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Let $a,b,r,s$ be given constants. Prove that the arithmetic progressions $\{ax + r : x \in \Bbb Z\}$ and $\{by + s : y \in \Bbb Z\}$ intersect if and only if $\gcd(a, b) \mid (s − r)$.

I supposed that the two intersect. Which gives $ax+r=by+s$. I believe that there's a point where they're equal, thus, some $x$ and $y$ exist such that $ax+r=by+s$ is true. Rearranging the equation yields you with $ax-by=s-r$. I know that $ax-by=\gcd(a,b)$. Now, I am just lost on how to compute the gcd(a,b).

Thank you for your time.

pjs36
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Arandomuser
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2 Answers2

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You've pretty much done it. You just can't say that "$ax-by=\gcd(a,b)$" — that's not necessarily true. But it is always true $\gcd(a,b)$ divides any expression of the form $ap+bq$ (with integer $p$ and $q$, of course). Therefore, you've obtained that $\gcd(a,b)\mid ax-by=s-r$, finishing your proof in this direction. Note that you still have to prove the other direction of "if and only if".

zipirovich
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You almost have it. You don't need to compute $\gcd(a,b)$. Note instead that $$\gcd(a,b)\mid (ax-by)$$ and therefore $$\gcd(a,b)\mid (r-s)$$

Backwards implication is somewhat more difficult to prove. You will need, for example, the Bezout's identity.

ajotatxe
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