Repeated Terms

Let the last term of $m^2$ be $l_m$ th term. We note that $l_m = \displaystyle \sum_{i=1}^m i^2 = \frac {m(m+1)(2m+1)}6$ . The $k$ th term is given by $m$ of the smallest $l_m$ larger than $k$ . That is:

$\begin{aligned} \frac {m(m+1)(2m+1)}6 & \ge k & \small \color{#3D99F6} \text{For }k = 1716 \\ m(m+1)(2m+1) & \ge 6\times 1716 \\ 2m^3 & \approx 6 \times 1716 \\ m & \approx \sqrt[3]{5148} \\ & \approx 17.267 \end{aligned}$

We note that $l_{17} = \dfrac {17(17+1)(2(17)+1)}6 = 1785 > 1716$ and $l_{16} = \dfrac {16(16+1)(2(16)+1)}6 = 1496 < 1716$ . Therefore, the 1716th term is $17^2$ and the required answer is $\boxed{17}$ .