# primes...

• Feb 28th 2010, 02:06 AM
flower3
primes...
find all primes of the form $n^4 -m^4$
• Feb 28th 2010, 02:15 AM
tonio
Quote:

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

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

Tonio
• Feb 28th 2010, 03:15 AM
Bacterius
So, to give a little hint, how many primes can you think of that satisfy $n^4-m^4=(n-m)(n+m)(n^2+m^2)$ ? And what condition can you put on $n$ and $m$ so that the expression is prime ? (remember the basic definition of a prime number) Conclude.
• Feb 28th 2010, 09:11 AM
pollardrho06
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, 09:12 AM
pollardrho06
Quote:

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

Tonio

Tonio pretty much nailed it.
• Feb 28th 2010, 10:10 AM
Bacterius
Quote:

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.
• Feb 28th 2010, 10:13 AM
Drexel28
This thread takes me on a stroll down memory lane.
• Feb 28th 2010, 01:12 PM
tonio
Quote:

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
• Feb 28th 2010, 06:53 PM
Bacterius
It's actually the same guy!? (Whew)
Geez, now that's impressive...