Finding all rational values of $x$ for which $256(x + 15\{x\})\{x\} = 2021$ Find all rational numbers $x$ that satisfy $$256(x + 15\{x\})\{x\} = 2021$$ where $\{x\}$ is the fractional part of $x.$

I first substituted in $x - \lfloor x \rfloor = \{x\},$ which gave me $$256(16x - 15\lfloor x \rfloor)(x - \lfloor x \rfloor) = 2021$$ after simplifying. However, from here, I got stuck.
 A: Hint: Let $\{x\} = a/b$ in lowest terms, $x = y + a/b$ where $y$ is an integer.
Then $x + 15 \{x\} = y + 16 a/b$ and the equation becomes
$$ 256 \left( b y + 16 a\right) \frac{a}{b^2} = 2021 $$
Conclude that $b$ must be a power of $2$...
A: Assuming that $x=\lfloor x\rfloor+\{x\}$, this is equivalent to
$$
(\lfloor x\rfloor+16\{x\})\{x\}=\frac{2021}{256}
$$
Let $n=\lfloor x\rfloor\in\mathbb{Z}$ and $t=\{x\}\in[0,1)$. Then we need to solve
$$
16t^2+nt-\frac{2021}{256}=0
$$
which by the quadratic formula gives
$$
t=\frac{-2n\pm\sqrt{4n^2+2021}}{64}
$$
To ensure that $t\ge0$, we need to take the positive square root.
If $\sqrt{4n^2+2021}\in\mathbb{Q}$, then $\sqrt{4n^2+2021}\in\mathbb{Z}$. So we must have a positive $m\in\mathbb{Z}$ so that
$$
(m-2n)(m+2n)=43\cdot47=1\cdot2021
$$
This leaves us with
$$\require{cancel}
\begin{array}{l}
&&t=\frac{m-2n}{64}&x=n+t\\\hline
n=1&m=45&t=\frac{43}{64}&x=\frac{107}{64}\\
n=-1&m=45&t=\frac{47}{64}&x=-\frac{17}{64}\\
n=505&m=1011&t=\frac1{64}&x=\frac{32321}{64}\\[-3pt]
n=-505&m=1011&\hspace{-1.5mm}\cancel{t=\frac{2021}{64}}
\end{array}
$$
A: Here my attempt 
$\mathrm{256}\left({x}+\mathrm{15}\left\{{x}\right\}\right)\left\{{x}\right\}=\mathrm{2021} \\ $
$\mathrm{256}\left(\left[{x}\right]+\mathrm{16}\left\{{x}\right\}\right)\left\{{x}\right\}=\mathrm{2021} \\ $
$\left[{x}\right]=\frac{\mathrm{2021}}{\mathrm{256}\left\{{x}\right\}}−\mathrm{16}\left\{{x}\right\} \\ $
$\left\{{x}\right\}=\frac{{n}}{\:\sqrt{\mathrm{16}.\mathrm{256}}}=\frac{{n}}{\mathrm{64}}\:,\:{n}\in{Z}^{+}  \\ $
$\left[{x}\right]=\frac{\mathrm{2021}/{n}−{n}}{\mathrm{4}},\:{n}\in\left\{\mathrm{1},\mathrm{43},\mathrm{47}\right\} \\ $
${x}=\frac{\mathrm{2021}/{n}−{n}}{\mathrm{4}}+\frac{{n}}{\mathrm{64}} \\ $
${x}\in\left\{\mathrm{505}\frac{\mathrm{1}}{\mathrm{64}},\:\mathrm{1}\frac{\mathrm{43}}{\mathrm{64}},\:-\frac{\mathrm{17}}{\mathrm{64}}\right\} \\ $
Hope this acceptable...

