How to prove a GCD question?

Question

Prove for all positive integers d, m, n. If d=gcd(m, n) then gcd(m, nk)=gcd(m, dk).

So I messed around with it, and I started from the conclusion. I made gcd(m, nk)=x, gcd(m, dk)=y. Then by Bezout's Lemma ms+nkt=x, ms+dkt=y, s and t are integers. I think I can make these 2 equations and change it to the form of d=gcd(m, n). Maybe by dividing? But If I divide it by d, then ms/d+nkt/d=x/d, then this equation might not be integers anymore. I don't know that the next step is. Thank you for looking this over for me.

This is copy-paste from here. Search this site for duplicates, too. — Dietrich Burde, Nov 06 '20 at 16:32
See this answer in the dupe. Or reduce to the coprime case $,(m,n)=1,$ by cancelling $,(m,n),$ throughout, as explained here. We have many quetions on this and related topics. Please search first. — Bill Dubuque, Nov 06 '20 at 22:33

Calvin Lin · Accepted Answer · 2020-11-06T17:33:02.543

0

Hint: Use the GCD in prime factorization form.
If $m = \prod p_i ^ {m_i}, n = \prod p_i^{n_i}$, then $ d = \prod p_i^{\min(m_i, n_i)}$.

The result follows by showing that

$$ \min (m_i, n_i + k_i) = \min (m_i, \min (m_i, n_i) + k_i)$$

A direct way of showing this is to condition on $ m_i \geq n_i$ (so $\min(m_i, n_i) = n_i$) and $ m_i < n_i$ (so both sides equal $m_i$).

edited Nov 06 '20 at 17:33

answered Nov 06 '20 at 17:20

Calvin Lin

68,864

Thanks. Is there a way to do it without the prime factorization form? – Trai Nov 06 '20 at 18:53

How to prove a GCD question?

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