TITLE: For which $\lambda \in \mathbb{N}$ the equation $x^3 - \lambda x - 2 =0$ has only rational roots?
QUESTION [2 upvotes]: 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.

REPLY [1 votes]: 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.
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