Problem 1388

If $a$ is a prime number $p$, that is, $a=p$, this theorem holds. If $a$ is a composite number, then by Lemma 11, $a$ can be decomposed into the product of prime factors. Let
$$a=p_{1} p_{2} \cdots p_{n}, \quad n \geqslant 2$$

where $p_{1}, p_{2}, \cdots, p_{n}$ are all prime numbers. Suppose $a$ can be decomposed into another form of the product of prime factors, that is,
$$a=q_{1} q_{2} \cdots q_{m}, \quad m \geqslant 2$$

where $q_{1}, q_{2}, \cdots, q_{m}$ are all prime numbers, then we get
$$p_{1} p_{2} \cdots p_{n}=q_{1} q_{2} \cdots q_{m}$$

Since $p_{1} \mid p_{1} p_{2} \cdots p_{n}$ and equation (3), we have $p_{1} \mid q_{1} q_{2} \cdots q_{m}$. Since $q_{1}$, $q_{2}, \cdots, q_{m}$ are all prime numbers, by Lemma 19, we get: $p_{1}$ must equal one of the $q_{r}$, and $x$ is one of $1,2, \cdots, m$. Let $p_{1}=q_{1}$, then from equation (3) we get
$$p_{2} \cdots p_{n}=q_{2} \cdots q_{m}$$

When $n=2$, since $p_{2} \cdots p_{n}=p_{2}$, we have $m=2, q_{2}=p_{2}$. Now suppose $n \geqslant 3$, since $p_{2} \mid p_{2} \cdots p_{n}$ and equation (4), we have $p_{2} \mid q_{2} \cdots q_{m}$. Since $q_{2}, \cdots, q_{m}$ are all prime numbers, by Lemma 19, we get: $p_{2}$ must equal one of the $q_{s}$, and $s$ is one of $2, \cdots, m$. Let $p_{2}=q_{2}$, then from equation (4) we get
$$p_{3} \cdots p_{n}=q_{3} \cdots q_{m}$$

When $n=3$, since $p_{3} \cdots p_{n}=p_{3}$, we have $m=3, q_{3}=p_{3}$. If $n \geqslant 4$, using the same method, because those $p$ and those $q$ are always one-to-one and equal, after canceling out, we must get $p_{n}=q_{m}$, which means $m=n$. Therefore, $a$ can only be decomposed into the product of prime factors in one way, disregarding the order of the prime factors.

From this theorem, we know that if the same prime factors are combined into their powers, then any integer $a>1$ can only be decomposed into one form:
$$a=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{n^{n}}^{a_{n}}, \quad n \geqslant 1$$

Here $p_{1}, p_{2}, \cdots, p_{n}$ are distinct prime numbers, and $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ are all positive integers. We call
$$a=p_{1}^{a_{1}} \cdot p_{2}^{\alpha_{2}} \cdots \cdot p_{n^{4}}^{\alpha_{4}^{4}}$$

the standard factorization of $a$.