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Let $a$ and $b$ two relatively prime numbers. We set $N_{a,b}:=\{a+kb, \, k \in \mathbb{N} \}$. Suppose that there exists a multiple of $b$, namely $hb \in N_{c,d}$ for some relatively prime numbers $c,d$. It is obvious that $b$ and $d$ have no common factors. But I have to show that there exists two numbers $t,s \in \mathbb{N}$ such that $c-a=tb-sd$.

I have no clue, can you help me please?

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From $hb \in N_{c,d}$, you have $$ hb = c+\hat k d $$ for some $\hat k$. Since, as you say, $(b,d)=1$, there exist $m$ and $n$ such that $$ a = mb+nd $$ and so $$c-a = hb-\hat k d - mb - nd = (h-m)b-(\hat k+n)d.$$

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  • $\begingroup$ @TheWanderer Hey there! Was my answer helpful? $\endgroup$ – Matthew Conroy Nov 2 '17 at 22:31

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