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

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Homework Statement


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


Homework Equations





The Attempt at a Solution


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

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

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

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

(3) [tex]a^2+b^2=1[/tex] and [tex]a+b=1[/tex], which gives [tex]a^2+b^2+2ab=1[/tex] and then [tex]2ab=0[/tex]. Again, we are left with [tex]a=1[/tex] and [tex]b=0[/tex], 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 [tex]a^4+b^4=p[/tex] instead of what I currently have up there. >_< Oops!
 
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Discussion

Do a and b have to be distinct? Because a=1=b gives p=2, which is prime.
 
Hi,
Jeffreydk, if a=1=b then p=0 from the original equation.

foxjwill,
on part 2,
I don't see how your algebra got you [tex]a^2+b^2-2ab=1[/tex]
I agree with your result up to a point, though.
Here's what I did:
[tex]a^2+b^2=1[/tex] and [tex]a-b=1 \Longrightarrow a=b+1[/tex]
then plug that into the first one:
[tex](b+1)^2+b^2=1[/tex]
[tex]b^2+2b+1+b^2=1[/tex]
which eventually gives 2 solutions:
[tex]b=0[/tex] and [tex]b=-1[/tex]
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:
happyg1 said:
Hi,
Jeffreydk, if a=1=b then p=0 from the original 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

happyg1 said:
foxjwill,
on part 2,
I don't see how your algebra got you [tex]a^2+b^2-2ab=1[/tex]
CC

I got that by squaring [tex]a-b=1[/tex]
 
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