Two numbers being coprime implies other two numbers being coprime

Question

Let $b$, $m$, $p$ be positive integers such that $b > p$. And suppose that $mb+p$ and $b$ are coprime. Then I have to show that $mb+p$ and $b-p$ are also coprime.

I tried myself but it seems trickier than I expected, as most number theoretic problems are.... Could anyone please help me?

A common divisor divides $b$ and $mb+p$, hence also $p$. So ... — Dietrich Burde, Aug 21 '19 at 21:20
So with $p=5$, $b=7$, $m=1$, the claim is that if $12$ and $7$ are coprime, then so are $12$ and $2$? — Hagen von Eitzen, Aug 21 '19 at 21:22
More broadly, if $m,p,b$ are all odd then $mb+p$ and $b-p$ are both even so you can generate lot of counterexamples. — lulu, Aug 21 '19 at 21:25

Bill Dubuque · Answer 1 · 2019-08-25T13:16:42.963

2

$1 = (mb\!+\!p,b) = (p,b)\,$ by Euclidean reductions, and similarly $\ \,\phantom{\^{|^|}}(mb\!+\!p,b\!-\!p) = (mp\!+\!p,b\!-\!p) = (m\!+\!1,b\!-\!p),\,$ by $\,(p,b\!-\!p)= (p,b) = 1$

But it is easy to make $(m\!+\!1,b\!-\!p)>1,\,$ e.g. choose coprime $b,p$ with $b\!-\!p>1$ and $m\!+\!1 = b\!-\!p,\,$ e.g. $\,b,p = 7,5,\ m = b\!-\!p\!-\!1 = 1\,$ (Hagen's counterexample). There are infinitely many more.

Therefore $ $ if $\,(mb\!+\!p,b) = 1\,$ then $\,(mb\!+\!p,b\!-\!p)=1\iff (m\!+\!1,b\!-\!p)=1$

Update $ $ Per request in comments below is further detail.

By Euclid $\ y'\equiv y\pmod{\! x}\,\Rightarrow\, (x,y') = (x,y)$

$\!\bmod b\!-\!p\!:\,\ b\equiv p\,\Rightarrow\, mb\!+\!p\equiv mp\!+\!p,\,$ so by above

we deduce $\ (b\!-\!p,\, mb\!+\!p) = (b\!-\!p,\,mp\!+\!p)= (b\!-\!p,\,p(m\!+\!1))$

but $\,(b\!-\!p,p)= (b,p) = 1\,$ so $\,(b\!-\!p,\,p(m\!+\!1)) = (b\!-\!p,\,m\!+\!1)$

edited Aug 25 '19 at 13:16

answered Aug 21 '19 at 21:42

Bill Dubuque

272,048

Can you explain the equalities after 'and similarly'? – Michael Munta Aug 23 '19 at 15:16
@Michael See the Update. – Bill Dubuque Aug 23 '19 at 15:44
Ok, can you please also explain the last line? – Michael Munta Aug 23 '19 at 17:06
@Michael It's case $,(a,p)=1,$ in $,(a,pn) = (a,(a,p)n),,$ by $,(a,pn) = (a,an,pn) = (a,(a,p)n)$ $\ \ \ $ – Bill Dubuque Aug 23 '19 at 17:15
Why is $(a,pn) = (a,an,pn)$? – Michael Munta Aug 24 '19 at 13:25
@Michael $\ d\mid a,b\iff d\mid a,an,b,\ $ by $,d\mid a,\Rightarrow, d\mid an\ \ $ – Bill Dubuque Aug 24 '19 at 13:30
One more thing, $(b-p,p) =(b,p)$? – Michael Munta Aug 25 '19 at 12:26
@Michaeal That too is "By Euclid..." since $, b-p\equiv b\pmod{p}.\ $ See here for a proof. This is the congruence version of the modular reduction step in the Euclidean algorithm. – Bill Dubuque Aug 25 '19 at 13:16

Two numbers being coprime implies other two numbers being coprime

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