Are there solutions to the following simultaneous equations?

Question

I have been looking into solutions to simultaneous equations of the form $$\sum_{r=1}^{n}a_{r} = \sum_{r=1}^{n}b_{r}$$ $$\sum_{r=1}^{n}a_{r}^{2} = \sum_{r=1}^{n}b_{r}^{2}$$ where $\forall i$, $a_{i}$ and $b_{i}$ are positive integers and no $a_{i}=b_{j}$, $\forall i, j$. This has infinitely many solutions for example $$n^{2}+(n+k+4)^{2}+(n+2k+5)^{2} \equiv (n+1)^{2}+(n+k+2)^{2}+(n+2k+6)^{2}$$ $$n+(n+k+4)+(n+2k+5) \equiv (n+1)+(n+k+2)+(n+2k+6)$$

I was wondering if there are solutions to the simultaneous equations $$\sum_{r=1}^{n}a_{r} = \sum_{r=1}^{n}b_{r}$$ $$\sum_{r=1}^{n}a_{r}^{2} = \sum_{r=1}^{n}b_{r}^{2}$$ $$\sum_{r=1}^{n}a_{r}^{3} = \sum_{r=1}^{n}b_{r}^{3}$$ or possibly for even higher powers.

Edit: Could you give example solutions?

Answer 1

Yes, you can do it for arbitrarily high powers. We can move the $b$s to the left and just ask for solutions to $\sum (\pm 1)a_r^k=0$. We can note that $(n+3)-(n+2)-(n+1)+n=0$ gives a solution to the first equation, then that $(n+3)^2-(n+2)^2-(n+1)^2+n^2=4$, so $-(n+7)^2+(n+6)^2+(n+5)^2-(n+4)^2+(n+3)^2-(n+2)^2-(n+1)^2+n^2=0$ gives a solution to the first two. Extending to $16$ terms, changing the signs the same way gives $$(n+15)^3-(n+14)^3-(n+13)^3+(n+12)^3-(n+11)^3+(n+10)^3+(n+9)^2-(n+8)^3-(n+7)^3+(n+6)^3+(n+5)^3-(n+4)^3+(n+3)^3-(n+2)^3-(n+1)^3+n^3=0$$ will allow us to satisfy the first three and so on.

Answer 2

If you remove the integer constraint, the given problem has a straightforward solution. Consider $p(x)$ and $p(x)+q(x)$, where $\deg p=n$ and $\deg q < n-k$. By Vieta's and Newton's formulas, the power sums of the roots of $p$ and $p+q$ are the same, up to $\sum_{\zeta}\zeta^k$.

On the other hand, if we manage to find a couple of polynomials $p,q$ such that all the roots of $p$ and $p+q$ are rational, we also have a solution of the integer problem by clearing denominators.

The set of equations $$x_1+x_2+\ldots+x_n = X_1+X_2+\ldots+X_n,$$ $$x_1^2+x_2^2+\ldots+x_n^2 = X_1^2+X_2^2+\ldots+X_n^2$$ $$x_1^3+x_2^3+\ldots+x_n^3 = X_1^3+X_2^3+\ldots+X_n^3$$ $$ \ldots $$ $$x_1^m+x_2^m+\ldots+x_n^m = X_1^m+X_2^m+\ldots+X_n^m$$ assuming $m<n$, defines an algebraic variety over $\mathbb{Q}^{2n}$ with an infinity of trivial rational points, $(x_1,x_2,\ldots,x_n)=(X_1,X_2,\ldots,X_n)\in\mathbb{Q}^n$. With the standard tools of algebraic geometry it should not be difficult to prove that this implies the existence of an infinity of non-trivial rational points, i.e. the possibility to pick $p$ and $q$ as in the previous paragraph.