I have the following question:
I don't see any counterexample, but assuming that there does exist one, can we add more similar polynomial identities, maybe qubes or something to prevent such counterexamples?
There are lots of solutions, and many of them are not really hard to find.
Select some positive and negative integers, none of which is zero or the additive inverse of any other in the set, having a zero sum overall. I will use $\{1,2,-3,4,-5,-6,7\}$. The sum of squares is of course $140$ in this case.
Now consider the additive inverses as a second set. By construction this has sum zero and the same sum of squares as above: $\{-1,-2,3,-4,5,6,-7\}$ again has sum zero and sum of squares $140$.
Add the same constant to both sets to get a positive solution. Reordering the numbers after performing this uniform addition, with $8$ as the increment, gives the two sets
$\{2,3,5,9,10,12,15\}$
$\{1,4,6,7,11,13,14\}$
both having sum $56$ and sum of squares $588$.
For the given problem the minimal number of elements is given by $3$ numbers $$ (41,42,110) \text{ and } (20,77,96) $$ Is an example.
The set of positive numbers with the smallest sum is instead given by $$ (1,5,6) \text{ and } (2,3,7) $$
This is the case with $k=2$ of the Prouhet–Tarry–Escott problem
It is known for, that for $2 \le k \le 9$ and $k=11$ we can find two sets with $n=k+1$ numbers that satisfy the equation. In general there is always a solution in which $n=2^k$ (see link for details)
Update: As said in the comments is actually sufficient to consider $n=\frac{1}{2}k(k+1)+1$ (see references in the comments)
Assume that
$$\sum_i m_i=\sum_i n_i,\\\sum_i m_i^2=\sum_i n_i^2.$$
Then for any integer $k$,
$$\sum_i(m_i+k)=\sum_i(n_i+k),$$$$\sum_i(m_i+k)^2=\sum_im_i^2+2k\sum_im_i+\sum_i k^2\\=\sum_in_i^2+2k\sum_in_i+\sum_i k^2=\sum_i(n_i+k)^2.$$
So if you find one solution [such as $(-2,2,3)/(-1,0,4)$], you can derive an infinity of them.
assuming that there does exist one, can we add more similar polynomial identities, maybe qubes or something to prevent such counterexamples?
Not polynomial identities, due to the pigeonhole principle.
Consider sets of (nonzero) natural numbers whose largest element is no more than $n$. There are $2^{n-1}$ such sets.
Their sum can be at most $\frac{n(n+1)}{2}$, the $n$th triangular number. The sum of their squares can be at most $\frac{n(n+1)(2n+1)}{6}$, the $n$th square pyramidal number. [And, in general, for any polynomial, the sum of all the first $n$ values is given by a polynomial of degree one higher].
The total number of pairs of sums and sums-of-squares is the product of these two polynomials, $\frac{n(n+1)}{2}\cdot\frac{n(n+1)(2n+1)}{6}$, a polynomial of degree 5. For sufficiently large $n$, this polynomial is less than $2^n$, so by the Pigeonhole Principle there are two such sets with the same sum and sum of squares.
[And, in general, for any number of polynomials of any degree...]
It's possible to uniquely identify sets from their sums if you don't limit yourself to polynomials. For example, given the value of $2^{m_1}+2^{m_2}+\ldots+2^{m_N}$, you can uniquely determine the $m_i$ as long as they're distinct.
Non-constructive counting proof that a counterexample exists. First for the original question, then for the follow-up. The basic idea is that if we consider a bound the elements of the set and increase it, the number of possible sets grows much much faster than the number of possible values for the sum and sum of squares.
First a simplification: We don't need to require the sets to be disjoint, only different. If we have two different sets $S$ and $S'$, then $S \setminus S'$ and $S' \setminus S$ are disjoint nonempty sets of equal size that have the same sum and sum of squares.
Let $S$ be a set of exactly $N$ natural numbers at most $M$.
The sum of $S$ is at most $NM$. The sum of squares of $S$ is at most $N M^2$. Therefore there are at most $(NM+1)(NM^2+1)$ pairs (sum, sum of squares). There are $\binom{M+1}{N}$ different such sets.
Since $\binom{M+1}{4}$ is a 4th degree polynomial in $M$, it must be larger than $(4M+1)(4M^2+1)$ for large values of $M$. Therefore there exists an $M$ for which there are two different sets $S$ and $S'$ of size $4$ that have the same sum and sum of squares.
I don't see any counterexample, but assuming that there does exist one, can we add more similar polynomial identities, maybe qubes or something to prevent such counterexamples?
Similar approach works for all finite amounts of "polynomial identities".
To be exact, let's say you have $k$ polynomials $p_1, \ldots, p_k$ with integer values, and the question is whether there are two different sets of equal size such that $(\sum_{s \in S} p_1(s), \ldots, \sum_{s \in S} p_k(s))$ has the same value of those sets.
Let $l$ be the largest degree of the $p_i$'s. There exists a constant $C$ such that if $s \leq M$, then $|p_i(s)| \leq C M^l$.
So now, if $S$ has $N$ elements that are at most $M$, then $|\sum_{s \in S} p_i(s)| \leq N C M^l$ for all $p_i$. So one polynomial can have at most $(2 N C M^l +1 )$ different values. So the tuple of all $k$ polynomials can have at most $(2 N C M^l +1)^k$ different values.
But there are $\binom{M+1}{N}$ such sets. Choosing $N = kl+1$, this is a polynomial of $M$ with degree $kl+1$. So you can always pick a large enough $M$ to ensure this is more than $(2 (kl+1) C M^l)^k$.
