Are There Any Primes That Satisfy a^4-b^4=p for Integers a and b?

[foxjwill]

Homework Statement

Find all primes p such that \exists a,b \in \mathbf{Z} such that a^4-b^4=p.

Homework Equations

The Attempt at a Solution

For simplicity, we can limit a and b to the positive integers.

Factoring, we have p=(a^2+b^2)(a-b)(a+b). By the unique factorization theorem, we are limited to three cases:

(1) a+b=1 and a-b=1, which gives a=1 and b=0, so p must be 1. But since 1 is not a prime, case 1 is eliminated.

(2) a^2+b^2=1 and a-b=1, which gives a^2+b^2-2ab=1 and then -2ab=0. Again, we are left with a=1 and b=0, so case 2 is eliminated.

(3) a^2+b^2=1 and a+b=1, which gives a^2+b^2+2ab=1 and then 2ab=0. Again, we are left with a=1 and b=0, so case 3 is eliminated.

Therefore, no primes satisfy the equation. Q.E.D.



Is my proof valid? If it is, is there a "more elegant" proof?


edit: I accidentally put the question as a^4+b^4=p instead of what I currently have up there. >_< Oops!

 

[jeffreydk]

Do a and b have to be distinct? Because a=1=b gives p=2, which is prime.

 

[happyg1]

Hi,
Jeffreydk, if a=1=b then p=0 from the origional equation.

foxjwill,
on part 2,
I don't see how your algebra got you a^2+b^2-2ab=1
I agree with your result up to a point, though.
Here's what I did:
a^2+b^2=1 and a-b=1 \Longrightarrow a=b+1
then plug that into the first one:
(b+1)^2+b^2=1
b^2+2b+1+b^2=1
which eventually gives 2 solutions:
b=0 and b=-1
both of which still don't make p prime when you solve for a...you just have to make sure that you cover all possibilities.

I didn't work out the 3rd part, but I bet you get 2 results out of it as well.

CC

 

[foxjwill]

Hi,
Jeffreydk, if a=1=b then p=0 from the origional equation.

yes, but he gave this remark before I had corrected my typo. So, when he saw it, a=b=1 did give p=2

foxjwill,
on part 2,
I don't see how your algebra got you a^2+b^2-2ab=1
CC

I got that by squaring a-b=1

 

[happyg1]

I did not see the edit...sorry.

I took it as a system of equations. Your method is also valid, but not quite complete.
-1 has to be included to be completely rigorous...no matter what, no primes will satisfy the thing :)
CC