Triangular numbers of the form $x^4+y^4$

Question

Recall that triangular numbers are those $T(n)=n(n+1)/2$ with $n\in\mathbb N=\{0,1,2,\ldots\}$. Fermat ever proved that the equation $x^4+y^4=z^2$ has no positive integer solution. So I think it's natural to investigate triangular numbers of the form $x^4+y^4$ with $x,y\in\mathbb N$. Clearly, $$T(0)=0^4+0^4\ \ \mbox{and}\ \ T(1)=1^4+0^4.$$ Via a computer, I find that $$15^4+28^4=665281=T(1153).$$

Question. Is $T(1153)=665281$ the only triangular number greater than one and of the form $x^4+y^4$ with $x,y\in\mathbb N$?

My computation indicates that $\{x^4+y^4:\ x,y=0,\ldots,5100\}$ contains no other triangular numbers greater than one. I guess that the above question has a positive answer. Can one prove this?

Your comments are welcome!

Answer 1

An exhaustive search up to $10^7$ finds that the only solutions of $x^4+y^4=T(n)$ in integers with $0 \leq x \leq y \leq 10^7$ are: $$ 0^4 + 0^4 = T(0), \quad 0^4 + 1^4 = T(1), \quad 15^4 + 28^4 = T(1153), $$ noted by OP Zhi-Wei Sun; $$ 3300^4 + 7712^4 = T(85508608), $$ already found by Tomita; and the new example $$ 6848279^4 + 6896460^4 = T(94462407145153). $$ Such searches can be done efficiently using the gp function hyperellratpoints, recently ported into gp from Stahlke and Stoll's C program ratpoints; the search up to $10^7$ took a few hours.

Probably there are infinitely many solutions, but quite sparse, because one expects the number of solutions with $0 \leq x,y \leq N$ to be asymptotically proportional to $$ \sum_{x=0}^N \sum_{y=0}^N \left(8(x^4+y^4)+1\right)^{-1/2} $$ which grows as a multiple of $\log N$. Proving anything like that seems intractable by present-day methods. This is typical of Diophantine equations leading to "log-K3 surfaces" such as $x^4 + y^4 = n(n+1)/2$.

Answer 2

New solution: $57331396^4+ 54099540^4=T(6224076560720896)$

$x^4 + y^4 = \frac{\large{n(n+1)}}{\large{2}}$
The problem can be reduced to finding the integer points on quartic as follows.
Let $Y=2n+1$ and $t=y$, then we get $Y^2 = 8x^4 + 8t^4 + 1$ in $x$ with parameter $t$.
Searching the integer points $(x,Y)$ with $t \lt 100000$, we found the integer point $(x,y,Y)=(3300, 7712, 171017217).$

Hence we get $(x,y,n)=(3300, 7712, 85508608).$
$3300^4+7712^4=T(85508608).$