primes?

**flower3** · January 13th 2010, 08:19 AM

find all primes of the form $n^4 - m^4$

**Drexel28** · January 13th 2010, 08:29 AM

Originally Posted by flower3

find all primes of the form $n^4 - m^4$

What???? Maybe, I am misinterpreting you, or I don't understand what's going on here, but if $p=n^4-m^4$ then $p=\left(n^2+m^2\right)\left(n^2-m^2\right)$

**Dinkydoe** · January 13th 2010, 08:41 AM

Maybe the question is: Show there don't exist primes of the form $n^4-m^4$ ?

**Drexel28** · January 13th 2010, 08:41 AM

Originally Posted by Dinkydoe

Maybe the question is: Show there don't exist primes of the form $n^4-m^4$ ?

That would be a pretty easy question.

**Drexel28** · January 13th 2010, 08:46 AM

There should be a subtle remark made here though. A prime is a natural number, so if $p=n^4-m^4$ we must have that $n^4>m^4\implies n>m$ so that we know that $n^2-m^2$ is a positive divisor.

**Dinkydoe** · January 13th 2010, 09:16 AM

Aside from that: $(n^2-m^2) \neq 1$ for any n,m: n> m what would be required if p is prime.

**Drexel28** · January 13th 2010, 09:19 AM

Originally Posted by Dinkydoe

Aside from that: $(n^2-m^2) \neq 1$ for any n,m: n> m what would be required if p is prime.

You cannot have that $n^2-m^2=1$ since $n^2-m^2=\left(n-m\right)\cdot\left(n+m\right)$ and since $n+m>1$ it follows that $n-m<1$ .

**Moo** · January 13th 2010, 10:48 AM

I don't understand why you're debating oO
So the answer's just : there's none !

**Drexel28** · January 13th 2010, 10:51 AM

Originally Posted by Moo

I don't understand why you're debating oO
So the answer's just : there's none !

I agree. It was just two things had to resolved, clearly the factorization disproves that it can be prime unless $n^2-m^2$ is one or less than zero. Both of which were disproven.

**mr fantastic** · January 13th 2010, 06:37 PM

I would prefer no more replies in this thread until the OP responds. Thankyou.

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