I have polynomials $A,B\in\mathbb{C}[t]$ that are relatively prime and non-constant such that $A^4-B^4$ is the square of a polynomial in $\mathbb{C}[t]$
Consider the polynomials $A_1, B_1, C_1\in\mathbb{C}[t]$ given by $$A-B=A_1^2\\ A+B=B_1^2\\ A^2+B^2=C_1^2 $$ (In the context of the problem, I know that these squares are justified). I want to show $A_1$ and $B_1$ are non-constant. It's pretty easy to see they can't both be constant, but if I assume just one of them is constant, I keep going in circles trying to get a contradiction. Any help is much appreciated.
I've figured out a solution, but I'm wondering if maybe there was an easier way:
Assume for sake of contradiction that $A_1$ is constant. Then $A-B$ is a constant, hence all of their coefficients must be the same except for their constant terms, so write $A = tP + a_0$ and $B = tP+b_0$ for appropriate polynomial $P\in\mathbb{C}[t]$. Then:
$\begin{align*} A^2+B^2 &=\ (tP+a_0)^2+(tP+b_0)^2\\ &=\ 2t^2P^2 + 2tP(a_0+b_0) + a_0^2+b_0^2 = C_1^2 \end{align*}$
Dividing both sides by 2 to make things cleaner:
$\begin{align*} t^2P^2 + tP(a_0+b_0) + \frac{a_0^2+b_0^2}{2} &= \left(\frac{C_1}{\sqrt{2}}\right)^2 \\ \left(tP+\frac{a_0+b_0}{2} \right)^2-\frac{(a_0+b_0)^2}{4}+\frac{a_0^2+b_0^2}{2} &=\left(\frac{C_1}{\sqrt{2}}\right)^2 \end{align*}$
The left side is a square nonconstant polynomial plus some extra constant terms, and the right hand side is a nonconstant square polynomial which means that the extra constant term on the left hand side must be 0 (this is a pretty easy fact to show).
So
$\begin{align*} \frac{2a_0^2+2b_0^2 -(a_0+b_0)^2}{4} &= 0\\ \frac{(a_0-b_0)^2}{4}&= 0 \Rightarrow a_0=b_0 \end{align*}$
So since $a_0 = b_0$, we get that $A=B$ contradicting them being relatively prime.