= Solution
A reduced crystallographic <root system> in a real <Euclidean space> is a finite spanning subset $\Phi$ not containing zero, with $\Phi\cap\mathbb R\alpha=\{\alpha,-\alpha\}$, closed under $s_\alpha(v)=v-2(v,\alpha)\alpha/(\alpha,\alpha)$, and satisfying $2(\beta,\alpha)/(\alpha,\alpha)\in\mathbb Z$ for all roots. A <base of a root system> is a vector-space <basis> $\Delta\subseteq\Phi$ such that every root has <integer> coefficients of one sign in that <basis>.
Choose $h$ not orthogonal to any root and call roots with $(h,\alpha)>0$ positive. Let $\Delta$ be the positive roots that cannot be written as a sum of two positive roots. Repeated decomposition terminates because each summand has strictly smaller positive $h$-value in a finite set. Thus every positive root is a nonnegative <integer> combination of $\Delta$.
For distinct $\alpha,\beta\in\Delta$, $(\alpha,\beta)\le0$: otherwise the allowed difference-root lemma would make either $\alpha-\beta$ or $\beta-\alpha$ positive, decomposing one of them. In a real <linear relation> among $\Delta$, move negative coefficients to the other side, obtaining $u=v$ as positive combinations with disjoint supports. Then $\|u\|^2=(u,v)\le0$, so $u=0$. But a nonempty positive combination has $(h,u)>0$, a contradiction. Thus the elements are <linearly independent>. They span every root and hence $E$, proving existence of a <base of a root system>. The difference-root lemma is used only for distinct roots; that qualification excludes the zero difference.
For the displayed set, there are eight coordinate roots, twenty-four roots with two nonzero unit entries, and sixteen half-coordinate roots: forty-eight in total. It spans $\mathbb R^4$, is reduced, and has squared root lengths one and two. Signed coordinate <permutations> preserve it. <Reflections> in coordinate roots are sign changes, and <reflections> in roots $e_i\pm e_j$ are signed coordinate <permutations>. Every half-coordinate root is obtained by a signed <permutation> from $h=(1/2,1/2,1/2,1/2)$, so it remains to check its <reflection>
$$
s_h(v)=v-\frac{\sum_i v_i}{2}(1,1,1,1).
$$
It sends $e_i$ to $e_i-h$, a half-coordinate root; it fixes $e_i-e_j$ and sends $e_i+e_j$ to the negative sum of the other two coordinate <vectors>. For a half-coordinate root, the sum of its four signs is $0,\pm2$ or $\pm4$; the formula sends it, respectively, to itself, a coordinate root, or its negative when parallel to $h$. Thus <reflection> closure holds. For a long root, the required Cartan <integer> is an <integer> dot product; for a coordinate root it is twice a half-integer; for $h$ it is $\sum_i\beta_i\in\mathbb Z$. Symmetry gives the remaining cases. Therefore this is the <F4 root system>.
In the printed base order, the <Gram matrix> is
$$
G=\begin{pmatrix}2&-1&0&0\\-1&2&-1&0\\0&-1&1&-1/2\\0&0&-1/2&1\end{pmatrix}.
$$
Using the convention $a_{ij}=2(\alpha_i,\alpha_j)/(\alpha_j,\alpha_j)$, the <Cartan matrix> is
$$
\boxed{A=\begin{pmatrix}2&-1&0&0\\-1&2&-2&0\\0&-1&2&-1\\0&0&-1&2\end{pmatrix}.}
$$
The row-coroot convention transposes this <matrix>. The <Dynkin diagram> is the chain $1-2\Rightarrow3-4$, with the double bond pointing toward the shorter third root:
\Image[/past-exam-of-the-lie-algebras-and-representation-theory-of-symmetric-groups-course-of-the-university-of-oxford/2007-f4-dynkin.png]
{title=The <F4 Dynkin diagram> in the printed simple-root order; the arrow points toward the shorter root}
The associated symmetric positive definite form, normalized as a squared length, is
$$
\boxed{Q(x)=2x_1^2+2x_2^2+x_3^2+x_4^2-2x_1x_2-2x_2x_3-x_3x_4.}
$$
Indeed it equals
$$
(x_4/2)^2+(x_1-x_4/2)^2+(-x_1+x_2-x_4/2)^2+(-x_2+x_3-x_4/2)^2,
$$
which is strictly positive for nonzero $x$. The alternative Gram normalization $Q/2$ is the same positive definite form up to a positive scalar.
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