Problem 859

Proof: By definition, $a$ must have a divisor $d \geqslant 2$. Let the set $T$ consist of all divisors $d \geqslant 2$ of $a$. By the principle of the smallest natural number, the set $T$ must have the smallest natural number, denoted as $p$. $p$ must be an irreducible number. Otherwise, if $p \geqslant 2$ is a composite number, by Theorem 3(i), $p$ must have a divisor $d^{\prime}: 2 \leqslant d^{\prime}<p$. Clearly, $d^{\prime}$ belongs to $T$, which contradicts the minimality of $p$. Proof completed.