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I am self studying Elementary number theory from David Burton Book. While trying exercises of Section 2.5 , I am unable to think about this problem

If a and b are relatively prime positive integers, prove that Diophantine equation ax-by =c has infinitely many solutions in the positive integers.

Can someone please help in how to solve this problem!!

Bill Dubuque
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By Bezout's Lemma, we have a solution to $ax+by=1$ when $\gcd(a,b)=1$. Then: $$ax-b(-y)=1 $$$$ \implies ax-b(-y)+kab-kab=1 $$$$\implies a(x+kb)-b(ka-y)=1$$$$\implies a(cx+ckb)-b(cka-cy)=c$$ Now, $k$ can take whatever integral value we want it to take. Thus, we have infinitely many values $aX-bY=c$, where a family of infinitely many solutions are: $$(X,Y)=(cx+ckb,cka-cy)$$

Haran
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  • but how to prove rigorously that infinitely many X and Y are positive? –  Jan 20 '20 at 14:29
  • if I apply positivity conditions on X, Y then k>( -b/x ) and k>(y/a) and since x, y are infinitely many solutions of ax+by=c, so I have infinitely many +ve solutions if k >max ( -b/x, y/a) . Is my argument right? –  Jan 20 '20 at 14:40
  • I didn't understand what you typed above. The way you prove that $X$ and $Y$ are positive infinitely many times is that you make $cka$ and $ckb$ sufficiently large and positive. If $c$ is positive, make $k$ sufficiently large and positive. If $c$ is negative, make $k$ sufficiently large (magnitude) and negative. – Haran Jan 20 '20 at 15:39
  • as (cx+ckb), (cka-cy) are solutions so applying positivity conditions on them is both of them must be >0 . Then my previous comment follows. Is it correct? –  Jan 20 '20 at 15:51
  • Yes @InvisibleMan ! – Haran Jan 20 '20 at 16:04
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Since $a,b$ are relatively prime we can find integers $u,v$ such that $$au+bv=1.$$

Then $x=cu+tb$ and $y=-cv+ta$ where $t$ can be any integer are solutions.