# B Proof of a lemma of BÉZOUT’S THEOREM

1. Sep 11, 2016

### SamitC

Hi,
One silly thing is bothering me. As per one lemma, If a, b, and c are positive integers such that gcd(a, b) = 1 and a | bc, then a | c. This is intuitively obvious. i.e.
Since GCD is 1 'a' does not divide 'b'. Now, 'a' divides 'bc' so, 'a' divides 'c'. Proved.

What is bothering me is : suppose bc/a = s. Then as = bc. Thus a = (b/s) c ..... (1)
Now, if c/a is an integer so is s/b. Which means b/s is not an integer. Putting this in (1) - how 'a' divides 'c'?
Thanks in advance

2. Sep 11, 2016

### Staff: Mentor

So what? Where does b/s appear?

Maybe it is easier to understand with a numerical example:
a=5, b=3, c=10
5 | 30 is true, 5 | 10 is true as well.
bc/a = s gives us s = 30/5 = 6.
s/b=2, and b/s=1/2 is not an integer. So what?

3. Sep 12, 2016

### SamitC

Thanks for your reply.
Sorry...don't know why i asked this question..... a (s/b) = c ....i dont know why I was thinking the other way.
Anyways...thanks

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