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About the notation: Denotes $(a,b)$ the greatest common divisor between $a$ and $b$.

Now, about the exercise, what I did was the following:

Since $(m, n) = (s, n) = 1$,

$ma + nb = sc + nd = 1$ then,

$ma + nb-sc-nd = 1$

$ma-sc + (b-d) n = 1$

let $b-d = j$ then the previous equality I can write it like this: $ma-sc + jn = 1$

I want to get to that $r (ms) + jn = 1$ but I can not think of how to continue to do it.

Besides, is it okay what I did until now?

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    $\begingroup$ $ma+nb-sc-nd=0$ how could you you equate the expression with $1$? $\endgroup$ – Dastan Jun 11 '18 at 3:52
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First lets assume $(m,n)=(s,n)=1$ then there are $x_1,x_2,y_1,y_2\in\mathbb{Z}$ such that $mx_1+ny_1=1$ and $sx_2+ny_2=1$ multiplying them we get $msx+ny=1$ for some integers $x,y$.

Now the other way i.e. lets assume $(ms,n)=1$ and let $(m,n)=k$ then $k|m\Rightarrow k|ms$ but $k|n$ thus $k|(ms,n)$ so $k=1$. Similarly we can show that $(s,n)=1$.

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  • $\begingroup$ The other condition of the if and only if how should it start? and thank you very much for answering! @Vulthuryol $\endgroup$ – Ayesca Jun 11 '18 at 4:09
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    $\begingroup$ You are welcome. The second line in my answer starts by assuming that $(ms,n)=1$. Will edit. $\endgroup$ – user428700 Jun 11 '18 at 4:10

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