Let $S_k(N)=\{(a_1,\cdots,a_k)\mid N=\sum_1^k a_i^2\}$ be subset of $\mathbb Z^k$, then $|S_k(N)|=r_k(N)$, since $r_k$ is sum of squares function.
There are several well-known explict fomulars such as...
$$r_2(N)=4\sum_{d\mid N\\2\not\mid d}(-1)^{\frac{(d-1)}2}\\r_4(N)=8\sum_{d\mid N\\4\not\mid d}d=8\sigma(N)-32\chi_{\mathbb N}\left(\frac N4\right)\cdot\sigma\left(\frac N4\right)\\ r_6(N)={4\sum_{d\mid N}d^2\left(4\left(-4\mid\frac Nd\right)-\left(-4\mid d\right)\right)}\,((X\mid Y)\text{ is Kronecker symbol})\\ r_8(N)=16\sum_{d\mid N}(-1)^{N-d}d^3$$
But I cannot find non-recursive explict fomulars about $r_5(N)$.
There exists recursive explict fomular using Legendre symbol about $r_5(N)$, since $N=2^\lambda\cdot\prod_{p}p^{\lambda_p}$ and $N'$ is square-free part of $N$, and $\epsilon_5(N')=\left\{\begin{matrix}0&N'\equiv1\pmod8\\\frac{16}{7}&N'\equiv5\pmod8\\4&\text{otherwise}\end{matrix}\right.$, shown as following.
$${r_5(N)=r_5(N')\left[\frac{2^{3\lfloor\frac\lambda2\rfloor+3}-1}{7}-\epsilon_5(N')\frac{2^{3\lfloor\frac\lambda2\rfloor}-1}{7}\right]}{\times\prod_p\left[\frac{2^{3\lfloor\frac{\lambda_p}2\rfloor+3}-1}{7}-p\left(\frac {N'}{p}\right)\frac{2^{3\lfloor\frac{\lambda_p}2\rfloor+3}-1}{7}\right]}$$
But it is too hard to use for calculate $r_5(N)$ for arbitrary $N$, so it is not my interest(unless there are expression for $r_5(N)$ when $N$ is squarefree). Is there are another way?