Quartic Polynomial

Since $4(x^4 - 3x^3 + 4x^2 - 2x + 1) \; = \; (2x^2 - 3x + 3)^2 - 5(x-1)^2$ the roots of the quartic are $\begin{array}{rclcrcl} a & = & \tfrac12\phi^2 + \tfrac12i5^{\frac14}\phi^{-\frac12} & \hspace{1cm}& b & = & \tfrac12\phi^2 - \tfrac12i5^{\frac14}\phi^{-\frac12} \\ c & = & \tfrac12\phi^{-2} + \tfrac12i5^{\frac14}\phi^{\frac12} & & d & = & \tfrac12\phi^{-2} - \tfrac12i5^{\frac14}\phi^{\frac12} \end{array}$ where $\phi = \tfrac12(\sqrt{5}+1)$ . Thus $a+b = \phi^2$ , $c+d = \phi^{-2}$ , so that $\frac{1}{a+b} + \frac{1}{c+d} \; = \; \phi^2 + \phi^{-2} \; = \; 3$ We also note that $a + c \; = \; \tfrac12(\phi^2 + \phi^{-2}) + \tfrac12i5^{\frac14}(\phi^{\frac12} + \phi^{-\frac12}) \; = \; \tfrac32 + \tfrac12i5^{\frac14}(\phi^{\frac12} + \phi^{-\frac12})$ while $b+d$ is its complex conjugate, and hence $\frac{1}{a+c} + \frac{1}{b+d} \; = \; 2\mathfrak{Re} \frac{1}{a+c} \; = \; \frac{12}{9 + \sqrt{5}(\phi^{\frac12} + \phi^{-\frac12})^2} \; = \; \frac{3(7-\sqrt{5})}{22}$ and, similarly $\frac{1}{a+d} + \frac{1}{b+c} \; = \; 2\mathfrak{Re}\frac{1}{a+d} \; = \; \frac{12}{9 + \sqrt{5}(\phi^{\frac12} - \phi^{-\frac12})^2} \; = \; \frac{3(7 + \sqrt{5})}{22}$ making the desired sum equal to $3 + \frac{3(7-\sqrt{5})}{22} + \frac{3(7+\sqrt{5})}{22} \; = \; \frac{54}{11}$ giving the answer $54 + 11 = \boxed{65}$ .

The polynomial $x^{4}-3x^{3}+4x^{2}-2x+1$ can be written as

$x^{4}-3x^{3}+4x^{2}-2x+1 = (x^{2}-Mx+N)(x^{2}-Rx+S)$ .

Looking at the coefficient of $x^{3}$ , we have $-M-R = -3$ . That is, $R = 3-M$ .

Looking at the coefficient of $x^{2}$ , we have $N+MR+S = 4$ . That is, $N+S = M^{2}-3M+4$ .

Looking at the coefficient of $x$ , we have $-NR-MS = -2$ . After some rearranging, we have $N = \dfrac{M^{3}-3M^{2}+4M-2}{2M-3}$ .

Looking at the constant coefficient, we have $NS = 1$ . That is, $S = \dfrac{1}{N}$ .

Then $N+\dfrac{1}{N} = M^{2}-3M+4$ .

$\dfrac{M^{3}-3M^{2}+4M-2}{2M-3}+\dfrac{2M-3}{M^{3}-3M^{2}+4M-2} = M^{2}-3M+4$

$(M^{3}-3M^{2}+4M-2)^{2}+(2M-3)^{2} = (M^{3}-3M^{2}+4M-2)(M^{2}-3M+4)(2M-3)$

After some rearranging, we have $M^{6}-9M^{5}+35M^{4}-75M^{3}+90M^{2}-54M+11 = 0$ .

This is a 6-degree polynomial whose roots are the sums of any two roots of the 4-degree polynomial in the first line.

That is, the roots of the equation $M^{6}-9M^{5}+35M^{4}-75M^{3}+90M^{2}-54M+11 = 0$ are $a+b$ , $a+c$ , $a+d$ , $b+c$ , $b+d$ , and $c+d$ .

Now we want to find the value of $\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{a+d}+\dfrac{1}{b+c}+\dfrac{1}{b+d}+\dfrac{1}{c+d} = \dfrac{p}{q}$ .

The required value is $\dfrac{(-1) \times \text{the coefficient of } M}{\text{the constant coefficient}}$ .

That is, $\dfrac{p}{q} = \dfrac{54}{11}$ .

Therefore, $p+q = 54+11 = 65$ .

$\text{FullSimplify}\left[\frac{1}{2} \text{Plus}\text{@@}\text{Table}\left[\frac{1}{\text{Plus}\text{@@}s},\left\{s,\text{Permutations}\left[x\text{/.}\, \text{Solve}\left[x^4-3 x^3+4 x^2-2 x+1=0\right],\{2\}\right]\right\}\right]\right] \Longrightarrow \frac{54}{11}$