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 
 A: 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 $\{ x^n \} + \{ y^n \} = \{ z^n \}$.
A: Hint
$$\{80.8^3\}+\{36.9^3\}=\{24.1^3\}$$
$$\{8.8^3\}+\{3.9^3\}=\{3.1^3\}$$
A: Consider any prime $p$ such that  $p \equiv_3 2 \\$.
Claim: Every number in $\mathbb{F}_{p^3 \setminus pZ}$ is a cubic residue.
Proof: Consider primitive root $g$ modulo $p^3$,by its definition it generates all of  $\mathbb{F}_{p^3 \setminus pZ}$.Let $h=g^3$ as$(3,\phi(p^3))=1$ $\implies$ $h$ is also a primitive root  and hence as all elements of the above set can be written as its powers we reach the conclusion.(all equalities in the above proof should be regarded modulo $p$)
Now pick three numbers $a,b,c \in \mathbb{F}_{p^3 \setminus pZ}$  such that $a=b+c$ (this should be regarded as an equal sign in $N$ rather than in the mentioned field).By above Claim there $\exists a_1,b_1,c_1 \in \mathbb{N}$ such that $a_1^3\equiv_{p^3} a$ and anologous for $b,c$.We have :
$$\left \{\frac{a_1^3}{p^3}\right \}=\frac{a}{p^3}$$
From here the conclusion follows.
