Floor Equation

Before applying brute force, let us simplify matters. We must have $\begin{aligned} \lfloor \sqrt{4n+2019} \rfloor & \le \; \sqrt{n} + \sqrt{n+1} \\ \sqrt{4n+2019} & \le \; 1 + \lfloor \sqrt{4n+2019} \rfloor \; \le \; \sqrt{n} + \sqrt{n+1} + 1 \; \le \; 2\sqrt{n+1} + 1 \\ 4n + 2019 & \le \; 4n+5 + 4\sqrt{n+1} \end{aligned}$ and hence $n \ge \big(\tfrac{2014}{4}\big)^2 - 1 = 253511.25$ , so that $n \ge 253512$ . This cuts the number of tests until we find the answer of $\boxed{254520}$ down to size.

This was the simplest solution I could come up with, but while playing around with the code it was interesting to see how close these two get for so long before they finally equal out.

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<?php
$i = 1;
while( floor(sqrt($i) + sqrt($i+1)) != floor(sqrt((4*$i) + 2019)) ){
    $i++;
}
echo $i; //254520
?>

Brute force search: Smallest solution was 254520. This is a computer solution.