If $|x| \le 8$, find the number of real solutions to equation $x^2 + x - \lfloor x \rfloor - \lfloor x^2 \rfloor = 1$

Question

Define $\{x\} = x − ⌊x⌋$, where $⌊x⌋$ denotes the largest integer not exceeding $x$. If $|x| ≤ 8$, find the number of real solutions to the equation $$\{x\} + \left\{ x^2 \right\} = 1$$

Answer 1

My approch:- $$x - \lfloor x \rfloor + x^2 - \lfloor x^2 \rfloor = 1$$ $$x + x^2 - \lfloor x \rfloor - \lfloor x^2 \rfloor - 1 = 0$$

Note:- There are no integer solution because when $x \in \Bbb Z, x^2 = \lfloor x^2 \rfloor$ and $x = \lfloor x \rfloor \implies x - \lfloor x \rfloor + x^2 - \lfloor x^2 \rfloor \ne 1$

Let $\lfloor x \rfloor = n \hspace{2.5cm} \{0\le n \le 7, \hspace{3mm}n \in \Bbb Z\}$
$\implies n \le x \lt n+1$

For $0 \le x \le 8$,
$n^2 \le \lfloor x^2 \rfloor \lt (n+1)^2$

$x = \cfrac{-1+\sqrt{1+4(\lfloor x \rfloor + \lfloor x^2 \rfloor + 1)}}{2} \hspace{1cm} \because x \ge 0$
Taking lower bound of $\lfloor x^2 \rfloor = n^2$

$x = \cfrac{-1+\sqrt{1+4(n + n^2 + 1)}}{2} = \cfrac{\sqrt{(2n + 1)^2 + 4} -1}{2} > n$

Taking upper bound of $\lfloor x^2 \rfloor = (n+1)^2$

$x = \cfrac{-1+\sqrt{1+4(n + (n+1)^2 + 1)}}{2} = \cfrac{\sqrt{(2n + 3)^2 - 1} -1}{2} < n+1$

This means, For every value of $\; n^2 \le \lfloor x^2 \rfloor \lt (n+1)^2$ where $0\le n \le 7,\; \exists\; $ a unique $x $ such that it satisfy $n \le x \lt n+1$

So for $0 \le x \le 8$,
$\sum_{n=0}^7 (n+1)^2 - n^2 = 64$ solutions.

Try for $\{-8 \le x \lt 0\}$ yourself $\rightarrow 49$ solutions.

Total solutions:- $113$