prime numbers problem

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

Quote by Carl140

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

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

de_brook

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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Carl140	prime numbers problem Tweet 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 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

Hurkyl Recognitions: Gold Member Science Advisor Staff Emeritus	What if r divides the denominator?

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.