Can Natural Numbers a, b, c with a Dividing bc Imply a Divides c?

[nmego12345]

Homework Statement

1. If a,b and c are natural numbers and a, b are coprime and a divides bc then prove that a divides c
2. Prove that the lcm of a,b is ab / gcd(a,b)

Homework Equations

if a is a divisor of b then a = mb for a natural number m
if a prime p is a divisor of ab then p is a divisor of a or a divisor of b

The Attempt at a Solution

1.since a is a divisor of bc so am = bc (m is a natural number)
so a = (c)(b/m)
so a/b = c/m
Ok since a,b are coprime so a/b = a number that is not natural
since a/b = c/m so c/m = a number that is not natural so c,m are coprime
back to a = (c)(b/m)
since a = (c)(b/m) which is a natural number, so bc must be a multiplie of m
since c isn't a multiplie of m, b must be so
so b is coprime with m
now a/c = b/m
since b is coprime with m
a is coprime with c
Q.E.D
(Wanna check if my approach is correct or not)

2.Prove that the lcm of a,b is ab/gcd(a,b)
let a = xm , b = ym (m = gcd(a,b))

ab/gcd(a,b) = xmym/m = xmy
It is divisible by a and b so it satisfies being a multiplie
here I gave up.

 

[haruspex]

c/m = a number that is not natural so c,m are coprime

That does not follow. 6/4 is not an integer, but 6 and 4 are not coprime.
Consider some prime divisor of a.