Squary Polynomial

We note that:

$\begin{cases} f \left( \dfrac{1}{2} \right) = \dfrac{25}{16} = \dfrac{(2^2+1)^2}{2^4} \\ f \left( \dfrac{1}{3} \right) = \dfrac{100}{81} = \dfrac{(3^2+1)^2}{3^4} \\ f \left( \dfrac{1}{4} \right) = \dfrac{289}{256} = \dfrac{(4^2+1)^2}{4^4} \\ f \left( \dfrac{1}{5} \right) = \dfrac{676}{625} = \dfrac{(5^2+1)^2}{5^4} \end{cases}$

$\begin{aligned} f(x) & = \frac{\left(\frac{1}{x^2} +1\right)^2}{\frac{1}{x^4}} \\ & = \frac{\left(\frac{1}{x^2} +1\right)^2}{\frac{1}{x^4}} \times \frac{x^4}{x^4} \\ \Rightarrow f(x) & = \left(1+x^2 \right)^2 \quad \text{a monic quartic polynomial} \\ f(10) & = \left(101 \right)^2 = 10201 \\ f(9) & = \left(82 \right)^2 = 6724 \end{aligned}$

Therefore,

$\begin{aligned} \phi (f(10)+f(9)) & = \phi (16925) \\ & = \phi (5^2 \dot{} 677) \\ & = 5^2 \dot{} 677 \left(1-\frac{1}{5}\right) \left(1-\frac{1}{677}\right) \\ & = \boxed{13520} \end{aligned}$