1. ## primes...

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

2. Originally Posted by flower3
find all primes of the form $\displaystyle n^4 -m^4$

$\displaystyle n^4-m^4=(n-m)(n+m)(n^2+m^2)$ ...

Tonio

3. So, to give a little hint, how many primes can you think of that satisfy $\displaystyle n^4-m^4=(n-m)(n+m)(n^2+m^2)$ ? And what condition can you put on $\displaystyle n$ and $\displaystyle m$ so that the expression is prime ? (remember the basic definition of a prime number) Conclude.

4. There aren't any primes of the form a^4 - b^4

To see this,

let n = a^4 - b^4

factor...

...
...

and you get a factor of (a + b) in the product which is neither 1 nor n and certainly divides n.

Done!

Regards,

Shahz.

5. Originally Posted by tonio
$\displaystyle n^4-m^4=(n-m)(n+m)(n^2+m^2)$ ...

Tonio
Tonio pretty much nailed it.

6. Originally Posted by pollardrho06
Tonio pretty much nailed it.
What I wanted Flower3 to think about was that there "might" have been primes of that form, and I wanted him (or most likely her) to investigate how this could have been a prime. The observation should be that if two of the factors are equal to 1, then the number is prime. And the conclusion would be that no two factors can be simultaneously equal to one, hence there is no prime of that form ...

I thought that was pedagogic.

7. This thread takes me on a stroll down memory lane.

8. Originally Posted by Drexel28
This thread takes me on a stroll down memory lane.

Unbelievable and annoying: same poster, same question , 1.5 months apart. No doubt, some would be better off knitting handkerchiefs....really!

Tonio

9. It's actually the same guy!?
Geez, now that's impressive...