TITLE: The diophantine eq. $x^4 +y^4 +1=z^2$
QUESTION [25 upvotes]: This question is an exact duplicate of the question
Does the equation $x^4+y^4+1=z^2$ have a non-trivial solution?
posted by Tito Piezas III on math.stackexchange.com.

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?

REPLY [4 votes]: This is not an answer but just a probabilistic (i.e. heuristic) argument which is too long for a comment.
I want to argue that the equation $x^4+y^4+1=z^2$ is likely to have non-trivial solutions and that the fact that there are none with $y<10^8$ isn't exactly a good evidence for lack of solutions. First, looking modulo 4 and 5 we notice that $x, y$ must both be divisible by 10. Also note that modulo other primes there doesn't seem to be any unusual obstructions, i.e. roughly half of the values of $\{x^4+y^4+1; x, y \in \mathbb{Z}/p\mathbb{Z}\}$ are quadratic residues modulo $p$.
We then write our equation as $10^4(x^4+y^4)+1=z^2$. A random number of order $n$ has probability roughly $\tfrac{1}{2\sqrt{n}}$ of being a square but here we know that it is a square modulo $10^4$, so the conditional probability is roughly $4/\sqrt{n}$. So for fixed $y$ the expected number of $x$ which would make $10^4(x^4+y^4)+1$ a perfect square is roughly
$$
\sum_{x=1}^\infty \frac{1}{25\sqrt{x^4+y^4}}\sim \frac{2}{27y},
$$
the last comparison is asymptotic for large $y$ (the exact constant is, of course, not $2/27$) but the value of the sum is somewhat smaller for small $y$.
The sum $\sum_{y=1}^\infty \frac{2}{27y}$ diverges (which suggests there are infinitely many solutions) but if you sum only over $y\leq 10^8$ the sum is still only roughly $1.4$