For a prime q and positive integers x, y, z, w , q divides yz, q divides xy and q doesn't divide yw implies that q^2 divides xz

Question

I'm attempting to prove that the above statement is true, but am having a bit of trouble.

So far I have

$qk = yz$

$qj = xy$

I'm not really sure to go from here, but I tried the following

I can then rearrange for $z,x$

$x = \frac {qk}{y}$

$z = \frac {qj}{y}$

And then combine such that

$xz = \frac {q^2(kj)}{y}$

If it wasn't for the y on the denominator, the proof would be finished, but I can't really see a way to get rid of it, which of course makes me think I've gone around doing this the wrong way.

What would be a better way to prove this statement?

Hint : A prime p divides xy iff p divides x or p divides y. – Akash Yadav Apr 01 '20 at 08:01 — Akash Yadav, Apr 01 '20 at 08:01

score 0 · Accepted Answer · answered Apr 01 '20 at 08:02

0

$q\not\mid yw$, thus $q\not\mid y$. Thus $q\mid xy$ implies $q\mid x$ and $q\mid yz$ implies $q\mid z$. Thus $q^2\mid xz$.

answered Apr 01 '20 at 08:02

joriki

238,052

For a prime q and positive integers x, y, z, w , q divides yz, q divides xy and q doesn't divide yw implies that q^2 divides xz

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