How can I prove or disprove that
$$ \forall i \in \{1, \ldots,k\} ,\forall j \in \{1,\ldots,m\}: a_i,b_j\in \mathbb{Z},n\in \mathbb{N},n>0$$
if $$a_0^n+a_1^n+...+a_k^n=b_0^n+b_1^n+...+b_m^n$$
then
$$\sum_{i=0}^ka_i-\sum_{j=0}^mb_j \equiv0\pmod n$$
For example $n=2,k=1,m=0$ then $$a_0^2+a_1^2=b_0^2$$ we have Pythagorean triples. With writing parametric solution we can prove it easily :
$$a_0=2uv$$ $$a_1=u^2-v^2$$ $$b_0=u^2+v^2$$ where $u,v\in \mathbb{Z}$
$$a_0+a_1-b_0=2uv+(u^2-v^2)-(u^2+v^2)=2uv-2v^2\equiv 0 \pmod 2$$
if $n=3$ I found some examples like $3^3+4^3+5^3=6^3$ which $3+4+5-6=6\equiv0\pmod3$ but no prove
Assume that $a_0^p+a_1^p +\cdots +a_{k}^p = b_0^p+b_1^p+\cdots+ b_{m}^p$ for some $k,m \in \mathbb{N}$ and $p$ is a prime number.
Then by Fermat's little theorem we know that $x^p=x \mod p$, so from the above we know that $a_0^p+a_1^p +\cdots +a_{k}^p = b_0^p+b_1^p+\cdots+ b_{m}^p \mod p$ and combine this with Fermat's little theorem we arrive at $a_0+a_1+\cdots+a_k= b_0+b_1+\cdots + b_m \mod p$ rearranging that we get $a_0+a_1+\cdots +a_k -b_0 -b_1- \cdots -b_m=0 \mod p$.
Which prove the conjecture for prime numbers.
For the general case let $n=p_1^{e_1} p_2^{e_2} \cdots$ , so going through the same reasoning above we arrive at $n = 1 \mod \phi(d)$ for some $d|n$, its obvious that this condition will not always apply.
For example : $n=7*11=77$, $\phi(77)=(7-1)(11-1) = 60$, $\phi(11)=10$,$\phi(7)=6$.
So $77 = 17 \mod 60$ and $77 = 7 \mod 10$ and $77 = 5 \mod 6$ , so in general it will not be true that $a_0+a_1+\cdots +a_k-b_0-b_1-\cdots -b_m= 0 \mod 77$ or $a_0+a_1+\cdots +a_k-b_0-b_1-\cdots -b_m= 0 \mod 11$ or $a_0+a_1+\cdots +a_k-b_0-b_1-\cdots -b_m= 0 \mod 7$.
For concrete example : $1^{77}+1^{77}+\cdots +1^{77}=2^{77}$, which is just $2^{77}*1^{77}=2^{77}$ so we have to check that $2-2^{77} = 0 \mod d$ for some $d|77$, and $d>1$.
We get that $2-2^{77}=61 \not=0 \mod 77$ and $2-2^{77}=6 \not=0 \mod 11$ and $2-2^{77}=5 \not=0 \mod 7$.