Prove that $x^3 = x^2$ has exactly two solutions

Question

I saw someone proving the uniqueness like this:

Suppose for a value $x=z$ such that $z\neq 0$ and $z\neq 1$, and $x^3=x^2$. Thus $z^3=z^2$. Since $z\neq 0$, multiplying both sides by $1/z^2$, we get $z=1$. Thus it contradicts with the assumption that $z\neq 1$. So we can conclude that $0$ and $1$ are the only possible values for $x$. $\square$

It obviously proves that $z=1$ is possible, but how does it prove that $z=0$ is possible? I feel like the fact that $z\neq 0$ and they divide by $z^2$ has something to do with it.
I understand that dividing by a variable excludes the case where it's value is $0$, which might discard correct answers. Does dividing by a variable always mean that one solution is where it's equal to $0$?

Technically the negation of $z\neq0 \land z\neq1$ is $z=0 \lor z=1$, which is true since we have proven $z=1$ is possible.

Answer 1

On one hand we know that $1^3=1^2$ and $0^3=0^2$ so $0$ and $1$ are solutions.

On the other hand $$x^3=x^2\implies x^3-x^2=0\implies x^2(x-1)=0$$ Therefore, if $x$ is a solution then $x$ must be either $0$ or $1$ and so those are the only two solutions.

Answer 2

It obviously proves that $z=1$ is possible, but how does it prove that $z=0$ is possible?

No, that is not what the proof you cite is proving. The proof is not saying that $z=1$ is possible, nor is it saying that $z=0$ is possible. In fact, the proof you cite in your second paragraph is the proof of the following statement:

If $x^3 = x^2$, then $x=0$ or $x=1$.

This statement is proven using contradiction, i.e. by assuming that $x\neq 0$ and $x\neq 1$ and $x^3=x^2$, and deriving a nonsensical statement out of the assumptions.

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Probably, the proof does not explicitly state that $0$ and $1$ are solution because verifying that is trivially simple. Nevertheless, I would write that in the beginning of the proof just to be sure.

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In fact, I would rewrite the proof significantly. For one, introducing two variables, $x$ and $z$, for the same thing, is just confusing. Second, the proof of contradiction is not really necessary and is just obfuscating the point. Finally, as mentioned, the proof really only says that there are at most two solutions, not that there are exactly $2$ solutions.

So this is how I would write it.

Because $0^3=0=0^2$ and $1^3=1=1^2$, it is clear that $0$ and $1$ are both solutions to $x^3=x^2$.

Now, we will prove that no other solution exists. We can prove this by showing that any solution to the equation is equal to either $0$ or $1$.

Let $x$ be some solution to the equation. Then, $x^3=x^2$, which means $x^3-x^2=0$, which further simplifies to $x^2(x-1)=0$. Because we know that the product of two numbers is $0$ if and only if one of the two numbers is $0$, this means that either $x^2=0$ or $x-1=0$. In the first case, we have $x\cdot x=0$ which is only possible if $x=0$. In the second case, $x-1=0$ implies $x=1$. So, if $x^3=x^2$, then $x=0$ or $x=1$, QED.

Answer 3

Well, we can do it like this: $$x^3=x^2$$ $$x^2\cdot x=x^2\cdot1$$

Now, dividing both sides by $x^2$, we get that $x=1$ is one solution. But how is $x=0$ also a solution?

Aha! We can't divide by $0$. This is why dividing with $x^2$ doesn't work if $x=0$ – that's dividing by zero, which leads to impossible situations like $1=2$ and others (as you probably know).

In case you already knew that...

Does dividing by a variable always mean that one solution is where it's equal to $0$?

In this case, yes! Both sides being divisible by $x$ means $0$ is one of the solutions – like a polynomial without a constant.

Answer 4

$\newcommand{\fitch}[2]{\begin{array}{|l}#1\\\hline#2\end{array}}$ The uniqueness part is just showing $\lnot\exists c\!\in\!\mathbb{R} (c^3 = c^2 \land c\neq 0 \land c\neq 1)$. This is just some sketch, but it captures how another solution to $x^3 = x^2$ that's not zero or one leads to a contradiction.

$\fitch{\exists c\!\in\!\mathbb{R} (c^3 = c^2 \land c\neq 0 \land c\neq 1)}{\text{Let } z\!\in\!\mathbb{R} \text{ such that } z^3 = z^2 \land z\neq 0 \land z\neq 1. \\ z^3 = z^2. \\ z^3 - z^2 = 0. \\ z^2 (z - 1) = 0. \\ z^2 = 0 \lor z -1 = 0. \\ \fitch{z^2 = 0}{z = 0. \\ \bot.} \\ \fitch{z - 1 = 0}{z = 1. \\ \bot.} \\ \bot.} \\ \lnot\exists c\!\in\!\mathbb{R} (c^3 = c^2 \land c\neq 0 \land c\neq 1).$