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Fix an algebraically closed field, for completeness say the complex numbers.

Consider a univariate polynomial $p$. We will say that $p$ can be "squarified" if there is a univariate non-constant rational function $q$ such that $p(q(x))$ is the square of some univariate rational function $r(x)$.

Question: Which polynomials can be squarified? More specifically, are there polynomials in the coefficients, whose solution sets characterize exactly the polynomials that can be squarified?

Below I list what I know. Basically, I think degrees 1-3 are well understood, possibly even degree 4. But I have no idea what happens for degree 5 and up.

All constant polynomials can be squarified

If $p(x)=a$, regardless of $q$, $p(q(x))=a=(\sqrt{a})^2$.

All degree-1 polynomials can be squarified

We will use the fact that a linear change of variables does not effect squarifiability. That is, $p(x)$ is squarifiable if and only if $p(ax+b)$ is squarifiable.

If $p$ is linear, by a change of variables, we can assume $p(x)=x$. So we just set $q(x)=x^2$, and then $p(q(x))=x^2$

All degree-2 polynomials can be squarified

As explained in another question (Characterize polynomials $p,q$ such that $p(q(x))$ is a perfect square), for a polynomial $p$ that is not already a perfect square, there is no polynomial $q$ that makes $p(q(x))$ a perfect square. However, there is a solution using a rational $q$.

By performing a linear change of variables, we can assume any degree-2 $p$ is of the form $p(x)=x^2+a$ (here, we needed the fact that the field is closed to remove the leading coefficient). Then you can set $q(x)=(4x^2-a)/(4x)$. We have:

$$p(q(x)) = q(x)^2+a = \frac{16 x^4-8 a x^2+a^2}{16x^2}+a = \frac{16 x^4+8 a x^2+a^2}{16x^2} = \left(\frac{4x^2+a}{4x}\right)^2$$

Not all degree 3 polynomials can be squarified(?)

Consider a degree 3 polynomial $p$. If it's discriminant is 0, then it has a repeated root. In this case, $p$ can be squarified. Indeed, if $p(x)=c(x-a)^2(x-b)$, we can choose $q(x)=x^2+b$. Then $p(q(x))=c(q(x)-a)^2 x^2$, which is a square.

I believe the above is tight: that is if the discriminant is non-zero for a degree-3 polynomial, then $p$ cannot be squarified. The reason is that $y^2=p(x)$ is an elliptic curve, and squarifying $p$ gives a rational parametrization of at least a part of the curve. If I understand correctly, an elliptic curve with non-zero discriminant cannot be parametrized. There's potentially a slight gap, as squarifying may not give a parametrization of the whole curve but only part of it.

Degree 4?

Let $p(x)$ be a degree 4 polynomial. If the discriminant is 0, then $p$ can be squarified. Indeed, if the discriminant is 0, then there is a repeated root, and we can write $p(x)=(x-a)^2 p_0(x)$ for a degree-2 polynomial $p_0$. We apply the degree 2 construction to get $q$ which squarifies $p_0$, and this $q$ will also scarify $p$.

Is this tight? Can all degree 4 polynomials be squarified, or only those with a vanishing discriminant?

Degree 5 and up?

For degree 5 and up, the situation becomes even less clear. Having a discriminant of 0 no longer appears sufficient to squarify. But can it be shown to be necessary? I suspect some generalization of the discriminant may give necessary and sufficient conditions, but I'm not sure where to look.

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Slight nitpick: for the degree $2$ remarks you need to assume the characteristic is not $2$ - and unless you want to be subtle, the following argument requires this too (I'll also require $k$ is algebraically closed - but this is easier to remove).

I think that your argument does continue - let's treat all the cases together (but note that what I'm doing is really just a continuation of your degree $3$ idea!).

Consider the curve $C : y^2 = f(x)$ in $\mathbb{A}^2$. We might as well assume that $f(x)$ is squarefree (i.e., no repeated roots), or we can simply change variables (by dividing $y$ by this factor) in order to assume the degree of $f$ is smaller.

Clearly $f(x)$ may be (as you put it) "squareified" if and only if there exists a rational map $$\phi : \mathbb{A}^1 \to C$$ but (by taking projective closures) we see that by the Riemann-Hurwitz theorem this implies that $C$ must have genus $0$ (precisely, this is since $-2 \geq deg(\phi) (2g(C) -2)$ ) - N.B. there are many ways to see this.

So this is just now a genus calculation for $C : y^2 = f(x)$. But recall these genera are well known (since $f(x)$ has no repeated roots) and that $g = \lceil n/2 \rceil -1$. We conclude that $C$ may be parametrised if and only if $n \leq 2 $

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