Solving the Prime Numbers Problem: Proving p=q with p and q as Prime Numbers

[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.

 

[John Creighto]

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

 

[Hurkyl]

What if r divides the denominator?

 

[Carl140]

What's r?

 

[Hurkyl]

What's r?

Some number that happens to divide the denominator.

 

[Carl140]

I still don't get it, sorry. Can you please explain a little bit more?

 

[John Creighto]

I think I have a solution but I won't post it without moderator approval.

 

[John Creighto]

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.

 

[Kurret]

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.

 

[de_brook]

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

 

[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