If $m \equiv m'$ mod $n$, then gcd$(m,n) = 1$ $\Longleftrightarrow$ gcd$(m',n) = 1$.

Question

How to prove the following statement:

I have never learned number theory before, for me it's already hard to prove g.c.d of two integers is 1, not to mention the whole statement above.

These kinds of facts jump out frequently in my abstract algebra course. Is there any good introductory number theory book that has clear and illuminating proof for these "basic" facts? Then I can remember the proof technique and these "basic" results by rote.

Your question has a similar answer here: https://math.stackexchange.com/questions/381310/proving-n-equiv-p-k-longleftrightarrow-gcdp-k-gcdn-k?rq=1 — Valent, Jan 31 '21 at 02:58

score 2 · Accepted Answer · answered Jan 31 '21 at 03:21

2

Actually the problem has a simple solution, if you are allowed to use the following result:

$$\ gcd(m,n)=1\ \mathrm{iff}\ \exists k_1,k_2\in\mathbb{Z}\ \mathrm{such\ that}\ k_1m+k_2n=1 $$ since $m\equiv m'(\mod n)$ then $m=m'+k_3n$ for some $k_3\in\mathbb{Z}$ replacing we have $$1=k_1m+k_2n= k_1(m'+k_3n)+k_2n=k_1m'+(k_2+k_3)n $$ which shows that $$ gcd(m',n)=1$$

You can try the other way.

answered Jan 31 '21 at 03:21

Valent

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Thanks! A very simple answer for me. – Hamilton Jan 31 '21 at 05:03

If $m \equiv m'$ mod $n$, then gcd$(m,n) = 1$ $\Longleftrightarrow$ gcd$(m',n) = 1$.

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