could you help me with the following assignment?
Let $K$ be a field with characteristic $0$ and $D \in K[X] \setminus K$.
We write $rad(f)$ for the radical of a polynomial, the product of all monic irreducible polynomials dividing $f$. Then
$$
f^2 - Dg^2 \ = \ 1 \quad \text{ with } \quad f, g \in K[X]
\quad \text{ and } \quad g \neq 0
$$
Can't be solved if $\quad \deg(D) \ > \ 2 \deg(rad(D)) -2\quad $
My own thoughts
First I rewrote the equality: $(f-1)(f+1) \ = \ Dg^2$. If $h|f-1$ and $h|f+1$, then $h|(f+1)-(f-1)$, so $h$ is a unit, and since the radical is monic, we know that $rad((f-1)(f+1)) \ = \ rad(f-1)rad(f+1)$. So the equality we get is: $$ rad(f-1) \cdot rad(f+1) \quad = \quad rad(Dg^2) $$ And this is where I got stuck. Could you give me a hint?
I try a solution, but complicated, and of course to be verified. Also, I have supposed that your field $K$ is algebraically closed. For a polynomial $Q$, I note $d(Q)$ its degree.
A) First, let $P$ a non constant polynomial. Then we have the following inequality: $$1+d(P)\leq d(rad(P-1))+d(rad(P+1))$$ To see why, put $$P-1=c_1\prod_{a\in A_1}(x-a)^{n_a}\prod_{a\in A_2}(x-a)$$ Here $A_1$ is the set of multiple zeros of $P-1$, $n_a$ is the multiplicity of $a$ as a zero of $P-1$. The set $A_2$ is the set of simples zeros of $P-1$.
In the same way, we put $$P+1=c_2\prod_{a\in B_1}(x-a)^{n_a}\prod_{a\in B_2}(x-a)$$
Now we have $d(rad(P-1))=|A_1|+|A_2|$ and $d(rad(P+1))=|B_1|+|B_2|$. Let $a\in A_1$. As $a$ is a multiple zero of $P-1$, we get that $P^{\prime}(a)=0$, and that $n_a=1+q_a$, where $q_a$ is the multiplicity of $a$ as zero of $P^{\prime}$.
We have hence:
$$d(P)=d(P-1)=\sum_{a\in A_1}(1+q_a)+|A_2|=|A_1|+|A_2|+\sum_{a\in A_1}q_a=d(rad(P-1))+\sum_{a\in A_1}q_a$$
We have also:
$$d(P)=d(P+1)=d(rad(P+1))+\sum_{a\in B_1}q_a$$
We add these two equalities (note that $A_1$ and $B_1$ are disjoints subsets of zeros of $P^{\prime}$):
$$2d(P)=d(rad(P-1))+d(rad(P+1))+\sum_{a\in A_1\cup B_1} q_a\leq d(rad(P-1))+d(rad(P+1))+d(P^{\prime})$$ and we are done.
B) Now suppose that there exists $f,g$ such that $f^2-Dg^2=1$. We note that $2d(f)=d(D)+2d(g)$, we start with your equality: $rad(f-1)rad(f+1)=rad(Dg^2)=rad(Dg)$, and we have:
$$2d(rad(f-1))+2d(rad(f+1))=2d(rad(Dg))\leq 2d(rad(D))+2deg(g)$$ Hence $$ 2d(rad(f-1))+2d(rad(f+1))\leq 2d(rad(D))+2d(f)-d(D)$$ Now $$d(D)\leq 2d(rad(D))-2(d(rad(f-1))+d(rad(f+1))-d(f))$$ If we apply A) to the polynomial $f$, we see that we have proved that
$$d(D)\leq 2d(rad(D))-2$$ and we are done.