Divisibility and p's and q's

[lordy12]

If 3p^2 = q^2 and p and q are integers, how do I prove that 3 is a common divisor for p and q?
My attempt: q^2 is divisible by 3, so q is divisible by 3. I can't prove that p is divisible by 3.

 

[Werg22]

Any integer that is not a perfect square has an irrational square root. What can you do with that?

 

[JasonRox]

Any integer that is not a perfect square has an irrational square root. What can you do with that?

Personally, I have no idea what you can do with that.

 

[JasonRox]

If 3p^2 = q^2 and p and q are integers, how do I prove that 3 is a common divisor for p and q?
My attempt: q^2 is divisible by 3, so q is divisible by 3. I can't prove that p is divisible by 3.

Now, just re-write it as p^2 = q^2/3. What does that tell you about p?

Remember, p is an integer so p^2 is also an integer. And q is divisible by 3, but there is only one 3.

Finish it off from there.

 

[Werg22]

Lol, yeah maybe it's not very relevant. Let me repent:

A power has the same prime factors than its root.

 

[tim_lou]

3|q, rewrite q=3q'

 

[lordy12]

oops...I completely disregarded the fact that p and q are integers.

sorry