Inspired by an undisclosed old problem

While it may not seem apparent at first, the inverse tangent function and Vieta's formulas are deeply intertwined. For those of you who haven't heard of Vieta's formulas, they are equations that relate the roots of a polynomial to that polynomial's coefficients and are extremely helpful in solving a variety of algebraic problems. To give you an idea on how I solved the problem, I'd like to start out by showing you a simpler example.

$\tan^{-1}(A)+\tan^{-1}(B) =\tan^{-1}\big(\frac{A+B}{1-AB}\big)$

This is known as the arctangent addition formula and it can be easily proven by taking the formula $\tan(X+Y)=\frac{\tan X+\tan Y}{1-\tan X\tan Y}$ and letting $X=\tan^{-1}(A)$ and $Y=\tan^{-1}(B)$ .

For a polynomial of the form $x^2+Sx+P$ with roots $A$ and $B$ , Vieta's formulas tell us the following:

$A+B=-S$

$AB=P$

Look at our formula for arctangent addition. We have $A+B$ and $AB$ on the RHS, so we can just plug in $-S$ and $P$ respectively to find our answer.

$\tan^{-1}(A)+\tan^{-1}(B)=\tan^{-1}\big(\frac{-S}{1-P}\big)$

This would be the solution for any 2nd degree polynomial, and obviously our equation is a much more complicated 5th degree polynomial, but our arctangent addition formula can be generalized to nth degree polynomials through the substitution of Vieta's formulas. Below, I present to you an equation that I have derived that is such a generalization:

For any $n$ th degree polynomial with coefficients $a_0.....a_n$ and roots $x_1.....x_n$ :

$\displaystyle\sum_{i=1}^n \tan^{-1}x_i = \tan^{-1}\Bigg(\displaystyle\frac{\displaystyle\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}a_{2k+1}(-1)^{k+1}}{\displaystyle\sum_{k=0}^{\lceil\frac{n-1}{2}\rceil}a_{2k}(-1)^{k}}\Bigg)$

Where $\lfloor x \rfloor$ and $\lceil x \rceil$ denote the floor and ceiling functions, respectively.

For a proof of this formula, please refer to this note .

I understand that this formula seems like a bit much to handle at first, but once broken down it's not too complicated to understand. What this formula says is that the sum of the arctangents of the roots of a polynomial can be written as one arctangent involving just the coefficients of the polynomial. The sums on the RHS can be calculated as follows:

1. Give a number to each coefficient from 0 to $n$ .

2. Take the alternating sum ( $a_0-a_2+a_4-a_6+...$ ) of all the even numbered coefficients.

3. Take the alternating sum of all the odd numbered coefficients and then negate your answer.

4. Take your answer from number 3 and divide it by your answer from number 2.

And that's it! This will give you the number inside your inverse tangent function on the RHS of the equation. Finally, we apply this formula to the question at hand:

$\displaystyle\sum_{m=1}^5 \tan^{-1}(x_m) = \tan^{-1}\Bigg(\frac{-(800-18+(-12))}{199-750+(-83)}\Bigg) = \tan^{-1}\bigg(\frac{385}{317}\bigg)$

And finally we take the tangent of this to get our answer:

$\tan\Bigg(\tan^{-1}\bigg(\frac{385}{317}\bigg)\Bigg) = \frac{385}{317}$

$\frac{385}{317}$ is irreducible so we're done. Taking our values for $a$ and $b$ and subtracting them gives us our final answer, $385-317=\boxed{68}$ .

My solution is more direct.

$\begin{aligned} \tan \left( \sum_{m=1}^\color{#D61F06}{2} \tan^{-1} x_m \right) & = \frac{x_1+x_2}{1-x_1x_2} \\ \tan \left( \sum_{m=1}^3 \tan^{-1} x_m \right) & = \frac{\frac{x_1+x_2}{1-x_1x_2}+x_3}{1-\frac{(x_1+x_2)x_3}{1-x_1x_2}} = \frac{x_1+x_2+x_3 -x_1x_2x_3}{1-(x_1x_2+x_2x_3+x_3x_1)} \\ & = \frac{\displaystyle \sum_{cyc}^3 x_1 - x_1x_2x_3}{1-\displaystyle \sum_{cyc}^3 x_1x_2} \\ \tan \left( \sum_{m=1}^4 \tan^{-1} x_m \right) & = \frac {\frac { \sum x_1 - x_1x_2x_3}{1-\sum x_1x_2} + x_4} {1 - \frac { (\sum x_1 - x_1x_2x_3)x_4}{1-\sum x_1x_2}} = \frac{\displaystyle \sum_{cyc}^4 x_1 - \sum_{cyc}^4 x_1x_2x_3}{1-\displaystyle \sum_{cyc}^4 x_1x_2 + x_1x_2x_3x_4} \\ \tan \left( \sum_{m=1}^5 \tan^{-1} x_m \right) & = \frac{ \frac{\sum x_1 - \sum x_1x_2x_3}{1 - \sum x_1x_2 + x_1x_2x_3x_4} + x_5 } {1 - \frac{(\sum x_1 - \sum x_1x_2x_3)x_5}{1 - \sum x_1x_2 + x_1x_2x_3x_4} } \\ & = \frac{\displaystyle \sum_{cyc}^5 x_1 - \sum_{cyc}^5 x_1x_2x_3 + x_1x_2x_3x_4x_5}{1-\displaystyle \sum_{cyc}^5 x_1x_2 - \sum_{cyc}^5 x_1x_2x_3x_4} \\ & = \frac{-\frac{800}{199} + \frac{18}{199} + \frac{12}{199}}{1 - \frac{750}{199} - \frac{83}{199}} = \frac{-770}{-634} = \frac {385}{317} \end{aligned}$

$\Rightarrow a - b = 385 - 317 = \boxed{68}$

(This might not work because $x_1, x_2, ..., x_5$ might be complex but this is interesting nonetheless)

Observe that
$\begin{aligned} \sum_{m = 1}^5 \tan^{-1} (x_m) = \sum_{m = 1}^5 \arg ( 1 + ix_m ) & = \arg \left (\prod_{m = 1}^5 (1+ix_m) \right ) \\ & = \arg\left (-i\prod_{m = 1}^5 (i - x_m) \right ) \end{aligned}$

Of course, $\prod_{m = 1}^5 (x - x_i) = \frac{\text{ the polynomial in the problem }}{199}$ , so we can plug in $x = i$ to get that the argument of our desired tangent is actually $\arg \frac{-634 - 770i}{199}$ or $\tan^{-1} \frac{385}{317}$ . Taking tangent we get the desired fraction and we can subtract to get $68$ .