## prime numbers problem

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.

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 Blog Entries: 3 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
 Recognitions: Gold Member Science Advisor Staff Emeritus What if r divides the denominator?

## prime numbers problem

What's r?

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 Quote by Carl140 What's r?
Some number that happens to divide the denominator.

 I still don't get it, sorry. Can you please explain a little bit more?
 Blog Entries: 3 I think I have a solution but I won't post it without moderator approval.
 Blog Entries: 3 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.
 I got it. Hint: Use the conjugate rule. For solution Spoiler (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.

 Quote by Carl140 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.
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

 Quote by de_brook 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
the prime is p = q

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