Does the equation $x^k+y^k-z^k-w^k=3\ (k>3)$ have a solution over $\mathbb N$?

Question

Clearly, $$3=0^2+2^2-1^2-0^2\ \ \mbox{and}\ \ 3=4^3 +4^3-5^3-0^3.$$

Question. Let $k>3$ be an integer. Does the equation $$ x^k+y^k-z^k-w^k=3\quad \ (x,y,z,w\in\mathbb N=\{0,1,2,\ldots\})\tag{1}$$ have a solution?

I have checked that the equation $(1)$ has no solution with $\max\{z,w\}\le 500$ for $k=4,5$. Also, for $k=6,7$, the equation $(1)$ has no solution with $\max\{z,w\}\le 300$. I conjecture that $(1)$ has no solution for each integer $k>3$.

By the way, I also conjecture that any integer can be written as $x^3+y^3-z^3-w^3$ with $x,y,z,w\in\mathbb N$ (see https://oeis.org/A351338). This is stronger than Sierpinski's conjecture that each integer is a sum of four integer cubes.

Your comments are welcome!

Answer 1

There are no solutions for $k>2$ even.

For $k$ even, $n^k$ is congruent to $1$ mod $8$ as soon as $n$ is odd. If $k>2$, then $n^k$ is conruent to $0$ mod $8$ as soon as $n$ is even.

So $x^k + y^k - z^k -w^k$ is congruent mod $8$ to a sum of two terms that are $0$ or $1$ and two terms that are $0$ or $-1$, i.e. to $-2, -1, 0,1$ or $2$.

Since none of these is $3$, there are no solutions modulo $8$, and thus no integer solutions.