The background of this question is this: Fermat proved that the equation, $$x^4+y^4 = z^2$$
has no solution in the positive integers. If we consider the near-miss, $$x^4+y^4-1 = z^2$$
then this has plenty (in fact, an infinity, as it can be solved by a Pell equation). But J. Cullen, by exhaustive search, found that the other near-miss, $$x^4+y^4+1 = z^2$$
has none with $0 < x,y < 10^6$.
Does the third equation really have none at all, or are the solutions just enormous?
Edit: The question asks about a different equation than I thought when I read it first time. I will leave this anyway because some people reading the question may be interested.
On the website http://sites.google.com/site/tpiezas/009 there is the identity listed:
$$(17p^2-12pq-13q^2)^4 + (17p^2+12pq-13q^2)^4 = (289p^4+14p^2q^2-239q^4)^2 + (17p^2-q^2)^4$$
That checks out.
One can solve the Pell equation $q^2 - 17 p^2 = \pm 1$, e.g. (p,q) = (0,1),(1,4),(8,33),(65,268),... which lead to infinitely many solutions of the Diophantine equation:
I don't know whether that is all solutions! I suspect so. Thanks to Henry, this is not all solutions!