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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

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1  
If you have found one solution, then you have found infintely many solution. This is because for any $x = \frac{m}{n} \in \mathbb{Q}$ and $k \in \mathbb{Z}$, $\{ x^3 \} = \{ (kn^2 + x)^3 \}$. –  achille hui Jan 24 at 9:13
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Yes,But Now  I don't have found one solution,Thank you –  math110 Jan 24 at 9:19
    
@achille: But those are essentially the same solution. It also makes sense to ask for solutions that have different values of $\{x^3\},\{y^3\},$ and $\{z^3\}$. –  TonyK Jan 24 at 10:12
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By brute force, looks like there are ton of solutions.$$\begin{array}{rcl} \left\{(33/4)^3\right\} + \left\{(1/2)^3\right\} &=& \left\{(57/4)^3\right\}\\ \left\{(23/4)^4\right\} + \left\{(1/2)^4\right\} &=& \left\{(11/4)^4\right\}\\ \left\{(623/4)^5\right\} + \left\{(1/2)^5\right\}&=& \left\{(15/4)^5\right\}\\ \left\{(647/4)^6\right\} + \left\{(1/2)^6\right\}&=& \left\{(39/4)^6\right\}\\ \left\{(485/4)^7\right\} + \left\{(1/2)^7\right\} &=& \left\{(357/4)^7\right\}\\ \left\{(21/4)^8\right\} + \left\{(1/2)^8\right\} &=& \left\{(181/4)^8\right\}\\ \end{array}$$ –  achille hui Jan 24 at 12:13
    
$x^3=\Big([x]+\{x\}\Big)^3=\underbrace{[x]^3}_\text{Int.}+\{x\}^3+3[x]\{x\}\Big(‌​[x]+\{x\}\Big)\iff\{x^3\}=\bigg\{\{x\}^3+3x[x]\{x\}\bigg\}$ –  Lucian Jan 24 at 16:28

2 Answers 2

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 $\{ x^n \} + \{ y^n \} = \{ z^n \}$.

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Hint $$\{80.8^3\}+\{36.9^3\}=\{24.1^3\}$$

$$\{8.8^3\}+\{3.9^3\}=\{3.1^3\}$$

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1  
$LHS = 1.001 \ne 0.001 = RHS$. –  achille hui Jan 24 at 9:55
    
@RHS, Sorry, I made some error, this is fixed now. –  Xoff Jan 24 at 10:06
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$0.269 + 0.552 \ne 0.981$. –  TonyK Jan 24 at 10:09
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Wonder who is @RHS... ;) –  Balarka Sen Jan 24 at 10:09
    
Sorry, sorry, I have two left hands this morning !!! Shame on me :) –  Xoff Jan 24 at 10:12

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