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 irrational but then do I have to show more? $x+1=1$ which gives me $x=0$ and since $x$ cannot equal to $0$ as this would make the statement false (everything is $0$). Is it enough to simply state this falsity or is there another way to express it?


  • $\begingroup$ Unfortunately no. $\endgroup$ – OR. Oct 17 '13 at 1:25
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    $\begingroup$ You cannot say that $x^2=1$ or $x+1=1$ from that factorisation. That only works if the right hand side is zero. $\endgroup$ – Bennett Gardiner Oct 17 '13 at 1:28
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    $\begingroup$ Assume $p/q$ is a solution, with $p$ and $q$ relatively prime. Then $(p/q)^3+(p/q)^2-1=0$, i.e. $p^3+qp^2+q^3=0$. The first two terms are multiples of $p$ so the last, $q^3$, must be divisible by $p$. But since $p$ and $q$ are relatively prime then $p=\pm1$. Looking now at the last two terms we get that $q=\pm1$ too. We can check that $1^3+1^2-1\neq0$ and that $(-1)^3+(-1)^2-1\neq0$. $\endgroup$ – OR. Oct 17 '13 at 1:28
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    $\begingroup$ $\sqrt{1}$ is not irrational. $\endgroup$ – Michael Hardy Oct 17 '13 at 2:35
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    $\begingroup$ Your solution would make sense if $x\in\mathbb N$. $\endgroup$ – Lazar Ljubenović Oct 23 '13 at 20:01

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

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    $\begingroup$ What's with all the upvotes, friends? I barely did anything. This better not end up in my top answers. ;) $\endgroup$ – Bruno Joyal Oct 17 '13 at 1:41
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    $\begingroup$ The upvotes are for the simplicity. You can't get any better than a 1 sentence answer. $\endgroup$ – tylerc0816 Oct 17 '13 at 1:50
  • $\begingroup$ @tylerc0816 Fair enough. I'm not really complaining. ;) $\endgroup$ – Bruno Joyal Oct 17 '13 at 1:51
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    $\begingroup$ It’s really the only answer to be given, and you won the lottery by being first. Bask in it. $\endgroup$ – Lubin Oct 17 '13 at 1:54
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    $\begingroup$ I'm +1ing it just so it will be in your top answers. =p J/K - it really is a strait forward, easy to understand, and complete answer (with a link to external sources and everything!) $\endgroup$ – EtherDragon Oct 17 '13 at 17:41

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. $$ \mbox{Then,}\quad x \not\in {\mathbb Q} $$



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.


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