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Prime numbers problem

  1. Mar 10, 2009 #1
    Hello,

    I can't get this small contest problem. How do you solve this kind of problem?

    Let p and q be prime numbers such that (p^2+q^2)/(p+q) is an integer.
    Prove p=q.
     
  2. jcsd
  3. Mar 10, 2009 #2
    I can't figure it out.

    p^2+q^2=n1*(p+q)

    p(p-n1)+q(q-n1)=0

    (p-n1)=n2*(q-n1)

    p-n1+n2 n1=n2 q
     
  4. Mar 10, 2009 #3

    Hurkyl

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    What if r divides the denominator?
     
  5. Mar 10, 2009 #4
    What's r?
     
  6. Mar 10, 2009 #5

    Hurkyl

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    Some number that happens to divide the denominator.
     
  7. Mar 10, 2009 #6
    I still don't get it, sorry. Can you please explain a little bit more?
     
  8. Mar 10, 2009 #7
    I think I have a solution but I won't post it without moderator approval.
     
  9. Mar 11, 2009 #8
    Okay, here is my hint. What theorem might be helpful to show what integer values of p+q will satisfy the following equation?

    (p+q)^2-m(p+q)-2pq=0

    Where m is an integer, p is prime and q is prime.
     
  10. Mar 11, 2009 #9
    I got it.
    Hint: Use the conjugate rule.


    For solution
    (p^2+q^2)/(p+q)=(p^2-q^2)+2q^2)(p+q)=p-q+q^2/((p+q)/2)

    but q^2 is only divisible by 1,q,q^2. (p+q)/2 is obv not equal to 1. if it is equal to q, p=q and if it is equal to q^2, q|p and then p=q since they are prime.
     
    Last edited: Mar 11, 2009
  11. Mar 13, 2009 #10
    This statement can be more generalised as follows;

    Let p and q be prime numbers then (p^2+q^2)/(p+q) is a prime if and only if p = q
     
  12. Mar 13, 2009 #11
    the prime is p = q
     
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