# Determining integer coefficients of polynomial satisfying one condition, using computer

$P_5(w) = c_0 + c_1w + c_2w^2 +...+ c_5w^5$, where $c_0, ... , c_5$are integers that I want to determine.

$Q_5(w) = w^2 (\frac{d}{dw}( \frac{P_5(w)}{w}) = -c_0 + c_2w^2 + 2c_3w^3 + 3c_4w^4 +4c_5w^5$, another deg 5 polynomial.

I have that every root (over $\mathbb{C}$) of \begin{equation} (Q_5(w))^2 + 16Q_5(w)(1 + w)^3 w^2 - 80(1 + w)^2 w^3 P_5(w) = 0 \end{equation} is also a root of \begin{equation} (24)^3 (1 + w)^5 w^4 [4P_5 (w) w - Q_5(w) ( 1 - w)]^3 + 108(Q_5(w))^5 = 0 \end{equation}

Is there a way to solve this using an algebra software?

• Every root (...) is also a root of (...) : with multiplicity? (In which case, you are stating that the latter polynomial is a multiple of the former). – Pietro Majer Sep 25 '12 at 20:37
• Maybe a few words about the origin of this question would be nice. The question looks a little like the search for a degree $5$ cover of Riemann spheres with some ramification requirements. – Peter Mueller Sep 26 '12 at 13:41

• Apparently Renato doesn't consider multiplicities. As $f_1(w)$ has degree at most $10$, his condition then is that the remainder of $f_2(w)^{10}$ modulo $f_1(w)$ vanishes. I believe `set the rest to zero' means the condition coming from setting the coefficients of this remainder to $0$. By the way, a solution to the question is $P(w)=w((w+1)^4+c)$ for any $c$. Also, $P(w)=-w(w+1)^4$ is a solution. Still, it would be nice to learn more about the background of this question! – Peter Mueller Sep 27 '12 at 9:59
• Sorry, rest = reminder , as Peter said "the remainder of $f_2(w)^10$ modulo $f_1(w)$ vanishes". The origin of this question comes from Mirror Symmetry and computing the superpotential $W$, which is just function, in my case from $C^2$ to $C$, of the Landau-Ginzburg model for the complement of a divisor in $CP^2$. The Superpotential is a Laurent polynomial whose coefficients are given by a count of holomorphic discs, and monomials are related to a relative homotopy class of the group. I was able to prove that $W = z + \frac{2(1+w)^2}{z^2} + \frac{P_5(w)}{w z^5}$ – – Renato Sep 27 '12 at 22:21
• From Floer theory we have that if $(z,w)∈critW$ then $W^3(z,w)=27$. Solving for z in one of the partial derivatives of W, we get that this condition is equivalent to the roots of $f_1(w)$ being also roots of $f_2(w)$. – Renato Sep 27 '12 at 22:41