Find the 1973rd non-perfect square.

**
Clarification:
**
Non-perfect squares are positive integers which cannot be expressed as
$k^2$
, where
$k$
is an integer. For example: 2, 3, 5, 7, 8,... and 2 is the first non-perfect square.

The answer is 2017.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

Let $q(n)$ be the $n$ th non-perfect integer. Then the sequence of non-perfect integers up to $q(n)=k^2 +1$ , where $k^2$ is the $k$ th perfect square is as follows:

$\begin{aligned} \{ q(n) \} & = \underbrace{ {\color{#D61F06}\cancel 1}, 2, 3, {\color{#D61F06}\cancel 4}, 5, 6, 7, 8, {\color{#D61F06}\cancel 9}, 10, \cdots {\color{#D61F06}\cancel {k^2}}, k^2+1}_{k^2+1-k \text{ terms}} \\ \implies n & = k^2 - k + 1 \\ q(k^2-k+1) & = k^2 + 1 \end{aligned}$

Now the largest perfect square smaller than 1973 is $\lfloor \sqrt {1973}\rfloor =44$ . Then, we have:

$\begin{aligned} q(k^2-k+1) & = k^2+1 & \small \color{#3D99F6} \text {Putting }k=44 \\ q(44^2-44+1) & = 44^2+1 \\ q(1893) & = 1937 \\ q(1893+80) & = 1937+80 \\ q(1973) & = \boxed{2017} \end{aligned}$