Are there solutions to the following simultaneous equations? 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? 
 A: 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.
A: 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.
