Suppose d != 1 and d = gcd(m,r)...?

Question

Suppose d != 1 and d = gcd(m,r)

so there exists s and t such that

ds = m and dt = r.

gcd ((m/d), (r/d)) = 1.
so
gcd (s, t)=1

okay.

now my question is

is gcd (s, t) = (1/d)gcd(m,r)

?


proof?

Answer 1

Yes, this one is easier than you're making it out to be:

You've already got:

gcd(m, r) = d
gcd(s, t) = 1

Multiply the first equation by (1/d)

(1/d)gcd(m, r) = 1

So clearly:
gcd(s, t) = 1 = (1/d)gcd(m, r)
gcd(s, t) = (1/d)gcd(m, r)

Answer 2

yea i think so