Perfect Square's Factors Aren't Perfect?

If $a$ is a factor of $n$ ,then $\frac na$ is also a factor of $n$ ,so factors always come in pairs.

About perfect squares,because $n^2$ has a factor $n$ ,and the "co-factor" is also $\frac{n^2}n=n$ ,so perfect square has odd number of distinct factors.

Let our perfect square be $n^{2}$ . The number of distinct factors a number has can be determined by taking the sum of one plus the powers of each integer in its prime factorization as such. Let's say our number $n$ has an arbitrary number of prime factors denoted by $a_{n}$ with some arbitrary power $q_{n}$ . Thus our number $n^{2}$ is $(a_{1}^{q_{1}} a_{2}^{q_{2}} a_{3}^{q_{3}} \cdots a_{n}^{q_{n}})^{2} = a_{1}^{2q_{1}} a_{2}^{2q_{2}} a_{3}^{2q_{3}} \cdots a_{n}^{2q_{n}}$ Ergo, the number of factors of $n^{2}$ is $(2q_{1} + 1)(2q_{2} + 1)(2q_{3} + 1) \cdots (2q_{n} + 1)$ Realize that the amount of factors of $n^2$ is a product of strictly odd numbers. With no power of $2$ among the product, no matter the prime factorization, it can be concluded that a perfect square always has an odd number of distinct factors.