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

Printable View

- Feb 28th 2010, 01:06 AMflower3primes...
find all primes of the form $\displaystyle n^4 -m^4 $

- Feb 28th 2010, 01:15 AMtonio
- Feb 28th 2010, 02:15 AMBacterius
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.

- Feb 28th 2010, 08:11 AMpollardrho06
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. - Feb 28th 2010, 08:12 AMpollardrho06
- Feb 28th 2010, 09:10 AMBacterius
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. - Feb 28th 2010, 09:13 AMDrexel28
This thread takes me on a stroll down memory lane.

- Feb 28th 2010, 12:12 PMtonio
- Feb 28th 2010, 05:53 PMBacterius
It's actually the same guy!? (Whew)

Geez, now that's impressive...