Vieta's + Newton's = Trouble

Consider factoring the equation as x(5x^4 + 4x^3 + 3x^2 + 2x + 1) = 0 and consider the complicated polynomial.

By Vieta's Theorem: We consider the polynomial equation x^4 + (4/5)x^3 + (3/5)x^2 + (2/5)x + (1/5) = 0.

ab + ac + ad + bc + bd + cd = sum of roots taken 2 at a time = 3/5

abc + abd + acd + bcd = sum of roots taken 3 at a time = -2/5

abcd = sum of roots taken 4 at a time = product of roots = 1/5

We use Newton's Identity where: a(n)S(2) + a(n-1)S(1) + 2a(n-2) = 0

Since a(n) = leading coefficient = 5 S(1) = -4/5 (By vieta) a(n-1) = 4 a(n-2) = 3

Solving for S(2) gives -14/25. And so a^2 + b^2 + c^2 + d^2 - abcd = (-14/25) - (1/5) = (-19/25). And so... (2/5)/(-19/25) = -10/19. Hence, the answer.

First factor the polynomial $5x^5 + 4x^4 + 3x^3 + 2x^2 + x = 0$ into $x(5x^4 + 4x^3 + 3x^2 + 2x + 1 = 0)$ . We may divide the polynomial by $x$ because we only need the nonzero polynomials.

Let $A = 5$ , $B = 4$ , $C = 3$ , $D = 2$ , and $E = 1$ (based on the coefficients of the polynomial)

By Vieta's Theorem,

$a + b + c + d = -B/A = \frac{-4}{5}$

$ab + ac + ad + bc + bd + cd = C/A = \frac{3}{5}$

$abc + abd + acd + bcd = -D/A =\frac{-2}{5}$

$abcd = E/A = \frac{1}{5}$

We cannot substitute the values directly into the equation because we do not have the value of $a^2 + b^2 + c^2 + d^2$ . But $a^2 + b^2 + c^2 + d^2$ is the sum of the squares of a polynomial of degree $4$ . So we can use the identity $(a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd)$ .

Using the identity (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd) ,

$(a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd)$

$a^2 + b^2 + c^2 + d^2 = (a + b + c + d)^2 - 2(ab + ac + ad + bc + bd + cd)$ .

Therefore if we substitute $(a + b + c + d)^2 - 2(ab + ac + ad + bc + bd + cd)$ to $a^2 + b^2 + c^2 + d^2$ , we can now use Vieta's Theorem directly into the expression.

By substitution,

$\dfrac{ab + ac + ad + bc + bd + cd + abc + abd + acd + bcd + abcd}{a^2 + b^2 + c^2 + d^2 - abcd}$

$\dfrac{ab + ac + ad + bc + bd + cd + abc + abd + acd + bcd + abcd}{(a + b + c + d)^2 - 2(ab + ac + ad + bc + bd + cd) - abcd}$

$= \dfrac{\frac{3}{5} - \frac{-2}{5} + \frac{1}{5}}{(\frac{-4}{5})^2 - 2(\frac{3}{5}) - \frac{1}{5}}$

$= \dfrac{\frac{2}{5}}{\frac{16}{25} - \frac{7}{5}}$

$= \dfrac{\frac{2}{5}}{\frac{-19}{5}}$

$= \dfrac{-10}{19}$

Therefore $m = 10$ and $n = 19$

So $m + n = 10 + 19 = \boxed{29}$

I don't know how can such a problem reach level 5!!!