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The question is simple:

Do there exist two polynomials $P,Q\in\mathbb{R}[x]$ such that $$ \text{deg}(P^3-Q^2) = 1\ ?$$

One may assume that both $P$ and $Q$ are monic, and most naively one could consider the case that $\text{deg}(P) = 2$ and $\text{deg}(Q) = 3$. Then writing out the general expressions, which depend on $3+2 = 5$ variables in total, one obtains 4 constraints to make $x^i, i \in \{2,3,4,5\}$ vanish. The interesting outcome of this rather tedious excercise is that $P^3 = Q^2$ in this case, so this gives no solutions. More generally, setting $\text{deg}(P) = 2n$ and $\text{deg}(Q) = 3n$ for some $n \in \mathbb{N}$, one has precisely $5n$ variables and $6n-2$ terms to kill, so that for $n \geq 2$ one has at least as many constraints as variables, $P^3=Q^2$ always being a set of solutions. This leads me to conjecture that such polynomials do not exist, but I have no idea how to prove it. Does anyone have any suggestions?

user26857
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Mike Daas
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1 Answers1

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Clearly this can't hold if $P$ or $Q$ are constants. Suppose they are nonconstant.

By the ABC theorem, if you had nonconstant polynomials $P$, $Q$, and $R$ such that $R$ is linear and

$$ P^3 + (-Q^2) = R $$

then $3 \deg(P) = 2 \deg(Q)$ and you'd have

$$ 3 \deg(P) = \max\{ 3\deg(P), 2 \deg(Q), 1 \} \leq \deg\left( \mathrm{rad}(PQR) \right) - 1 \leq \deg(P) + \deg(Q) $$

This implies $\deg(Q) \geq 2 \deg(P)$, and thus $2\deg(Q) \geq 4 \deg(P)$, a contradiction!

Thus, the hypothesis is false: it is impossible for $P^3 - Q^2$ to be linear.

user26857
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    Well spotted! Though a priori $\gcd(P,Q,R) = R$ is also possibility, but then $R^2$ divides the LHS but not the RHS, so it is not possible – Mike Daas Jan 13 '18 at 16:46