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

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?

Thanks!

• Unfortunately no. – OR. Oct 17 '13 at 1:25
• You cannot say that $x^2=1$ or $x+1=1$ from that factorisation. That only works if the right hand side is zero. – Bennett Gardiner Oct 17 '13 at 1:28
• 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 $(-1)^3+(-1)^2-1\neq0$. – OR. Oct 17 '13 at 1:28
• $\sqrt{1}$ is not irrational. – Michael Hardy Oct 17 '13 at 2:35
• Your solution would make sense if $x\in\mathbb N$. – 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.

• What's with all the upvotes, friends? I barely did anything. This better not end up in my top answers. ;) – Bruno Joyal Oct 17 '13 at 1:41
• The upvotes are for the simplicity. You can't get any better than a 1 sentence answer. – tylerc0816 Oct 17 '13 at 1:50
• @tylerc0816 Fair enough. I'm not really complaining. ;) – Bruno Joyal Oct 17 '13 at 1:51
• It’s really the only answer to be given, and you won the lottery by being first. Bask in it. – Lubin Oct 17 '13 at 1:54
• 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!) – 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. $$\mbox{Then,}\quad x \not\in {\mathbb Q}$$

## 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 $A$).

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.