= Solution
We first prove the useful lemma that <dominant root-lattice highest weights have zero weight>. Write $Q=\mathbb Z\Phi$ for the <root lattice>, and let a <dominant integral weight> $\lambda\in Q$ be the <highest weight> of a finite-dimensional irreducible representation.
First, $\lambda$ is a nonnegative integral combination of <simple roots>. Indeed, write $\lambda=\lambda_+-\lambda_-$, separating its positive and negative coefficients in the <root basis>. The two parts have disjoint supports, and distinct <simple roots> have nonpositive <inner product>, so $(\lambda_+,\lambda_-)\leq0$. If $\lambda_-\ne0$, then
$$
(\lambda,\lambda_-)=(\lambda_+,\lambda_-)-(\lambda_-,\lambda_-)<0.
$$
On the other hand, dominance gives $(\lambda,\alpha_i)\geq0$ for each <simple root>, hence $(\lambda,\lambda_-)\geq0$. This contradiction proves $\lambda_-=0$.
Now suppose a nonzero weight $\mu=\sum_i m_i\alpha_i$ has all $m_i\in\mathbb Z_{\geq0}$. The identity
$$
0<(\mu,\mu)=\sum_i m_i(\mu,\alpha_i)
$$
shows that some $i$ satisfies $m_i>0$ and $\langle\mu,\alpha_i^\vee\rangle>0$. We use the standard <sl2 Lie algebra> fact that its <lowering operator> is injective on any positive $h$ eigenspace in a finite-dimensional representation. This follows from the <classification of finite-dimensional sl2 representations>: in each irreducible $V(n)$, the only weight killed by the <lowering operator> is the lowest weight $-n\leq0$.
Consequently a nonzero vector of weight $\mu$ lowers to a nonzero vector of weight $\mu-\alpha_i$. Its simple-root coefficients remain nonnegative and their sum decreases by one. Starting at $\lambda$, repeated lowering must therefore reach the zero weight. Notice that intermediate weights need not remain dominant; positivity of the chosen coroot pairing is enough at each step.
For a <minuscule representation>, every weight belongs to $W\lambda$, so the zero weight just obtained lies in this orbit. Every <Weyl group> element is invertible, and $w\lambda=0$ forces $\lambda=0$. This proves that <minuscule weights in the root lattice are zero>. Under the assumption $X=Q$, every possible <highest weight> lies in $Q$, hence
$$
\boxed{X=\mathbb Z\Phi\ \Longrightarrow\ \text{the only minuscule irreducible representation is }V(0),\text{ the trivial one}.}
$$
Here $V(0)$ is the one-dimensional <trivial Lie algebra representation>, by the classification of finite-dimensional irreducible <highest-weight representations>.
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