Real solution of the equation {x} = {x²} = {x³}

[juantheron]

Find all real solution of the equation $\{x\} = \{x^2\} = \{x^3\}$.

My Try: Let $\{x\} = \{x^2\} = \{x^3\} = k\;,$ where $k\in \mathbb{R}$ and $0\leq k<1$

Now we can write it as $x-\lfloor x \rfloor = x^2-\lfloor x^2 \rfloor = x^3-\lfloor x^2 \rfloor = k$

Now I did not Understand How can i proceed further.

Help me for solving above Question.

Thanks

 

[I like Serena]

Re: Real solution of the equation $\{x\} = \{x^2\} = \{x^3\}$

Find all real solution of the equation $\{x\} = \{x^2\} = \{x^3\}$.

My Try: Let $\{x\} = \{x^2\} = \{x^3\} = k\;,$ where $k\in \mathbb{R}$ and $0\leq k<1$

Now we can write it as $x-\lfloor x \rfloor = x^2-\lfloor x^2 \rfloor = x^3-\lfloor x^2 \rfloor = k$

Now I did not Understand How can i proceed further.

Help me for solving above Question.

Thanks

Let's define $x=i+f$ where $i$ is an integer and $0 \le f < 1$.

Then:
$$\{x\} = \{x^2\}$$
$$f = 2if + f^2$$
$$f^2 + f(2i-1) = 0$$
$$f = 0 \vee f = 1-2i$$
In other words, the only solutions occur when the fraction is 0.

... and all integers are solutions.

 

[Evgeny.Makarov]

Re: Real solution of the equation $\{x\} = \{x^2\} = \{x^3\}$

Let's define $x=i+f$ where $i$ is an integer and $0 \le f < 1$.

Then:
$$\{x\} = \{x^2\}$$
$$f = 2if + f^2$$

$\{x\} = \{x^2\}$ does not imply that $f^2 + f(2i-1) = 0$. I don't have suggestions at the moment, but there are many non-integer solutions.

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[I like Serena]

Re: Real solution of the equation $\{x\} = \{x^2\} = \{x^3\}$

$\{x\} = \{x^2\}$ does not imply that $f^2 + f(2i-1) = 0$. I don't have suggestions at the moment, but there are many non-integer solutions.

Ah. You're right.
I see my mistake - it's right in my first step.

A proper solution for $\{x\}=\{x^2\}$ follows from $1+x=x^2 \Rightarrow x^2-x-1=0 \Rightarrow x=\frac 1 2 (\sqrt 5 + 1) \approx 1.618$.
It's the golden number!

 

[Evgeny.Makarov]

It seems that $x^2=x+k$ for all $k\in\Bbb Z$ determines a solution of $\{x^2\}=\{x\}$, and solutions for $k=1,3,4,5,7,8,9,10,11,13,\dots$ are non-integer.

 

[kaliprasad]

It seems that $x^2=x+k$ for all $k\in\Bbb Z$ determines a solution of $\{x^2\}=\{x\}$, and solutions for $k=1,3,4,5,7,8,9,10,11,13,\dots$ are non-integer.

$x^2 = x + k$

then $k = x^2-x = x(x-1)$

so k can take all values and if it is n of the form n(n-1) then it is non - integer. n(n-1) shall give integers

as $n(n-1) + n = n^2$