With $a,\ b,\ m,\ n\in \mathbb{Z}$. Show that if ${\rm gcd}(m,n)=1$, then ${\rm gcd}(ma+nb,mn)={\rm gcd}(a,n){\rm gcd}(b,m)$

Question

I don't really know how to prove this, i suppose it must have something to do with the definition of greatest common divisor and Bézout's identity.

I looked at some other questions here and thought that i had to get to gcd(ma+nb,mn)=1 and that gcd(a,n)gcd(b,m)=1 too, but i'm not sure.

What happens when you use the values suggested by HK in the result you propose in your title? The point of that hint is that if you carry it out, it may help you with the main problem. In this case, it may show that you copied it wrong. — GEdgar, Oct 19 '18 at 00:11
Well it is true, i can see that, but it's for an specific case — Yaheli Felix, Oct 19 '18 at 00:14
@YaheliFelix The typo is that the coef of $b$ should be $n$ not $m$. — Bill Dubuque, Oct 19 '18 at 00:20

Bill Dubuque · Accepted Answer · 2022-03-08T15:34:23.370

By $\rm\color{#c00}H$ere $\color{#c00}{(m,n)\!=\!1}\Rightarrow (ma\!+\!nb,\color{#c00}{mn}) \overset{{\rm\color{#c00}H}}= (ma\!+\!nb,\color{#c00}m)(ma\!+\!nb,\color{#c00}n) = (b,m)(a,n)\ $ by

$$\begin{align}(\color{#0a0}ma+nb,\color{#0a0}m) &\overset{\rm\color{#f6f} R}= (nb,m) \overset{\rm\color{darkorange}L}= (b,m)\\[.2em] (ma+\color{#0af}nb,\ \color{#0af}n) &\overset{\rm\color{#f6f} R}= (ma,n) \overset{\rm\color{darkorange}L}= (a,n)\end{align}\qquad$$

where above we applied gcd $\rm\color{#0a0}{mod}\ \color{#0af}{reduction}$ = $\rm\color{#f6f} R,\ $ and $\ \rm\color{darkorange}L = $ Euclid's Lemma, by $\, \color{#c00}{(m,n)=1}$.

score 0 · Answer 2 · answered Oct 19 '18 at 00:42

I find it useful when doing calculations involving $\gcd(a,n)$ (or whatever variables) to use the notation:

Let $\gcd(a,n) = d$ and let $a = a'd; a'=\frac ad$ and $n = n'd; n' = \frac nd$ and we can know that that $a'$ and $n'$ are relatively prime.

Likewise let $\gcd(b,m) = e$ and let $b = b'e$ and $m = m'e$ and $b'$ and $m'$ are relatively prime.

Thus we want to prove that $\gcd(m'a'de + n'b'de, m'n'de) = de$

.....

Now you probably have proven at some point that $\gcd(fg, hg) = g\gcd(f,h)$.

So $\gcd(m'a'de + n'b'de, m'n'de) = \gcd(m'a' + n'b', m'n')de$ and we just need to prove $\gcd(m'a' + n' b', m'n') = 1$.

....

So $p|m'$ and $p|n'$.

But we were told $\gcd(m,n) = 1$ so there can't be any prime divisors that divide both $m',n'$.

So there are no prime divisors of $\gcd(m'a' + n' b', m'n')$. So $\gcd(m'a + n'b', m'n') = 1$.

With $a,\ b,\ m,\ n\in \mathbb{Z}$. Show that if ${\rm gcd}(m,n)=1$, then ${\rm gcd}(ma+nb,mn)={\rm gcd}(a,n){\rm gcd}(b,m)$

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