If b and c are coprime, how to show that $(bx+cy, bc) = (b,y) (c,x)$? I've tried showing that each side divides each other but I'm having trouble finding a suitable argument.
 A: Let $\,d = bx\!+\!cy.\,$ Then  $\ (b,d) = (b,cy)= (b,y)\ $ and $\ (c,d) = (c,bx)= (c,x)$
so  $\ (d,bc) = (d,(b,d)(c,d))= (d,(b,y)(c,x))=(b,y)(c,x)\ $ by $\,(b,y)(c,x)\mid d$
A: Let $p$ be a prime number that divides $(bx+cy,bc)$. Particularly $p\mid bc$ and since $b$ and $c$ are coprime then either $p\mid b$ or $p\mid c$.
Assume the former, i.e. $p\mid b$. Then $p\mid bx$ but we know that $p\mid (bx+cy)$ so $p\mid cy$. But $p\nmid c$ because $b$ and $c$ are coprime. So it follows that $p\mid y$, therefore $p\mid (b,y)$.
If $p\nmid b$ as we asumed, then with a similar reasoning we can deduce that $p\mid (c,x)$.
This means that $(bx+cy,bc)\mid (b,y)(c,x)$. *
Now let $q$ and $r$ be prime numbers that divide $(b,y)$ and $(c,x)$ repectively (and thus divide $b$ and $c$, respectively). Then $q\mid (bx+cy)$ and $r\mid (bx+cy)$. Note that $q\nmid (c,x)$ and $r\nmid (c,x)$ since $b$ and $c$ are coprime.
This means that $(b,y)(c,x)\mid (bx+cy,bc)$ **
The equations marked with * and ** combined result in
$$(bx+cy,bc) = (b,y)(c,x)$$
A: Hint: The right-hand-side certainly divides the left-hand-side.  Let $(b,y)=n$ and $(c,x)=m$.  Then $n$ divides both $b$ and $y$ and $m$ divides both $c$ and $x$.  Then $nm$ divides $bc$ because $n\mid b$ and $m\mid c$.  Moreover, $nm$ divides $bx+cy$ because $n\mid b$ and $n\mid y$ and $m\mid x$ and $m\mid c$, so each term factors into two pieces which are individually divisible by $n$ and $m$.
