Take $a_k = \left(\dfrac {n(n+1)(2n+1)} {6}\right)^4 k$, for $1\leq k \leq n$. Then $\sum_{k=1}^n a_k^2 = \left (\left (\dfrac {n(n+1)(2n+1)} {6}\right )^3\right )^3$ and $\sum_{k=1}^n a_k^3 = \left (\left (\dfrac {n(n+1)(2n+1)} {6}\right )^6 \dfrac {n(n+1)} {2}\right)^2$.

Mrdavid445 wrote: Determine whether or not there are any positive integral solutions of the simultaneous equations \begin{align*}x_1^2+x_2^2+\cdots+x_{1985}^2&=y^3,\\ x_1^3+x_2^3+\cdots+x_{1985}^3&=z^2\end{align*}with distinct integers $x_1$, $x_2$, $\ldots$, $x_{1985}$. Dangit this felt surprisingly hard or maybe I am just dumb.Anyway here's my solution similar to above:- Pick any distinct numbers $(x_1,x_2,\cdots x_n)$ and let $$a=x_1^2+x_2^2+\cdots x_n^2$$$$b=x_1^3+x_2^3+\cdots x_n^3$$Now I claim the numbers $(a^4b^3x_1,a^4b^3x_2,\cdots a^4b^3x_n)$ works.indeed $$\sum (x_i')^2=a^8b^6(\sum x_i^2)=(a^3b^2)^3$$$$\sum(x_i')^3=a^{12}b^9(\sum x_i^3)=(a^6b^5)^2$$So we are done.$\square$

This is one of those problems where reading a solution doesn't answer how you'd come up with the solution. If anyone is interested, my thought process followed this trajectory: I thought the answer was no. Got stuck, so I decided to try proving "yes" instead. I tried to solve the N=2 case instead of N=1985. Got stuck, and could satisfy the z equation but not the y equation (not easily at least). I had the idea that maybe this was one of those problems where if I solved a smaller case (say N=2), then I could use that as a gadget to construct a solution of size N+1. So the line of thinking here shifted from "find a solution" to "Suppose I have a solution (x1, ..., xn), ... what else can I construct?" For example, I thought that maybe I could construct a 2n solution with (x1, 2x1, x2, 2x2, ...) but that satisfied only the z equation. Rats! But if I have a solution for N=2, then n -> 2n is real! So this motivated me to spend a lot of time looking for an N=2 solution. Another line of thought I had was starting with (x1, ..., xn) and splitting xn into two other x's to give an (n+1) solution, but that didn't work. I tried 4 or 5 different tricks for N=2, didn't see anything. Stepped away for 30 minutes. Realized that I could manipulate the (1,2) semi-solution to preserve the z equation, but add an extra factor of 5 to the RHS of the y equation! (25,50) doesn't quite work, but (625, 625*2) does. That helped me solve N=2. Went down some dead ends with the gadget thing for a bit before realizing that N=2 generalized because sum of the first N cubes is a perfect square, and I could do the same scaling trick no matter the factorization of y. Nice!

Yes, consider $x_k = k \cdot \left(1^2+ 2^2 + \cdots + 1985^2\right)^4$. We have \[x_1^2 + x_2^2 + \cdots + x_{1985}^2 = (1^2 + 2^2 + \cdots + 1985^2) \cdot \left(\left(1^2+ 2^2 + \cdots + 1985^2\right)^4\right)^2 = \left(1^2+2^2+\cdots + 1985^2\right)^9,\]which is a perfect cube, and \[x_1^3+x_2^3 + \cdots + x_{1985}^3 = (1^3 + 2^3 + \cdots + 1985^3 ) \cdot \left(1^2 + 2^2 + \cdots + 1985 ^2 \right)^{12} = (1+2+\cdots + 1985)^2 \cdot \left(1^2 + 2^2 + \cdots + 1985^2 \right)^{12},\]which is a perfect square.

Let \begin{align*} 1^2 + 2^2 + \dots + 1985^2 &= b \\ 1^3 + 2^3 + \dots + 1985^3 &= c. \end{align*}Then take $x_i = i \cdot b^4c^3$.