Why are there at most two root lengths for a finite, simple, complex lie algebra? I know it is from the constraint that the $2(\alpha,\beta)/(\alpha,\alpha)$ is integer, but what is the argument?
Also, if $\alpha$ is a root, $k\alpha$ is also a root only when k is 1 or -1? How to prove this result?
1.)The length of a root $\alpha$ is given by $\sqrt{(\alpha,\alpha)}$. The condition $2(\alpha,\beta)/(\alpha,\alpha)\in \mathbb{Z}$ enforces that the angle $\theta$ between two roots $\alpha$ and $\beta$ is one of the following seven possibilities: $\frac{\pi}{2}, \frac{\pi}{3},\frac{2\pi}{3},\frac{\pi}{4},\frac{3\pi}{4}, \frac{\pi}{6},\frac{5\pi}{6}$. To see this study the equation $(\alpha,\beta)=\sqrt{(\alpha,\alpha)}\sqrt{(\beta,\beta)}\cos (\theta)$, so that $$ 4\cos^2(\theta)=2\frac{(\alpha,\beta)}{(\alpha,\alpha)}\cdot 2\frac{(\alpha,\beta)}{(\beta,\beta)}\in \{0,1,2,3\}, $$ because $0\le \cos^2(\theta)\le 4$, and $\alpha\neq \pm \beta$. Now for $\theta=\frac{\pi}{3},\frac{2\pi}{3}$ we see that the roots have equal length, and for the other angles there are two possible length. For $\theta=\frac{\pi}{4},\frac{3\pi}{4}$ the different roots have length proportional to $\sqrt{2}$, and for $\theta=\frac{\pi}{6},\frac{5\pi}{6}$ the factor is $\sqrt{3}$.
2.) By definition of an abstract root system, twice a root is not a root. So let $\beta=k\alpha$. We may assume that $|k|<1$, if not already $k=\pm 1$. By definition, $2\langle \beta,\alpha\rangle /\langle \alpha,\alpha\rangle$ is integral (this is again an axiom). This is only possible for $k=0$ or $k=\pm \frac{1}{2}$. The latter is impossible, because twice a root is not a root. It follows $k=0$, a contradiction. It follows that $k=\pm 1$.
If you are not assuming that the root system of a simple complex Lie algebra satisfies the axioms of an abstract root system, you have to show that the weight spaces $L_{\alpha}$ are $1$-dimensional, and $\dim L_{k\alpha}=0$ for all $k\ge 2$. This will use the Killing form of $L$, and a Diophantine argument for a dimension equation over $\mathbb{Z}$ concerning the weight spaces.