If $a^2$ divides $b^2$, then $a$ divides $b$ [duplicate]

By the fundamental theorem of arithemtic, you can write $a$ and $b$ as a product of primes, say $$ a=p_1^{\alpha_1}\cdots p_r^{\alpha_r},\qquad b=p_1^{\beta_1}\cdots p_r^{\beta_r} $$ where $\alpha_i,\beta_i\geq 0$. Allow the exponents to possibly be $0$ if such a prime $p_i$ occurs in the factorization of one integer but not the other.

So $a^2=p_1^{2\alpha_1}\cdots p_r^{2\alpha_r}$ and $b^2=p_1^{2\beta_1}\cdots p_r^{2\beta_r}$. Since $a^2\mid b^2$, by unique factorization, necessarily $2\alpha_i\leq 2\beta_i$ for each $i$. That implies $\alpha_i\leq\beta_i$ for all $i$, and so $a\mid b$.

To say that $a^2$ divides $b^2$ is to say that $n=b^2/a^2 = (b/a)^2$ is an integer. Now integers only have square roots which are integers or irrational. Since $b/a$ is rational, it must be an integer, which is to say that $a$ divides $b$.

Call a positive integer $y$ bad if for some positive $x$, we have $x^2\mid y^2$ but $x \nmid y$.

If there are bad positive integers, there is a smallest bad one, say $b$. Since $b$ is bad, there is a positive integer $a$ such that $a^2 \mid b^2$ but $a \nmid b$.

It is clear that $a \ne 1$. So some prime $p$ divides $a^2$. But if a prime divides the product $cd$, it divides $c$ or $d$ or both. Thus $p\mid a$.

Since $a^2\mid b^2$, we have $p\mid b^2$, and therefore $p\mid b$.

Let $a=pa_1$ and $b=pb_1$. Since $a^2\mid b^2$, we have $(pa_1)^2=q(pb_1)^2$ for some integer $q$, and therefore $a_1^2\mid b_1^2$.

But $a_1\nmid b_1$, since if it does, one can easily show that $a\mid b$.

So we have shown that $b_1$ is bad. It is clear that $b_1\lt b$, which contradicts the supposed minimality of $b$.

We conclude that there are no bad positive integers.

The proofs given use the Unique Factorization Theorem, or the existence of GCDs, or some equivalent, but the result is true even in places where there aren't any GCDs, so there must be a proof that doesn't rely on these properties. Here's one that works in the ring $O_K$ of integers in a number field $K$, whether there are GCDs or not.

If $a,b$ are in $O_K$ and $a^2$ divides $b^2$, then $b^2=a^2c$ for some $c$ in $O_K$. So $\sqrt c=b/a$ is in $K$. But $\sqrt c$ is a zero of $x^2-c$, a monic polynomial with algebraic integer coefficients, so $\sqrt c$ is an algebraic integer, so $\sqrt c$ is in $O_K$, so $a$ divides $b$.