If $f\in \mathbb{Z[x]}$ is a polynomial and $f(2^n)$ is a perfect square for all $n$, then there is a $g\in \mathbb{Z[x]}$ such that $f=g^2$

Question

If $f\in \mathbb{Z[x]}$ is a polynomial and $f(2^n)$ is a perfect square for all $n$, then there is a $g\in \mathbb{Z[x]}$ such that $f=g^2$

I understand that there might be an issue with the problem statement but I don't know anything further.

Here's a link that might help (related problem).

EDIT : I see the exact problem posted in AoPS long time ago. I couldn't understand the solution posted there.

I'm quoting user Math154's solution :

Multiplying by the square of $f$'s denominator if necessary, WLOG $f\in\mathbb{Z}[x]$ and $t_n=\sqrt{f(2^n)}\in\mathbb{Z}^+$ for all $n\ge1$.

First suppose that $d=\deg f$ is even, so $f(x)=cg(x)^2+h(x)$ for polynomials $g,h\in\mathbb{Q}[x]$ with $\deg{g}=d/2$ and $\deg{h}<d/2$ and a positive integer $c$ (it clearly can't be negative). Then $$t_n=\sqrt{f(2^n)}=\sqrt{cg(2^n)^2+h(2^n)}=\sqrt{c}g(2^n)+O(2^{-n})$$. Yet $g(2^n)=\sum_{k=0}^{d/2}c_k (2^k)^n$ is a rational power sum in $2^0,\ldots,2^{d/2}$, so because $(x-2^0)\cdots(x-2^{d/2})\in\mathbb{Z}[x]$ has coefficients independent of $n$, $\{t_n\}$ satisfies a linear recurrence of order $1+d/2$ with integer coefficients up to an error of $o(1)$. But $t_n\in\mathbb{Z}\forall{n\ge1}$, so for sufficiently large $n\ge N$, $\{t_n\}$ satisfies the linear recurrence exactly, i.e. there exists a polynomial $p(x)$ such that $t_n=p(2^n)$ for all $n\ge N$ and thus $f(x)=p(x)^2$ for all $x$. But $g$ is unique up to sign, so $p(x)^2=cq(x)^2$ and considering $x=2^n$ for sufficiently large $n$, $c$ must be a square, i.e. $p(x)\in\mathbb{Q}[x]$ as desired.

Now if $\deg f$ is odd, then by the previous case there exist polynomials $g_1,g_2\in\mathbb{Q}[x]$ such that $f(x^2)=g_1(x)^2$ and $f(2x^2)=g_2(x)^2$, so $g_1(x\sqrt{2})=\pm g_2(x)$, which contradicts rationality by equating coefficients.

Can I get a clear understanding of the solution? I don't think I get it well enough, at least from the $(x-2^0)(x-2^1)\cdots \in \mathbb{Z}[x]$ and the linear recurrence part. Also, at the initial stage, the user claims that $t_n=\sqrt{f(2^n)}=\sqrt{cg(2^n)^2+h(2^n)}=\sqrt{c}g(2^n)+O(2^{-n})$. How is it $O(2^{-n})$? Isn't it supposed to be $O\left(2^{\frac{n(d-2)}{4}}\right)$?

Answer 1

In this answer I address the two potential issues you point out at the end of your question.

[1] Also, at the initial stage, the user claims ...

The estimate from Mean Value Theorem is: $$ |f(x+h) - f(x)| \leq \sup_{y}|f'(y)| |h|$$ where the sup is over $y$ in between $x$ and $x+h$. Let "$ A\lesssim B$" mean that $A ≤ C B$ for some unimportant constant $C>0$. If $f$ is the square root function, $h = h(2^n)$, and $x=cg(2^n)^2 $, notice that $|h(2^n)|\lesssim 2^{n(d/2-1)}$, and $\sup_y |f'(y)|\lesssim \frac1{|f(x)|}\lesssim 2^{-nd/2}$, so we get the estimate $$ \left|\sqrt{ cg(2^n)^2 + h(2^n)} - \sqrt{c}g(2^n)\right| \lesssim \frac{2^{n(d/2 - 1)}}{2^{nd/2}} = 2^{-n}$$ as claimed.

[2] the linear recurrence part...

You will need to know the basic theorem for higher order linear recurrences. Here's one version:

Theorem. Suppose we had the "characteristic polynomial" expressed in the following two ways, $$c(x) = x^K - \alpha_1 x^{K-1} - \dots - \alpha_K = (x-\lambda_1)\dots(x-\lambda_K),$$ where $\alpha_i,\lambda_i$ are constants, then for any constants $C_1,\dots,C_K$, the sequence $$ a_n = \sum_{i=1}^K C_i \lambda_i^n$$ is a solution to the recurrence $$ a_n = \alpha_1 a_{n-1} + \dots + \alpha_K \alpha_{n-K} $$ Proof. Since the recurrence is linear, linear combinations of solutions are also solutions. So it suffices to check that each $ \lambda_i^n$ is a solution. But this is obvious, as $\lambda_i^{n-K}c(\lambda_i) = 0$.

To apply this: what the answer is trying to say is that $g(2^n)$ solves a recurrence exactly. Since $g(2^n)$ is some combination of $1,\dots,(2^{d/2})^n$, let $\lambda_i = 2^{i-1}$, $i=1,\dots,K=d/2+1$, the corresponding characteristic polynomial is $(x-1)\dots(x-2^{d/2})$ as the user hinted towards, and the recurrence could even be written down explicitly.