Proof that there is no rational solutions to the equation $x^3+2x-1=0$

Proof by contradiction: Assuming that there is a rational solution to the equation $x^3+2x-1=0$.

Let $x=a/b$ where $a$ and $b$ are coprime with $b$ not equal to zero.

Performing a substitution into the equation, it simplifies to $a^3+2ab^2-b^3=0$.

Three cases to consider (since $a$ and $b$ are coprime so they can't be both even):

Case 1: $a$ is even and $b$ is odd, then

$a^3$ is even
$2ab$ is even
$b^3$ is odd

So the LHS of the equation is odd and the RHS of the equation is even. Therefore, there is a contradiction.

Case 2: $a$ is odd and $b$ is even, then

$a^3$ is odd
$2ab$ is even
$b^3$ is even

So the LHS of the equation is odd and the RHS of the equation is even. Therefore, there is a contradiction.

Case 3: $a$ and $b$ are odd, then

$a^3$ is odd
$2ab$ is even
$b^3$ is odd

So the LHS of the equation is even and the RHS of the equation is even.

So does that means that there is a rational solution when $a$ and $b$ are odd? I am stuck with this. Can someone help me out? When we perform proof by contradiction, do we have to perform it for all cases? Thanks in advanced!

• You've shown that if there is a rational root, then the numerator and denominator of that root are both odd. Nothing more. – Patrick Stevens Feb 1 '18 at 11:07

By the rational root theorem, only $\pm1$ could possibly be rational roots of your polynomial, but they aren't. Therefore, it has no rational roots.

• Thanks Jose but I am not quite sure how the theorem works. This is our first lesson on proof by contradiction in highschool. – user450003 Feb 1 '18 at 11:28
• @user450003 Did you read the statement of the theorem? – José Carlos Santos Feb 1 '18 at 11:29
• Yes I did and I am still confused. – user450003 Feb 1 '18 at 11:55
• @user450003 What's the problem? It says that if there was a rational root $\frac ab$, with $a\in\mathbb Z$ and $b\in\mathbb N$, then $a\mid-1$ and $b\mid 1$, which means that $\frac ab=\pm1$. – José Carlos Santos Feb 1 '18 at 11:59
• I don't understand why a∣−1 and b∣1? – user450003 Feb 1 '18 at 12:23

Much easier using the following theorem: if $f(x) = a_nx^n + \cdots a_1 x + a_0\in\Bbb Z[x]$, then $$f(p/q) = 0 \hbox{ (p/q irreducible)}\implies p\vert a_0,q\vert a_n.$$

Rewrite the equation in the form $$x^3+2x=1$$. Suppose there is a rational solution $$m/n$$ with $$\gcd(m,n)=1$$. Such a solution can be chosen with $$n$$ positive integer. Then we get $$\frac{m^3}{n^3} +2\frac{m}{n} =1.$$

Multiplying by $$n^3$$ we get, $$m^3+2mn^2= n^3;\quad \mbox{equivalently}\quad m(m^2+2n^2) =n^3\qquad(*)$$

This shows $$m$$ is a factor of $$n^3$$. The hypothesis $$\gcd(m,n)=1$$ forces to conclude $$m=1$$. Now putting $$m=1$$, in equation $$(*)$$ we get $$1+2n^2=n^3$$ which can be rewritten as $$n^2(n-2)=1$$. This last equation contradicts the supposition that $$n$$ is a positive integer. So there are no rational roots to your equation.

$$a^3+2ab^2-b^3=0$$ implies $$a^3=(-2ab+b^2)b$$ and so $$b$$ divides $$a^3$$. Since $$a$$ and $$b$$ are coprime, we must have $$b=\pm 1$$. Then $$a(a^2+2)=\pm 1$$, which cannot happen because $$a^2+2\ge2$$ cannot divide $$1$$.

• Thanks for the help but I don't understand why b must be plus or minus 1. – user450003 Feb 1 '18 at 11:29
• @user450003, if $p$ is a prime dividing $b$, then $p$ divides $a$, but this cannot happen because $a$ and $b$ are coprime. – lhf Feb 1 '18 at 11:31
• @lhf Why didn't you look at the case $b = -1$? – john Sep 12 '19 at 16:43
• @john, fixed and simplified. Thanks for the push. – lhf Sep 12 '19 at 16:54