- #1
Diophantus
- 70
- 0
I forgot how to prove the following, can anyone jog my memory:
Let m,n be natural numbers s.t. gcd(m,n) = 1 and n|k and m|k for some natural number k.
Then nm|k.
I know I'll kick myself when I find out. Thanks.
Let m,n be natural numbers s.t. gcd(m,n) = 1 and n|k and m|k for some natural number k.
Then nm|k.
I know I'll kick myself when I find out. Thanks.