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? 

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.
