How prove this the equation $\{x^3\}+\{y^3\}=\{z^3\}$has infinitely many rational non-integers solutions, Show that the equation
$$\{x^3\}+\{y^3\}=\{z^3\}$$
has infinitely many rational non-integers solutions,Here,$\{a\}$ denotes the fractional part of $a$
I have solve this follow two problem
the equation  1:
$$\{x\}+\{y\}=\{z\}$$ 
has infinitely many rational non-integers solutions.
I take $$x=n+0.2,y=n+0.3,z=n+0.5,n\in N^{+}$$
the equation 2:
$$\{x^2\}+\{y^2\}=\{z^2\}$$ 
has infinitely many rational non-integers solutions.
then I take
$$x=10n+0.3,y=10n+0.4,z=10n+0.5$$
because
$$x^2=100n^2+6n+0.09,y^2=100n^2+8n+0.16,z^2=100n^2+10n+0.25$$
$$\Longrightarrow \{x^2\}=0.09,\{y^2\}=0.16,\{z^2\}=0.25$$
$$\Longrightarrow \{x^2\}+\{y^2\}=\{z^2\}$$
But for
the equation
$$\{x^3\}+\{y^3\}=\{z^3\}$$
has infinitely many rational non-integers solutions
I can't.Thank you,
and  I gues  this follow problem maybe is true.and maybe can prove it?
the equation
$$\{x^4\}+\{y^4\}=\{z^4\}$$
has infinitely many rational non-integers solutions?
the equation
$$\{x^5\}+\{y^5\}=\{z^5\}$$
has infinitely many rational non-integers solutions?
and so on 

For any $n > 2$, pick any $m > 1$ such that $\gcd(m,n) = 1$. Notice
$$\gcd(m,n) = 1 \implies \gcd(m^n,n) = 1$$
We can define a number $\lambda$ by 
$$\lambda = \text{mod}( n^{\varphi(m^n)-1}, m^n )$$
where $\varphi(x)$ is the Euler's totient function and $\lambda$ will satisfy $$\lambda n = 1 \pmod{m^n}$$ 
Let $k \in \mathbb{Z}_{+} \text{ s.t. } \lambda n = k m^n + 1$, we have:
$$\begin{align}
\left(\frac{1}{m^2} + \lambda m^{n-2}\right)^n
= & \frac{1}{m^{2n}} 
+ n \lambda \frac{m^{n-2}}{m^{2(n-1)}}
+ \underbrace{\binom{n}{2}\lambda^2 \frac{m^{2(n-2)}}{m^{2(n-2)}} + \cdots}_{\in \mathbb{Z}}\\
= & \frac{1}{m^{2n}} + \frac{1}{m^n} + \underbrace{k + \binom{n}{2}\lambda^2 \frac{m^{2(n-2)}}{m^{2(n-2)}} + \cdots}_{\in \mathbb{Z}}
\end{align}$$
This implies
$$\left\{\frac{1}{m^{2n}}\right\} +
\left\{\frac{1}{m^{n}}\right\}
= \left\{\left(\frac{1}{m^2} + \lambda m^{n-2}\right)^n\right\}
$$ 
For example, when $n = 3$, we can take $m = 2$,
$$\lambda = 3\quad\longrightarrow\quad
\left\{\frac{1}{4^3}\right\} + \left\{\frac{1}{2^3}\right\} = \left\{\left(\frac{25}{4}\right)^3\right\}$$
When $n = 4$, we can take $m = 2$,
$$\lambda = 61\quad\longrightarrow\quad
\left\{\frac{1}{9^4}\right\} + \left\{\frac{1}{3^4}\right\} = \left\{\left(\frac{4942}{9}\right)^4\right\}$$
When $n = 5$, we can take $m = 2$,
$$\lambda = 13\quad\longrightarrow\quad
\left\{\frac{1}{4^5}\right\} + \left\{\frac{1}{2^5}\right\} = \left\{\left(\frac{417}{4}\right)^5\right\}$$
Since for any $n > 2$, there are infinitely many $m$ relative prime to it. This implies there are infinitely many non-integral rational solutions for $\{ x^n \} + \{ y^n \} = \{ z^n \}$.
