Edit : The proof of what I'm asking can be found in Anthony Knapp's book Elliptic curves page 85.
Husemöller's book : On page 38 of the second edition and on page 37 of the first, you can find the following theorem
(4.1) Theorem : Let $E$ be an elliptic curve defined over a field $k$ by the equation $$y^2 = (x-\alpha)(x-\beta)(x-\gamma) = x^3 + ax^2 +bx + c \ \text{with} \ \alpha, \beta \ \text{and} \ \gamma \in k.$$ For $(x',y') \in E(k)$ there exist $(x,y) \in E(k)$ with $2(x,y)=(x',y')$ if and only if $x'-\alpha$, $x'-\beta$ and $x'-\gamma$ are squares in k.
For the proof, I understand it as "if $-\alpha$, $-\beta$ and $-\gamma$ are squares, then there exists $(x,y)$ such that $2(x,y) = (0,y')$". But I'm unable to see the other way around. That is, if there exists such an $(x,y)$, then $-\alpha$, $-\beta$ and $-\gamma$ should all be squares.
I'd like to point out that in the course of the proof, he rewrites $(x'-\alpha)$ as $\alpha$ etc... when looking for a solution to $2(x,y)=(0,y')$ on the translated curve.
I've tried "the sum of three square roots is not a rational" but $\sqrt{8} + \sqrt{18} -\sqrt{50} = 0$. I've tried looking at the four $\lambda$'s at the end. In particular using $\lambda_{1}$ and $\lambda_{2}$, summing them and to get a contradiction. But I have no way of knowing that both are rational (belong to the field $k$). I just know that at least one is rational as the existence of a rational $(x,y)$ implies a rational $\lambda$ in the duplication formula.
This is driving me crazy, please help.