Counting integers of the form $x^4 + y^4$ ?? I was intrested in counting Integers of the form $x^4 + y^4$ ( smaller than Some n) both with and without multiplicity.
The Sum of 2 squares analogue is more Well-known.
Gauss circle problem counts essentially these numbers.
Ramanujan-Landau constant relates here too.
A simple combinatorial argument for the bisquare case gives $$\text{density} ( x^4 + y^4) =\frac12 * \text{density}(x^4) * \text{density}(y^4) = \frac12 * n^{1/4} * n^{1/4} = \frac12 * n^{1/2}$$ Or the $K$th Number of that form grows like $4 K^2$.
However I Get stuck at getting better estimates !
The $4$ above is not optimal I think and I wonder about inf and sup of the value $C$ in $C K^2$.
By analogue of the Sum of squares I tried integrals such as
$$ \int 2( t^4 + x^4)^{1/4}\,dx  = t^2 * 2 * 2F_1(-\frac14,\frac14;\frac54;-1) = 2.084... t^2$$
But these values seem wrong.
$2.084$ is way to small , it seems C is more like $\sqrt{27}$ or such.
How to think about this ?
I assume a lim to a value $C$ but Maybe that is wrong ; lim sup might not be lim inf ??
 A: Indeed, this is very similar to Gauss circle problem. For any $R\gg\sqrt{2}$, the number $N(R)$ of lattice points in the region $x^4+y^4\leq R^4$ is bounded between the area enclosed by $x^4+y^4\leq (R-\sqrt{2})^4$ and the area enclosed by $x^4+y^4\leq (R+\sqrt{2})^4$.
Since by Euler's Beta function
$$ 4\int_{0}^{1}(1-x^4)^{1/4}\,dx = \tfrac{1}{2\sqrt{\pi}}\,\Gamma\left(\tfrac{1}{4}\right)^2 \tag{A}$$
we have:
$$ N(R) = \tfrac{1}{2\sqrt{\pi}}\,\Gamma\left(\tfrac{1}{4}\right)^2 R^2+O(R^{3/2}) \tag{B}$$
and the number of integers in $[0,N]$ which can be represented as $x^4+y^4$ for $(x,y)\in\mathbb{Z}^2$, counted according to their multiplicity, is $\tfrac{1}{2\sqrt{\pi}}\,\Gamma\left(\tfrac{1}{4}\right)^2 \sqrt{N}+O(N^{3/8}).$
If we disregard the multiplicity, there cannot be more that $(2n^{1/4})\cdot(2n^{1/4})= 4\sqrt{n}$ integer solutions of $x^4+y^4\leq n$. The exact order of magnitude should be off by a logarithmic factor raised to a very small power, by sieve methods. The smallest number expressible as the sum of two fourth powers in two different ways is
$$ 635318657 = 133^4+134^4 = 158^4+59^4 $$
as found by Euler.

Interesting remark: there is a gap between the count according to the multiplicity and the count disregarding the multiplicity. In particular, this shows that the function counting the number of representations as a sum of two fourth powers is unbounded. Of course, the same holds in the cubic case and proves the existence of taxicab numbers (and their quartic analogues) like $1729=12^3+1^2=10^3+9^3$. The existence of quintic taxicab numbers is unsolved.
