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Show that if $a$ and $b$ are relatively prime integers then $gcd(a+kb, b+ka)$ divides $k^2-1$

I am trying to two linear combination of $a+kb$ and $b+ka$

but can't write to as linear combination can you one help...

thank you....

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If $d,$divides both $a+bk,ak+b$

$d$ must divide $k(a+bk)-(ak+b)=b(k^2-1)$

and $k(ak+b)-(a+bk)=a(k^2-1)$

The idea is to eliminate $a,b$ one by one.

So, $d$ must divide $(a(k^2-1),b(k^2-1))=(k^2-1)(a,b)$

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  • $\begingroup$ ....thank you so much sir..for your time $\endgroup$ – Inverse Problem Feb 15 '18 at 4:10
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Since $a,b$ are relatively prime integers we have $$ax+by=1$$ for some integers $x,y$. Then any common factor of $a+kb$ and $b+ka$ is also a factor of $$(a+kb)(ky-x)+(b+ka)(kx-y)=k^2-1\ .$$

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  • $\begingroup$ @David......thank you sir... $\endgroup$ – Inverse Problem Feb 15 '18 at 4:10

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