<s>[INST] How do we solve $x^2 + \{x\}^2 = 33$ without computer? This is a problem taken from a group on Facebook. I wonder how to solve this without numerical process.
$x^2 + \{x\}^2 = 33\tag{1}$
My unfinished attempt:
$$\begin{align}
x^2 + \{x\}^2 &= 33\\
x^2 + \left(x - \lfloor x \rfloor\right)^2 &= 33\\
x^2 + x^2 - 2x \lfloor x \rfloor + \lfloor x \rfloor^2 &= 33\\
2x\left(x - \lfloor x \rfloor\right) + \lfloor x \rfloor^2 &= 33\\
2x\{x\} + \lfloor x \rfloor^2 = 33
\end{align}$$
I'm stuck at there. Also we know that the fractional part of $x$ is bounded i.e.:
$$0\leq\{x\}<1$$
From that, I can predict where the two solutions are placed at:
$$\begin{align}
0&\leq \{x\} < 1\\
0 &\leq \{x\}^2 < 1\\
x^2 &\leq x^2 + \{x\}^2 < x^2 + 1\\
x^2 &\leq 33 < x^2 + 1 \tag{$x^2 + \{x\}^2 = 33$}\\
S &= -\sqrt{33}\leq x \leq \sqrt{33} \quad \bigcap \quad  x< -\sqrt{32} \lor x > \sqrt{32}\\
\therefore S &= -\sqrt{33}\leq x < -4\sqrt2  \quad \bigcup \quad  4\sqrt{2} < x \leq \sqrt{33}\\
S &\approx -5.74456 \leq x < -5.65685  \quad \bigcup \quad  5.65685 < x \leq 5.74456
\end{align}$$
I don't know if that's even going to help. Anyway, here are the two solutions (to the original problem) that's given by Wolfram Alpha:
$$\begin{align}
x_1 &= \frac12 \left(3\sqrt{17} - 1\right) \approx -5.815\\
x_2 &= \frac12 \left(-1 - 3\sqrt{113}\right) \approx 5.685
\end{align}$$
You see $x_1$ isn't at the interval. I'm confused.
 [/INST] We have $0\le \{x\}^2 < 1$, so $x^2$ is between $32$ and $33$. This tells us that $\lfloor x \rfloor$ must be either $5$ or $-6$.
For the positive solution, write $x=5+y$ with $0\le y < 1$and
$$ (5+y)^2 + y^2 = 33 $$
which you can now solve for $y$ with standard methods.
The negative solution goes very similarly -- but with a different quadratic equation to solve.
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