1. ## primes?

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

2. Originally Posted by flower3
find all primes of the form $\displaystyle n^4 - m^4$
What???? Maybe, I am misinterpreting you, or I don't understand what's going on here, but if $\displaystyle p=n^4-m^4$ then $\displaystyle p=\left(n^2+m^2\right)\left(n^2-m^2\right)$

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

4. Originally Posted by Dinkydoe
Maybe the question is: Show there don't exist primes of the form $\displaystyle n^4-m^4$?
That would be a pretty easy question.

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

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

7. Originally Posted by Dinkydoe
Aside from that: $\displaystyle (n^2-m^2) \neq 1$ for any n,m: n> m what would be required if p is prime.
You cannot have that $\displaystyle n^2-m^2=1$ since $\displaystyle n^2-m^2=\left(n-m\right)\cdot\left(n+m\right)$ and since $\displaystyle n+m>1$ it follows that $\displaystyle n-m<1$.

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

9. 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 $\displaystyle n^2-m^2$ is one or less than zero. Both of which were disproven.

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