Let $a, b ,n \in Z$ with $n > 0$ and $a \equiv b \pmod n$.
Show that $\gcd(a,n) = \gcd(b,n) = d$.
I tried rewriting $b = a + ny$ for some $y \in Z$. Now if I have $d \mid a$ and $d \mid n$, then $d \mid b$. However I don't know how to continue.
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Do the reverse: if $d$ divides $b,n$ then... – abiessu Sep 09 '19 at 19:51
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If I write $a = b + ny$ , then $d \mid b$ and $d \mid n$ so $d \mid a$. So $\gcd(a,n) = \gcd(b,n)$, right? – meowmeowxw Sep 09 '19 at 19:57
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Suppose $k$ divides $a$ and $n$. Since $a\equiv b\pmod n$, there exists $\ell$ such that $a-b=\ell n$. Then $b=a-\ell n$, and since $k$ divides $a$ and $n$, it divides $b$. Similarly, if $k$ divides $b$ and $n$ it also divides $a$. Now conclude result from here.
pancini
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This shows that the set of divisors of $a$ and $n$ is the same as the set of divisors of $b$ and $n$. The greatest common divisor is just the biggest element in each set, but both sets are equal. – pancini Sep 09 '19 at 20:09
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