Let $n$ and $k$ be positive integers. Show that $\gcd(n, nk \pm 1) = 1$

Question

This is what I've begun with:

So lets say $x=n,y=nk+1,x,y,n\in\Bbb Z$ Lets say we have a integer, d that divides both x and y. So, $d\vert x$ and $d\vert y$ so if there is a gcd, then d must also divide y-x. $d\vert(y-x)$ but we said $x=n,y=nk+1$. Then, $d\vert(nk+1-n)$=$d\vert (n(k-1)+1)$. But I have no idea if I'm heading in the right direction? Any tips?

You can repeat your argument to obtain $d\mid (n(k-2)+1)$, …, $d\mid (n\cdot 0+1)$, so $d$ is a divisor of $1$. Or immediately use that $d\mid n$ implies $d\mid nk$. — Christoph, Mar 14 '21 at 10:30

score 0 · Answer 1 · answered Mar 14 '21 at 10:24

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Let $p\mid n$ and $p\mid nk+1$. Then $n=pq$ and $nk+1=pr$ for some $q,r\in\mathbb Z$. Therefore \begin{align*} nk+1=kpq+1=pr&\implies p(r-kq)=1\\ &\implies p\mid 1. \end{align*}

answered Mar 14 '21 at 10:24

Surb

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Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque Mar 14 '21 at 11:07

Let $n$ and $k$ be positive integers. Show that $\gcd(n, nk \pm 1) = 1$

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