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since $( b|c)$, $c = xb$ So, $\gcd(a,b) = \gcd(a + xb, b)$ Euclidean algorithm shows shows that for $a=bq + r$, $\gcd(a,b) = \gcd(b,r)$ This is where im stuck, I've approached this quesiton so many ways but can't figure out how to apply the algorithm for the proof.

Bill Dubuque
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Draken147
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2 Answers2

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Since $b \mid c$, $\exists k \in \mathbb Z$ such that $c=bk$. Now by Euclidean Algorithm, one always has $gcd(a,b)=gcd(a+bt,b)$ for any integer $t$. Now choose $k=t$.

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Suppose that $\gcd\;(a,b)=k$, then, $k|a$ and $k|b$. Clearly, $k|c$ as $b|c$. Now, this implies $k|(a+c)$ and $k|b$.

Now, we want to show if there exists an $m$ such that $m|(a+c)$ and $m|b$, then $m|k$, as $k$ is the $\gcd(a+c,b)$.

Now, $m|(a+c) \implies m|a$ and $m|b$, from this it shows that $m|\gcd(a,b)$. Hence $\gcd(a,b)=\gcd(a+c,b)$

Nitish Kumar
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  • Im a bit confused with what u meant by 'm s.t' ? – Draken147 Apr 13 '20 at 15:48
  • I have improved it. It means there are more divisors, not necessarily only $k$. If there exists such an $m$, then we show $k$ is the largest divisor and hence it is the g.c.d @Draken147 – Nitish Kumar Apr 13 '20 at 15:50
  • On the middle part of the proof, u said that 'm|k as k is the gcd of (a+c,b). Do u mean k is the gcd of a,b because we havent proved the that yet? it was only proved that k is a divisor of (a+c) and b – Draken147 Apr 13 '20 at 16:24
  • $k$ is the greatest we have shown this. By showing that any divisor $m$ of $a+c,b$, will have to divide $k$. – Nitish Kumar Apr 13 '20 at 16:37
  • oh ok thank you – Draken147 Apr 14 '20 at 07:51