Prove that $x^3 + x^2 = 1$ has no rational solutions?

Question

Is this enough for a proof?:

$$x^3+x^2 = 1$$

I would factor and get: $x^2(x+1) = 1$

I would show that $x = \sqrt1$, which is rational but then what else would I have to show? $x+1=1$ which gives me $x=0$ and since $x$ cannot equal to $0$ as this would make the statement false ($0$ times anything is $0$). Is it enough to simply state this falsity or is there another way to express it?

Thanks!

Answer 1

By the rational root theorem, a rational root would have to be $x=1$ or $x=-1$, but neither works.

Answer 2

Let's assume $x = p/q$. $p$ and $q$ integers without a common factor. Then, $$ p^{3} + p^{2}q = q^{3} $$

It's is only satisfied whenever $p$ and $q$ are simultaneously even. It contradicts the initial hypothesis that we can set $x = p/q$ where $p$ and $q$ has not common factors. $$ \mbox{Then,}\quad x \not\in {\mathbb Q} $$

Answer 3

Theorem

The solution satisfying the following equation $$ A \times B =0 $$ is $A=0$ (for any $B$) or $B=0$ (for any $A$).

You cannot apply the same pattern for the case in which the right hand side is not zero. Why? For example, $$ A\times B = 2 $$ If you choose $A=2$ then $B$ must be $1$ (rather than for any $B$). If you choose $B=2$ then $A$ must be $1$ (rather than for any $A$).

Back to your question

If you want to find the solution of $$x^2(x+1) =1$$ you have to make sure the right hand side equals to 0.

\begin{align*} x^2(x+1) &=1\\ x^3 + x^2 -1 &=0 \end{align*}

To prove the equation has no rational solution see this comment.