# Factorizing a bivariate polynomial

I have a bivariate polynomial for each $$n=0,1,2...$$ $$f_n(x,y)=\sum _{k=0}^n \frac{(-1)^k}{2 k+1} \binom{n}{k} \left(x ^2-y ^2\right)^{2 n-2 k}\left([y ( x^2 -1) +x(1 -y^2 )]^{2 k+1}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad-[y ( x^2 -1) -x(1 -y^2 )]^{2 k+1}\right)$$ looking at low orders explicitly in Mathematica reveals that this polynomial contains $$n+1$$ factors of $$(1-y^2)$$. I know that one factor comes from the difference of two odd powers, but cannot see where the other $$n$$ factors come from. The factorized expressions appear quite complex, for example: $$f_3=\frac{2}{35} \left(y ^2-1\right)^4 x \left(-35 y^6+35 \left(5 y ^2-1\right) x ^{10}+7 \left(15 y ^4-73 y ^2+3\right) x ^8+(y -1) (y +1) \left(5 y^4-388 y^2+5\right) x ^6-7 y ^2 \left(3 y ^4-73 y ^2+15\right) x ^4+35 y ^4 \left(y^2-5\right) x ^2+35 x ^{12}\right)$$

Assuming $$f_n$$ contains $$(1-y^2)^{n+1}$$, How can I find the formula for $$f_n$$ in its factorized form, for general $$n$$?

If it is any use, the series coefficients are related to the following integral: $$\sum _{k=0}^n \frac{(-1)^k}{2 k+1}\binom{n}{k} z^{2k+1}=\int_0^z(1-u^2)^n\mathrm{d}u$$

• This would be easier to read with $\xi,\eta$ replaced by $x,y$. – Matt F. Aug 29 '19 at 1:53

So, $$\begin{split} f_n(\xi,\eta) &= (\xi^2-\eta^2)^{2n+1} \left( F\big(\frac{\eta ( \xi^2 -1) +\xi(1 -\eta^2)}{\xi^2-\eta^2}\big) - F\big(\frac{\eta ( \xi^2 -1) -\xi(1 -\eta^2)}{\xi^2-\eta^2}\big)\right) \\ &= (\xi^2-\eta^2)^{2n+1} \left( F\big(\frac{\eta\xi + 1}{\xi+\eta}\big) - F\big(\frac{\eta\xi - 1}{\xi-\eta}\big)\right), \end{split}$$ where $$F(x) := \int_0^x (1-u^2)^n\,du.$$
To show that $$(1-\eta^2)^{n+1}$$ is a factor $$f_n(\xi,\eta)$$, it's enough to show that $$(1-\eta^2)^n$$ divides $$\begin{split} \frac{\partial}{\partial \eta} \left( F\big(\frac{\eta\xi + 1}{\xi+\eta}\big) - F\big(\frac{\eta\xi - 1}{\xi-\eta}\big) \right) &= \bigg(1-\big(\frac{\eta\xi + 1}{\xi+\eta}\big)^2\bigg)^n \frac{\xi^2-1}{(\xi+\eta)^2} - \bigg(1-\big(\frac{\eta\xi - 1}{\xi-\eta}\big)^2\bigg)^n\frac{\xi^2-1}{(\xi-\eta)^2} \\ &= (1-\eta^2)^n (\xi^2-1)^{n+1} \left( \frac{1}{(\xi+\eta)^{2n+2}} - \frac{1}{(\xi-\eta)^{2n+2}}\right), \end{split}$$ and we see that it does.
• I have trouble understanding the argument regarding the derivative - could you please elaborate on why then $f_n$ would be divisible by $(1-\eta^2)^{n+1}$? – Matt Majic Aug 29 '19 at 9:52
• @MattMajic: $\eta=\pm 1$ is a zero of $(F(...)-F(...))$ (the factor in parentheses of $f_n$). It has multiplicity $n+1$ iff it's a zero of $\frac{\partial}{\partial \eta}(F(...)-F(...))$ of multiplicity $n$. – Max Alekseyev Aug 29 '19 at 13:33