Four Variables

Let $f(x) = \dfrac{x_1}{x^{2}+1} + \dfrac{x_2}{x^{2}+2} + \dfrac{x_3}{x^{2}+3} + \dfrac{x_4}{x^{2}+4}$ .

Then $f(\pm 1) = 1$ , $f(\pm 2) = \dfrac{1}{4}$ , $f(\pm 3) = \dfrac{1}{9}$ , and $f(\pm 4) = \dfrac{1}{16}$ .

The expression to be determined is $f(5)$ .

Let $p(x) = (x^{2}+1)(x^{2}+2)(x^{2}+3)(x^{2}+4)$ and $q(x) = p(x)f(x)$ .

Then for $n = 1,2,3,4$ , we have $q(n) = p(n) \times \dfrac{1}{n^{2}}$ .

That is, $p(n) - n^{2}q(n) = 0$ .

We have $p(x)-x^{2}q(x) = 0$ is a 8-degree polynomial equation and its roots are $\pm 1$ , $\pm 2$ , $\pm 3$ , and $\pm 4$ .

That is, $p(x)-x^{2}q(x) = C(x^{2}-1)(x^{2}-4)(x^{2}-9)(x^{2}-16)$ , where $C$ is a real number.

Putting $x = 0$ , we have $C = \dfrac{p(0)}{(-1) \times (-4) \times (-9) \times (-16)} = \dfrac{1 \times 2 \times 3 \times 4}{1 \times 4 \times 9 \times 16} = \dfrac{1}{24}$ .

So $p(x)-x^{2}q(x) = \dfrac{1}{24}(x^{2}-1)(x^{2}-4)(x^{2}-9)(x^{2}-16)$ .

Dividing by $p(x)$ , we have

$1 - x^{2}f(x) = \dfrac{1}{24} \dfrac{(x^{2}-1)(x^{2}-4)(x^{2}-9)(x^{2}-16)}{(x^{2}+1)(x^{2}+2)(x^{2}+3)(x^{2}+4)}$ .

Putting $x = 5$ , we have

$1-25f(5) = \dfrac{1}{24} \dfrac{24 \times 21 \times 16 \times 9}{26 \times 27 \times 28 \times 29} = \dfrac{2}{377}$ .

So $\dfrac{x_1}{26} + \dfrac{x_2}{27} + \dfrac{x_3}{28} + \dfrac{x_4}{29} = f(5) = \dfrac{15}{377} = \dfrac{p}{q}$ .

Therefore, $p+q = 15+377 = 392$ .