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This was an exam question that I got totally wrong and am a bit question. Prove $x^3 + x + 1 = 0$ has no solutions. Prove by contradiction.

Assume: $x^3 +x +1 =0$ has at least one rational root.

So, what I attempted to do here was solve for $x$.

$\begin{align} x^3 +x &= -1\\ x(x^2 +1) &= -1\\ x= -1 &\vee x = \sqrt {-2}\end{align}$

I noticed in other answers I found online that the way to do this is to get to $a^3 + ab^2 + b^3 = 0$ and then go through each scenario for $a$ and $b$ (odd or even.) And this makes much more sense in retrospect. But can someone explain why solving for $x$ is not an appropriate starting point?

Sreekanth Karumanaghat
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Marla
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5 Answers5

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As an odd-degree polynomial, $x^3+x+1=0$ certainly has real solutions, but none that are rational. Applying the rational root theorem shows us that the only possible rational roots are $\pm1$, and substitution shows us that neither of these works.

It's worth noting that you can't conclude from $a\cdot b=-1$ that $a=-1$ or $b=-1$. For example, let $a=2,b=-\frac12$. The only time that we can conclude (for real numbers $a,b,c$) that $ab=c$ implies $a=c$ or $b=c$ is when $c=0$.

In particular, the only real solution to $x(x^2+1)=-1$ (or equivalently, $x^3+x+1=0$) is $$x=\sqrt[3]{\frac{\sqrt{93}-9}{18}}-\sqrt[3]{\frac{\sqrt{93}+9}{18}},$$ which isn't remotely obvious by your method. Who would even guess that? We can find this through Cardano's method.

sds
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Cameron Buie
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  • Right, I came up with the two solutions sqrt (-2) and -1. But isn't (-1) a false solution because (-1)^3 + (-1) +1 does not equal 0? – Marla Apr 11 '13 at 18:02
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    And the scarier thing is that (without the prior work showing us so) it isn't even remotely obvious that that solution is an irrational number, since Cardano's method gives us similarly scary formulas even when the solution is an integer! –  Apr 11 '13 at 18:14
  • @Marla: Yes, $-1$ is a false solution, as is any rational number. Even knowing that the real root has the above form doesn't make it clear that it's irrational, though, as Hurkyl points out. We pretty much have to use the rational root theorem or do a proof by contradiction. – Cameron Buie Apr 11 '13 at 18:14
  • @Hurkyl: Excellent point. – Cameron Buie Apr 11 '13 at 18:16
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    @Marla: As an example of one of those scary formulas from Cardano's method: $$4=\sqrt[3]{26+15\sqrt3}+\sqrt[3]{26-15\sqrt3}.$$ Crazy, huh? – Cameron Buie Apr 26 '13 at 22:57
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Just because $a\,b=-1$, you can't conclude that either $a$ or $b$ must equal $-1$.

Eckhard
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No solution or no rational solution? All odd degree polynomial have at least one real root. Solving for $x$ would be fine, but your solution is wrong. From $x(x^2+1)=-1$ you can't conclude $x=-1$ or $x=\sqrt{-2}$

Ross Millikan
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A few things to point out. First, if you know the rational root theorem, then it is easy to see that $\pm 1$ are the only possible candidates for rational roots.

That being said, it seems you are very confused by the procedure we usually use to separate out the roots of a polynomial.

Normally we take a polynomial and attempt to factor it. For example $$0=x^2 + 7x + 12 = (x+3)(x+4)$$ Now we are able to conclude that either $$(x+3)=0\ \ \text{or}\ \ \ (x+4)=0$$ This is not an arbitrary step, but rather it comes from the fact that when you're working with real numbers, if $ab=0$ then either $a=0$ or $b=0$. More formally, $\mathbb{R}$ has no zero divisors because it is a field.

To me, it looks like what you've tried to do is to pull the same trick where the right hand side is not zero, i.e. you tried to conclude something analagous to $$(x+3)(x+4)=1 \implies (x+3)=1\ \ \text{or}\ \ \ (x+4)=1$$ That is simply untrue. There is nothing which says that if $ab=c$ for $c\neq 0$ that $a=c$ or $b=c$.

EuYu
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The problem is that $ab = -1$ doesn't imply "$a = -1$ or "b = -1". If you knew $a,b$ were integers, you could infer that "$a=1$ and $b=-1$ or $a=-1$ and $b=1$" because $a$ must be a divisor of $-1$ and there are only two possibilities, but that's not the case here.

Your initial idea can be salvaged though; if you suppose $a/b$ is a rational root with $\gcd(a,b) = 1$, then your approach to the problem is to rewrite it as

$$ a (a^2 + b^2) = -b^3 $$

Now, both sides are integers and we can do something useful; we know that $a$ must be a divisor of $-b^3$; given our hypothesis there are only two possibilities for $a$. You can continue to make similar divisibility arguments until you have excluded all possibilities.