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Calculate real values of $x$ in $\{x\} = \{x^2\} = \{x^3\}$ , where $\{x\}$ is the fractional part of $x$.

My Attempt:

Let $\{x\} = \{x^2\} = \{x^3\} = k$. Because the fractional part of $X$ is given by $\{X\} = X-\lfloor X \rfloor$, we know the following to be true:

$$ 0\leq \{x\} <1\\ 0\leq \{x^2\} <1\\ 0\leq \{x^3\} <1\\ $$

Using the definition of the fractional part, our equation becomes

$$ x-\lfloor x \rfloor = x^2-\lfloor x^2 \rfloor = x^3-\lfloor x^3 \rfloor = k $$

I'm not sure how to proceed from here.

Fabrosi
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juantheron
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1 Answers1

8

For any $x$ such $\{x\} = \{x^2\}$, we have

$$x^2 - x = \lfloor x^2\rfloor + \{x^2\} - \lfloor x\rfloor - \{x\} = \lfloor x^2\rfloor - \lfloor x\rfloor \in \mathbb{Z}.$$

Likewise $x^3 - x \in \mathbb{Z}$ and $x^3 - x^2 \in \mathbb{Z}$. If $x^2 - x \neq 0$ (that is, $x \neq 0, 1$) then

$$x = \frac{x^3 - x^2}{x^2 - x} \in \mathbb{Q}.$$

Let $x = \frac{p}{q}$ where $(p, q) = 1$. Then $x^2 - x = \frac{p^2 - pq}{q^2} = \frac{p(p-q)}{q^2}$. As $x^2 - x \in \mathbb{Z}$, $q^2 \mid p(p-q)$, but $(p, q) = 1$, so $q^2 \mid p - q$. We can therefore write $p$ in the form $p = kq^2 + q$ where $k \in \mathbb{Z}\setminus\{0\}$. It follows that $(p, q) = q$, so we must have $q = 1$; hence $x \in \mathbb{Z}\setminus\{0, 1\}$.

Checking $x = 0$ and $x = 1$ separately, we find they are also solutions.

Therefore, $\{x\} = \{x^2\} = \{x^3\}$ if and only if $x$ is an integer.

Michael Albanese
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