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So far I have that

Since we known that (a,b) = 1 by definition we can write it as ax+by=1. We also know that (c,d) = 1, so cx+dy=1. Since they are both equal to 1 I can set them equal to each other where ax+by=cx+dy.

Now I am unsure what my next steps would be to prove this. It seems that it would only work if the gcd of every variable is 1. But again, I am unsure how to show that.

bella
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    There is no reason why the values of $x,y$ in $ax+by=1$ should be the same as the values of $x,y$ in $cx+dy=1$. You could write $ax_1+by_1=cx_2+dy_2$, but this is unlikely to be helpful. – David Feb 03 '16 at 03:42
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    As an additional hint to what was already provided below, for positive integers, if $x\mid y$ then $x\leq y$. If you also have $y\mid x$ then $y\leq x$. Now, if $x\leq y$ and $y\leq x$... – JMoravitz Feb 03 '16 at 03:54

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Hint: $a\mid ad$, so $a\mid bc$, so $a\mid c$ because...

See if you can provide the reason for the last step and then complete the proof.

David
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