$f^2=g^3+1$ only when $f,g\in\Bbb C$

Question

Trying to learn some algebraic geometry I found the following exercise, where the author claims I will get some insight in the field:

Prove that if $f$ and $g$ are polynomials (with complex coefficients) such that $$f^2=g^3+1\tag{1}$$ then $f$ and $g$ are both constant.

I want to know if my proof is correct, and if there is a faster way to get the result. Here we go:

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Write $g^3=f^2-1=(f+1)(f-1)$. If a polynomial $d$ divides either $f,g$ then $d\mid 1$, so $f,g$ have no common factor. Also, there is no non-constant polynomial that divides both $f+1,f-1$, as that polynomial ought to divide $2$. Therefore there are polynomials $u,v$ such that $$f+1=u^3$$$$f-1=v^3$$ This gives $2=u^3-v^3=(u-v)(u^2+uv+v^2)$. Therefore $u-v=c$ for some constant $c\in\Bbb C$. This gives $$(v+c)^3-1=v^3+1$$$$3cv^2+3c^2v+c^3-2=0$$ The last equation is only possible if $v$ is a constant, this forces $f,g$ to be constants as well.

Answer 1

From the equation $$ f^2=g^3+1 $$ it follows that

• $\gcd(f,g)=1\\[2pt]$
• $2\deg(f)=3\deg(g)$

Suppose $g$ is non-constant.

From $2\deg(f)=3\deg(g)$, we get $\deg(f) > \deg(g)$.

But then, taking formal derivatives, we get \begin{align*} & 2ff'=3g^2g' \\[4pt] \implies\;& f{\,\mid\,}g'\quad\bigl(\,\text{since $\gcd(f,g)=1$}\,\bigr) \\[4pt] \implies\;& \deg(f)\le\deg(g') \\[4pt] \implies\;& \deg(f) < \deg(g) \\[4pt] \end{align*} contradiction.

Answer 2

Not sure if you actually needed such a result for this problem which can be solved in various ways, but it seems that it is a particular case of the result due to Davenport, see

H. Davenport (1965), On Norske Vid. Selsk. Forrh. 38, pp. 86-87.

I wasn't able to find this paper, but there is a paper with the statement of Davenport's result (perhaps one has to check references therein).

In any case let me state the result and share the proof that I know:

Theorem (Davenport, 1965). Let $f$ and $g$ be coprime non-constant polynomials with complex coefficients. Then, if $f^2\neq g^3$, one has $$ \deg (f^2-g^3)\geq\frac{1}{2}\deg g+1. $$

Clearly, this solves your exercise because if $g$ is not a constant, then the degree of $f^2-g^3$ is at least $\frac{1}{2}\deg g+1>1$.

Proof of the theorem. Basically this fact is rather a simple consequence of the Mason-Stothers theorem which in turn can be regarded as a polynomial analogue of the famous $abc$-conjecture. However, Mason-Stothers theorem is not difficult (e. g. its proof is given even on the corresponding Wikipedia page).

For completeness, I won't use this fact and simply repeat the proof in this particular case. Denote $h=f^2-g^3$, clearly $f$, $g$ and $h$ are coprime. Assume that $\deg h<\frac{1}{2}\deg g+1$, then we have $3\deg f=2\deg g$. Differentiation gives us $$ h'=2ff'-3g^2g'. $$ Now consider the following expression (the Wronskian): $$ W(f^2,g^3)=(f^2)'g^3-f^2(g^3)'=2ff'g^3-3f^2g^2g'=fg^2(2f'g-3fg') $$ Then, from $g^3=f^2-h$ we get $$ W(f^2,g^3)=(f^2)'(f^2-h)-f^2(f^2-h)'=f^2h'-h(f^2)'=f(fh'-2f'h)=W(f^2,h). $$ Now note that $W(f^2,g^3)=W(f^2,h)$ is a non-zero polynomial (why?). It follows that $$ p:=g^2(2f'g-3fg')=fh'-2fh' $$ is a non-zero polynomial. However, $p$ is divisible by $g^2$ and has degree at most $\deg f+\deg h-1$. Therefore, $$ \deg f+\deg h-1\geq 2\deg g, $$ and using $\deg f=\frac{3}{2}\deg g$ we obtain the required inequality $\deg h\geq \frac{1}{2}\deg+1$. $\square$

Remark 1. The point of Mason-Stothers theorem is that if you have equality $f+g=h$ for coprime polynomials $f,g,h$, then it is "unlikely" that their product $fgh$ has small number of distinct roots. This is more or less what happening in the Davenport's theorem: all roots $f^2$ and $g^3$ have multiplicity and that forces the difference $f^2-g^3$ to have "large" degree.

Remark 2. This trick with Wronskian might look very artificial but it is not quite true. The expression $W(a,b)=a'b-ab'$ appears, for example, when one computes the derivative $\frac{a}{b}$, or (even better) when calculates the logarithmic derivative: $$ (\log (f/g))'=(\log f)'-(\log g)'=\frac{f'}{f}-\frac{g'}{g}=\frac{f'g-fg'}{fg}. $$ Now one can "see" that because logarithm is not too sensitive to multiplicities (i.e. $\log f^3$ is just $3\log f$), the Wronskian $W(f^3,g^2)$ is very "close" to $W(f,g)$. On the technical level this leads to certain cancellations and eventually to better bounds for the degrees of invloved polynomials. The proof of Mason-Stothers (due to N. Snyder) is more or less the same.