Prove $x^3 + x +1 = 0$ has no rational solutions.

Question

Use a suitable method to prove that there is no rational solutions to the equation: $$x^3 + x + 1 = 0$$ or give a counter-example to show otherwise.

I tried to do this question via polynomial factorization and contradiction.

Let $$f(x) = x^3 + x + 1 = 0$$

By polynomial factorisation by assuming there is a root $r$,

$$f(x) = (x-r)(x^2 + xr + r^2 + 1)$$

$$f(x) = (x^3 + rx^2 + xr^2 + x - rx^2 - xr^2 - r^3 - r)$$

$$f(x) = (x^3 + x - r^3 - r)$$

Comparing the coefficients:

$$r^3 - r = 1$$

From this, we can assume that root $r$ exists, and hence, we have proven by contradiction that a root exists.

I want to ask if my proof is correct.

Answer 1

(I do think the question is interesting: not because it's particularly insightful, but because it has a flaw commonly committed by newcomers)

Your proof is not correct, for two reasons:

• A minor computational error, that gives the illusion that you found out something.
• A major logical error.

Let's start with the minor one:

From $f(x) = x^3+x+1=(x-r)(x^2 + xr + r^2 + 1)=x^3 + x - r^3 - r$, you write that, by comparing coefficients, you get $r^3 - r = 1$. It's wrong.

Actually, by comparing coefficients, you have

$$x^3+x+1=x^3+x-r^3-r$$ $$1=-r^3-r$$ $$r^3+r+1=0$$

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Now, the major problem. You would like to find a proof by contradiction. Initially, you want to prove that the polynomial $x^3+x+1$ has no rational root. That is, any complex number $r$ such that $r^3+r+1=0$ satisfies $r\notin \Bbb Q$.

To find a proof by contradiction, you would have to assume $r$ is a root, and $r$ is rational, and find a contradiction.

Here, you only assumed that $r$ is a root. There is no contradiction to be found, as any polynomial with complex coefficients and degree $n>0$ has exactly $n$ complex roots. It has roots, hence there will be no contradiction assuming it has a root.

With your assumption, and after correcting the little computational mistake, you find out that $r^3+r+1=0$. But that's the assumption! You assumed $r$ is a root of $x^3+x+1$, i.e. exactly that $r^3+r+1=0$, and you find out that the same equality is true.

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So your proof by contradiction in this current state is moot. It's still possible to do it though, using the idea behind the more general rational root theorem.

Assume $r$ is a rational root of $x^3+x+1$. That is, there are coprime integers $a$ and $b$ such that $r=a/b$ and $r^3+r+1=0$. The important fact, here, is that a rational number is the quotient of two coprime integers.

Then:

$$\frac{a^3}{b^3}+\frac ab+1=0$$

$$a^3+ab^2+b^3=0$$

Now, we have easily a contradiction: since the sum is zero and two of the three terms are divisible by $b$, then the third must also be divisible by $b$. However, per our assumtpions, it's coprime to $b$.