Proof of a member of a set, algebra

Question

Being $x\in \Bbb Z$ I'm trying to find the elements $x$ of $S$, such that, $S=\{x\ |\ \frac{x}{x-1}\in \Bbb Z\} $, until now the only member I can find is $x=2$ and $x=0$, is there a way I can prove these are the only cases? Thanks! (:

Answer 1

If $x-1 \neq \pm 1$, then $x-1 \not \mid x$ as they are coprime. Hence those are the only two solutions.

Answer 2

For $\frac{x}{x-1}$ to be an integer, it is both necessary and sufficient that $x-1$ divide $x$. But this means that $x-1$ must divide $1$ (if this isn't clear, set $y = x-1$). This happens exactly when $x - 1 \in \{ 1, -1 \}$.

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Another way to see it:

$$ \frac{x}{x-1} = \frac{(x-1)+1}{x-1} = 1 + \frac{1}{x-1} $$

This is an integer iff $\frac{1}{x-1}$ is, i.e., if $x-1$ is $\pm 1$.

Answer 3

For x/(x-1) to be an integer, x-1 is a factor of x.

Since 2 is the least factor (except for one) for all possible factors of a number, we have the following two cases.

1. If x≥0, x-1 ≤ x/2, then x≤2
2. If x<0, x-1 ≥ x/2, then x≥2

Therefore, x=2.

Now, suppose one is the factor of x. |x-1|=1, x=0 or 2. #