Waring's problem over $\mathbb Q_{\ge0}$ Let $k$ be a positive integer. Note that $a/b=ab^{k-1}/b^k$ for any integers $a$ and $b>0$.
If every $n\in\mathbb N=\{0,1,2,\ldots\}$ can be written as $x_1^k+\cdots+x_{s}^k$ with $x_1,\ldots,x_s\in\mathbb N$, then each $r\in\mathbb Q_{\ge0}$ can be written as $x_1^k+\cdots+x_s^k$ with $x_1,\ldots,x_s\in\mathbb Q_{\ge0}$, where
$$\mathbb Q_{\ge0}=\{r\in\mathbb Q:\ r\ge0\}.$$
Instead of the classical Waring problem over $\mathbb N$, we may consider Waring's problem over $\mathbb Q_{\ge0}$. Let $s(k)$ be the least positive integer $s$ such that each $r\in\mathbb Q_{\ge0}$ can be written as $x_1^k+\ldots+x_s^k$ with $x_1,\ldots,x_s\in\mathbb Q_{\ge0}$. Then
$s(k)$ exists and moreover $s(k)\le g(k)$. It is interesting to find the exact value of $s(k)$.
By Theorems 233 and 234 in Hardy and Wright's book A Introduction to the Theory of Numbers, we have $s(3)=3$ (the inequality $s(3)\le 3$ was obtained by Richmond in 1923).
Question. Can one prove that $s(k)\ge k$ for each integer $k>2$? Is it true that $s(4)=5$?
Your comments are welcoeme!
 A: I can answer the last question: $s(4)$ equals $15$, not $5$. Variables below will denote integers.
First we show that $s(4)\leq 15$. We use the result of Davenport (1939) that there exists $m\geq 1$ such that every positive integer outside $\{16^h k:\text{$h\geq 0$ and $k\leq m$}\}$ is a sum of $15$ integral biquadrates. Now let $a,b\geq 1$ arbitrary, and let $n$ be an odd number such that $n^4>m$. Then $ab^3n^4$ is a sum of $15$ integral biquadrates, hence $a/b$ is a sum of $15$ rational biquadrates.
Now we show that $s(4)\geq 15$. It suffices to show that $15$ is not the sum of $14$ rational biquadrates. Equivalently, $15 n^4$ (for $n\geq 1$) is not the sum of $14$ integral biquadrates. Assume that $15n^4$ is the sum of $14$ integral biquadrates, with $n\geq 1$ minimal. We shall use twice that every integral biquadrate is $\equiv 0,1\pmod{16}$. If $n$ is even, then $15 n^4\equiv 0\pmod{16}$, so every biquadrate in the decomposition is even. That is, $15(n/2)^4$ is also the sum of $14$ integral biquadrates, contradicting the minimality of $n$. If $n$ is odd, then $15 n^4\equiv 15\pmod{16}$, while a sum of $14$ biquadrates is $\equiv 0,\dotsc,14\pmod{16}$. This is again a contradiction.
