An algebra problem by Wildan Bagus Wicaksono

Sum of third powers: $1^3+2^3+3^3+...+n^3=\left(\dfrac {n(n+1)}2\right)^2$

$\sqrt{\dfrac{1^3+2^3+3^3+...+n^3}n}=\sqrt n\times\frac{n+1}2$

$\sqrt{2017}\approx44.91$

The nearest number above that is $45$ , which happens to be odd, so that $\frac{n+1}2$ is an integer.

So the answer is $45^2=\boxed{2025}$

$\displaystyle \begin{aligned} m & = \sqrt{\frac{1^3 + 2^3 + 3^3 + \cdots + n^3}{n}} & \small \color{#3D99F6} \text{Recall the identity} \ \sum_{k \ = \ 1}^{n} k^3 = \left(\sum_{k \ = \ 1}^{n} k\right)^2 \\ & = \sqrt{\frac{\left(1 + 2 + 3 + \cdots + n\right)^2}{n}} \\ & = \left(1 + 2 + 3 + \cdots + n\right) \cdot \frac{1}{\sqrt{n}} \\ & = \frac{n(n + 1)}{2\sqrt{n}} \end{aligned}$

Observe that if $n$ is a square, then $\sqrt{n} \mid n$ , so it follows that $\sqrt{n} \mid n(n + 1)$ . Also notice that $2 \mid n(n + 1) \ \forall \ n \in \mathbb{N}$ : this can be easily proven by induction.

Thus, $\displaystyle \frac{n(n + 1)}{2\sqrt{n}} \in \mathbb{Z}^+ \iff \sqrt{n} \in \mathbb{Z}^+$ .

We can bound the squares $> 2017$ to $1600 = 40^2 < n^2 < 50^2 = 2500$ . With trial and error, we find that the smallest square $> 2017$ is $45^2 = \boxed{2025}$ .

$\begin{aligned} N & = \sqrt{\frac {\color{#3D99F6}1^3+2^3+3^3+\cdots +n^3}n} & \small \color{#3D99F6} \text{Note that }\sum_{k=1}^n k^3 = \left(\frac {n(n+1)}2 \right)^2 \\ & = \sqrt{\frac {\color{#3D99F6}n^2(n+1)^2}{{\color{#3D99F6}4}n}} \\ & = \frac {\sqrt n(n+1)}2 \end{aligned}$

For $N$ to be an integer, $n$ must be a perfect square. And the smallest perfect square $n > 2017$ is given by $n = \left \lceil \sqrt{2017} \right \rceil ^2$ $= \left \lceil 44.911 \right \rceil ^2$ $= 45^2$ $= \boxed{2025}$ . Then $N = \dfrac {45\cdot 2026}2 = 45585$ .