An irreducible root system is <simply laced root system>[simply laced] when every root has the same length, equivalently when its <Dynkin diagram> has no multiple edge.

If all roots have the same length, the two Cartan integers for $\alpha$ and $\beta$ are equal. The <root-system finiteness lemma> then makes their product either zero or one, so
$$
\langle\alpha,\beta^\vee\rangle\in\{0,\pm1\}
$$
for $\alpha\ne\pm\beta$.

Conversely, when all such Cartan integers lie in $\{0,\pm1\}$, any two nonorthogonal roots have Cartan integers of absolute value one in both directions. Their squared lengths are therefore equal. Irreducibility makes the graph joining nonorthogonal roots connected, so all roots have the same length. Thus the system is simply laced.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 102 / 4 / iii / Solution

The crystallographic axiom makes
$$
m=\langle\alpha,\beta^\vee\rangle,
\qquad
n=\langle\beta,\alpha^\vee\rangle
$$
integers. If $\theta$ is the angle between the roots, then
$$
mn=
\frac{2(\alpha,\beta)}{(\beta,\beta)}
\frac{2(\beta,\alpha)}{(\alpha,\alpha)}
=4\cos^2\theta.
$$
This is a nonnegative integer. Since $\beta\ne\pm\alpha$, the roots are not parallel, so $\cos^2\theta<1$. Therefore
$$
\boxed{mn\in\{0,1,2,3\}},
$$
which is the <root-system finiteness lemma>.

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 102 / 4 / i / Solution

For roots $\alpha\ne\pm\beta$,
$$
\langle\alpha,\beta^\vee\rangle
\langle\beta,\alpha^\vee\rangle
=4\cos^2\theta\in\{0,1,2,3\},
$$
where $\theta$ is their angle. Integrality gives an integer, and nonparallelism makes it strictly smaller than four.


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 Incoming links: Root-system finiteness lemma