Given $a,b,m,n \in \mathbb{Z}^+$, is it possible to always find a relatively prime pair of positive integers $x,y$ so that $ax+m$ and $by+n$ are not relatively prime?
For $a,b,m,n \in \mathbb{Z}^+$, does there always exist $x,y \in \mathbb{Z}^+$ with ${\rm gcd}(x,y) = 1$ but ${\rm gcd}(ax+m,by+n) > 1$?
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elementary-number-theory
greatest-common-divisor
1 Answers
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Yes, we can. Suppose $p$ is a prime number and not a divisor of any number in the problem. There exists an integer $x$ such that $p\mid ax+m$. Furthermore, there exists an integer $z$ such that $p\mid b(xz+1)+n$. Let $y=xz+1$.
To be precise, we are using Bezout's Theorem. 1) Because $gcd(a, p)=1$, there exists integers $x$ and $w$ such that $$ax+pw=-m,$$ which implies $p|ax+m$.
2) Note that $gcd(p, bx)=1$, there exists integers $z$ and $u$ such that $$(bx)z+pu = -(b + n),$$ which implies $p\mid b(xz+1)+ n$
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0Looks plausible to me. But can you explain a little bit more on how you came to the conclusion that there must be $ x $ for which $ p \mid ax+m $ ? I'm probably missing something obvious. Or some general theorem you can refer to? – 2017-01-10
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0@RutgerMoody Bezout's theorem – 2017-01-10
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0Thanks. $\gcd(p,a)=1 \implies $ we can find: $ pk -ax' = 1 \implies pkm -ax'm = m \implies ( x = x'm ) \implies p \mid ax + m $. That's what you mean right? – 2017-01-10
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0@RutgerMoody yes – 2017-01-10
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0And the xz+1? I see this makes x and y coprime. But calculation like above would result in $ n \mid y $ then if m,n not coprime neither would be x and y. How did you get the expression for y ? – 2017-01-10
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1@RutgerMoody added explanation above – 2017-01-10