For which $\lambda \in \mathbb{N}$ the equation $x^3 - \lambda x - 2 =0$ has only rational roots?

Question

For which $\lambda \in \mathbb{N}$ the equation $x^3 - \lambda x - 2 =0$ has only rational roots?

My attempt: I start to plug values for $\lambda$ but I only find that $\lambda = 3$ works, but is there any more? I really don't know any method that can solve this problem.

Answer 1

By the Rational Root Theorem, if $p/q$ is a rational root of $x^3 - \lambda x - 2\in \mathbb{Z}[x]$ with $\gcd(p,q)=1$ then $p$ divides $-2$ and $q$ divides $1$, that is $p/q\in\{1,-1,2,-2\}$. Now use these numbers to find the possible values for $\lambda$ and check each case.

Can you take it from here? What may we conclude?

Answer 2

Here is a wholly elementary approach.

Rewrite the equation $x(x^2-\lambda)=2$ and let $x=\frac ab$ in lowest terms so that $$a(a^2-\lambda b^2)=2b^3$$

Well if $b\gt 1$ it can't be a factor of either factor on the left (it is co-prime with $a$) - so we have $b=1, a=x$.

So $x(x^2-\lambda)=2$ is an equation in integers, and the only possible integer factors of $2$ are $\pm 1, \pm 2$ so $x$ has to be one of these.

Then $\lambda=\cfrac {x^3-2}{x}$

We obtain $(x,\lambda)=(-2,5); (-1,3); (1,-1); (2,3)$

$\lambda =-1$ is not allowed (but wouldn't work anyway)

For $\lambda =5$ we just have the single root $x=-2$, and this cannot be a triple root because the product of the roots of the cubic must be $2$ and not $-8$.

For $\lambda = 3$ we have the two roots $x=-1, x=2$ and from the sum of the roots (or product of roots) we have the three roots $-1, -1, 2$. So this is the only solution.