Advanced P14(BOMC 2)

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*Mahi*
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Advanced P14(BOMC 2)

Unread post by *Mahi* » Sun Apr 01, 2012 7:17 pm

Let $a,b$ be relatively prime positive integers. Show that \[ax+by=n\] has nonnegative integer solutions for all integers $n>ab-a-b$. What if $n=ab-a-b$?
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SANZEED
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Re: Advanced P14(BOMC 2)

Unread post by SANZEED » Mon Apr 02, 2012 7:09 pm

Hint:
Can we use linear expression and residue classes for the first part?
Taking mods kills the second part.
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Nadim Ul Abrar
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Re: Advanced P14(BOMC 2)

Unread post by Nadim Ul Abrar » Mon Apr 02, 2012 8:26 pm

part 2

First lets solve the equation $ax+by=ab-a-b$
Note that $x=x_0=-1,y=y_0=(a-1)$ is a solution .
So that equation has a general solution
$x=x_0+bt=-1+bt,y=y_0-at=a-1-at$ , $t \in \mathbb{Z}$

let $y$ is non negative , then of course $t \leq 0$.
But for $ t \leq 0$ , $x\leq 0$

let $x$ is non negative . then $t\geq 1$
but for $t\geq 1$ $y$ is negqtive .

So for this case x,y can't be nonnegative at a time .


please someone help me to prove the first part .. :'( hiks hiks
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