TITLE: Find the smallest $a\in\Bbb N$ if $a-\lfloor\sqrt a\rfloor^2=100$
QUESTION [8 upvotes]: I don't know how to solve this algebraically.
Sure, I could just brute-force this, but that wouldn't help me understand this equation. (It is not the solution that is so important, it is the way that got you there.)
How do I find the smallest $a\in\Bbb N$ for which this is true?
$$a-\lfloor\sqrt a\rfloor^2=100$$
What about an arbitrary number $b\in \Bbb N$?
$$a-\lfloor\sqrt a\rfloor^2=b$$

REPLY [2 votes]: Imagine $a$ running through the natural numbers. While
$$k^2\leq a<(k+1)^2,\quad {\rm i.e.,}\quad k^2\leq a\leq k^2+2k$$
for some $k\geq0$ the difference $a-\lfloor\sqrt{a}\rfloor^2=a-k^2$ increases from $0$ to $2k$. The first time this difference reaches $100$ is when $k=50$ and  $a=50^2+2\cdot 50=2600$.