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!



Triangular numbers of the form $x^4+y^4$
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
$x^4 + y^4 = n(n+1)/2$.


