# primes?

• January 13th 2010, 09:19 AM
flower3
primes?
find all primes of the form $n^4 - m^4$
• January 13th 2010, 09:29 AM
Drexel28
Quote:

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 $p=\left(n^2+m^2\right)\left(n^2-m^2\right)$
• January 13th 2010, 09:41 AM
Dinkydoe
Maybe the question is: Show there don't exist primes of the form $n^4-m^4$?
• January 13th 2010, 09:41 AM
Drexel28
Quote:

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.
• January 13th 2010, 09:46 AM
Drexel28
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.
• January 13th 2010, 10:16 AM
Dinkydoe
Aside from that: $(n^2-m^2) \neq 1$ for any n,m: n> m what would be required if p is prime.
• January 13th 2010, 10:19 AM
Drexel28
Quote:

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 $n-m<1$.
• January 13th 2010, 11:48 AM
Moo
I don't understand why you're debating oO
So the answer's just : there's none !
• January 13th 2010, 11:51 AM
Drexel28
Quote:

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.
• January 13th 2010, 07:37 PM
mr fantastic
I would prefer no more replies in this thread until the OP responds. Thankyou.