This is more of my own observations than a question.
Let $P(x)=ax^2+bx+c$. For convenience, assume $a,b,c$ are integral. The question is as follows. Suppose, $P(x)$ is a perfect square (for every integer, or for infinitely many integers, as $x$ sweeps through $\mathbb{R}$ or $\mathbb{Z}$). What can we say about $P(\cdot)$? Does it have to be a square of a linear polynomial?
Case 1: It is clear that, if $a>0$, then $P(x)$ is a perfect square (of an integer), for infinitely many $x\in\mathbb{R}$. Simply solve the associated quadratic.
Case 2: Suppose, $P(x)$ is a perfect square, for infinitely many $x\in\mathbb{Z}$. Is it necessarily true that, $P(x)$ itself must be a square of a degree-1 polynomial, that is, is it true that, $P(x)=(dx+e)^2$, for some $d,e\in\mathbb{R}$? The answer is no. Consider, $P(x)=2x^2+1$. From Pell's equations, it holds that, $2x^2+1=y^2$ infinitely often, and the pairs, $(x_n,y_n)$ can be generated via, $(2+\sqrt{3})^n = x_n+y_n\sqrt{3}$.
Case 3: Suppose, $P(x)$ is a perfect square, for every $x\in\mathbb{Z}$. Is it true that, $P(x)=(dx+e)^2$, for some monic polynomial? Now, it turns out that, the answer is yes. There is an elementary proof of this fact (albeit well-known).
