- #1

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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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- Thread starter Carl140
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- #1

- 49

- 0

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

- 489

- 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

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

- #3

Hurkyl

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What if *r* divides the denominator?

- #4

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What's r?

- #5

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

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

- #6

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

- #7

- 489

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

- #8

- 489

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

Where m is an integer, p is prime and q is prime.

- #9

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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.

Hint: Use the conjugate rule.

For solution

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:

- #10

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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

- #11

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the prime is p = qThis 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

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