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Divisibility and p's and q's

  • Thread starter lordy12
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
1,425
1
Any integer that is not a perfect square has an irrational square root. What can you do with that?
 
  • #3
JasonRox
Homework Helper
Gold Member
2,314
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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.
 
  • #4
JasonRox
Homework Helper
Gold Member
2,314
3
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. :wink:

Finish it off from there.
 
  • #5
1,425
1
Lol, yeah maybe it's not very relevant. Let me repent:

A power has the same prime factors than its root.
 
  • #6
682
1
3|q, rewrite q=3q'
 
  • #7
36
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oops...I completely disregarded the fact that p and q are integers.

sorry
 

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