)=1$ then $c$ has a multiplicative inverse $d \pmod n$ In that case your question is just $a\equiv b\pmod n \implies ad\equiv bd\pmod n$ which is certainly true. – lulu Feb 16 '21 at 22:05
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    Hint: if gcd$(c,n)=1$ then you can find $p, q$ such that $pc+nq=1$ – Mark Bennet Feb 16 '21 at 22:06
  • If you prefer to restrict to integers divisible by $c$, then write $a=cA, b=cB$, Then $cA\equiv cB\pmod n\iff A\equiv B\pmod n$ again by multiplying by the multiplicative inverse of $c$. – lulu Feb 16 '21 at 22:07
  • Immediate consequence of the linked Congruence Product Rule - simply scale $,a\equiv b,$ by $,c^{-1},,$ which exists by Bezout since $,\gcd(c,n) = 1.\ $ Note $,n/c, :=, n:! c^{-1}.,$ See here and here for more on modular fractions. – Bill Dubuque Feb 16 '21 at 22:23

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