Divisibility Of A Square Implies Divisibility of the Base For Squarefree Integers

Theorem

The proposition that \(b \mid a^{2}\) implies that \(b \mid a\) for all positive integers \(a\) if and only if \(b\) is squarefree integer.

Proof

To see why, consider the prime factorisations of \(a\) and \(b\) by the fundamental theorem of arithmetic:

\[ a = p_{1}^{r_{1}} \cdot p_{2}^{r_{2}} \cdot \dots \cdot p_{n}^{r_{n}}\]

\[ b = q_{1}^{s_{1}} \cdot q_{2}^{s_{2}} \cdot \dots \cdot q_{k}^{s_{k}}\]

For backwards implication, we assume that \(b\) is squarefree, that is \(s_{i} = 1\) for all \(i \leq k\). Then, expressing \(a^{2}\) as a product of powers of primes, we have:

\[ a^{2} = p_{1}^{2r_{1}} \cdot p_{2}^{2r_{2}} \cdot \dots \cdot p_{n}^{2r_{n}}.\]

So since all of the powers of \(a^{2}\) are at least \(2\), if we can divide through by \(b\) (match up each \(q_{i}\) with a \(p_{j}\)), then doing so only reduces each \(r_{i}\) by at most \(1\). Therefore, we can match up the factors in the exact same way for dividing by \(a\).

For implication the other way, we prove the contrapositive by assuming that \(b\) is not the product of distinct primes, that is, \(s_{i} > 1\) for at least one \(i \leq k\). Then, consider the counterexample of the value of \(a\), constructed by taking \(a^{2}\) to be the product of \(q_{i}\) up to the next even power of \(s_{i}\).

By doing so, \(\frac{a^{2}}{b}\) is a product of distinct primes, and hence \(a^{2} | b^{2}\) trivially. However, \(a \nmid b\) since whichever \(s_{i} > 1\) will not be greater than twice itself when rounded up to an even number. For example, \(2 \leq 4, 3 + 1 \leq 6, 4 \leq 8, \dots\).

A specific example of this for intuition is \(b = 12\), where \(12 = 3 \cdot 2^{2}\), and so with \(a^{2} = 3^{2} \cdot 2^{2} = 36\), \(a = 6\) and \(12 \nmid 6\).