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

Proof

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

For backwards implication, we assume that $b$ is squarefree, that is $s_i=1$ for all $i\le k$. Then, expressing $a^2$ as a product of powers of primes, we have:

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 $i\le k$. 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, $2\le 4,3+1\le 6,4\le 8,\dots$.

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 $12\nmid 6$.