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Let $p,q$ belong to $\mathbb{N}$ and are relatively prime to each other. If $\alpha,\beta$ belong to $\mathbb{N}$, are also relatively prime to each other,then are $(p\beta+q\alpha)$ and $q\beta$ always relatively prime ?

3 Answers 3

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The answer is clearly no: take $p=\alpha$, $q=\beta$. Then $p\beta+q\alpha=2pq$ and $q\beta=pq$.

If however $\beta$ is co-prime to $q$, then the answer is equally clearly yes, since no divisor of $q$ divides $p\beta$ and no divisor of $\beta$ divides $q\alpha$.

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Simply repeatedly apply $\rm\color{#C00}E = $ Euclid's lemma:

$\begin{align} 1 &= (pb+qa,qb)\\ \overset{\rm\color{#C00}E }\iff\ \ \ \ \ \ \ 1 &= (pb+qa,q) = (pb,q) \overset{\rm\color{#C00}E }= (b,q)\quad\ {\rm via}\quad\ (p,q) = 1\\ {\rm and}\ \ 1 &= (pb+qa,b) = (qa,b) \overset{\rm\color{#C00}E }= (q,b)\quad\ {\rm via}\quad\ (a,b) = 1\end{align}$

Therefore $\ \ 1 = (pb+qa,qb) \iff 1 = (q,b)$

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No, e.g. $p=3,q=5,\alpha=3,\beta=5$ is a counterexample.