# Find prime p s.t. a^4-b^4=p

1. Jul 19, 2008

### foxjwill

1. The problem statement, all variables and given/known data
Find all primes p such that $$\exists a,b \in \mathbf{Z}$$ such that $$a^4-b^4=p$$.

2. Relevant equations

3. 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!

2. Jul 19, 2008

### jeffreydk

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

3. Jul 19, 2008

### 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

Last edited: Jul 19, 2008
4. Jul 19, 2008

### foxjwill

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

I got that by squaring $$a-b=1$$

5. Jul 19, 2008

### 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