If $b $ divides $ad $ with$ \mathrm {gcd}(a,b)=1$, then $b $ divides $d $

Question

Let $a\in \mathbb {Z} $ and $a,b \in \mathbb {N} $. Suppose that $b $ divides $ad $ where $\mathrm {gcd} (a,b)=1$. What can you conclude from the above assumptions?

I have a feeling that we must have $b $ divides $d $, but I am unsure and have no intuition guiding me as to a formal way to express what I believe is the conclusion.

Yes, with these assumptions can conclude $b|d.$ A simple proof may be based on using $ax+by=1.$ — coffeemath, Nov 07 '16 at 17:20

score 1 · Accepted Answer · answered Nov 07 '16 at 17:20

1

Hint. Using Bezout's theorem, there exists $(u,v)\in\mathbb{Z}^2$ such that $au+bv=1$. Try multiplying by $d$.

answered Nov 07 '16 at 17:20

C. Falcon

19,553

If $b $ divides $ad $ with$ \mathrm {gcd}(a,b)=1$, then $b $ divides $d $

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