If $a$ and $b$ are coprime, i.e. $\gcd(a,b) = 1 = ax+by$, for certain integer $x$ and $y$, and $a = bc$, then $a \mid c$
I've tried multiply $c$ to both sides of $1 = ax+by$, but didn't work.
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1Are you sure that this strategy didn't work out? – user2345678 Oct 14 '17 at 23:03
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In fact with these conditions $b=\pm 1$, otherwise $bc$ has prime factors not in $a$. – Joffan Nov 08 '17 at 23:58
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Have a look at this question and other posts linked there. – Martin Sleziak Nov 09 '17 at 06:39
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1Possible duplicate of suppose $\gcd(a,b)= 1$ and $a$ divides $bc$. Show that $a$ must divide $c$. – Martin Sleziak Nov 09 '17 at 06:40
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$1 = ax + by = bcx + by = b(cx + y)$ so $b|1$ so $b = \pm 1$ so $a = \pm c$. So $\gcd (a,b)=\gcd(bc,b)=1$ so $b|1$ so $b = \pm 1$ so $a =\pm c$. – fleablood Nov 09 '17 at 07:58
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As $a=bc$, $a|bc$.
From Bezout's identity, there exist $x,y$ such that $xa+yb=1$. Multiply both sides by $c$ and get $$xac+ybc=c.$$ $xac$ is divisible by $a$, $ybc$ is divisible by $bc$, which is divisible (by hypothesis) by $a$. So $c$ must also be divisble by $a$.
man_in_green_shirt
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Still it should work if you use all the hypotheses: you obtain $c=acx+bcy$. Now, $a$ is supposed to divide $bc$, i.e. there exists $k\in \mathbf Z$ such that $bc=ak$, so $$c=acx+aky=a(cx+ky).$$
Bernard
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