Over all integers?

We want to use a good mod to try to prove this. A rule of thumb for powers of $k$ is to use a mod $n$ such that $\phi(n)=k$ or $\phi(n)=2k$ . After some working, we should find that $n=11$ satisfies. Let's use this.

By Fermat's Little Theorem , $x^{10} \equiv 0,1 \pmod{11}$ . Thus, $x^5 \equiv 0,\pm 1 \pmod{11}$ . Thus, we have

$\begin{aligned} a_1^5+a_2^5 + \ldots + a_n^5 &\equiv m_1+m_2+\ldots+m_n \pmod{11}, \quad m_i \in \{0,\pm 1\}\\ &\equiv 28 \pmod{11}\\ &\equiv -5 \pmod{11} \end{aligned}$

Thus, $n$ must be at least 5 for this to occur. In fact, we find that $n=5$ does work if we have $a_1=2$ , $a_2=a_3=a_4=a_5=-1$ . Thus, $n\geq 5$ .

Considering negative integers allowed, and since and odd power like 5 of a negative integer renders the sign negative, i.e., (-1)^5=-1. We need 28=32-4=2^5-1-1-1-1=2^5+4×(-1)^5, that is 5 integers.

For this type better consider a least +integer like 2 then a -integer -1 substitute you'll get least 'n' value