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

Question

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!

Answer 1

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.

Answer 2

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

Answer 3

$a^3+2ab^2-b^3=0$ 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 $1$.

Answer 4

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.