Vieta's formula relates the coefficients of polynomials to the sum and products of their roots. This can provide a shortcut to finding solutions in more complicated algebraic polynomials.

If the roots of \(p(x) = x^3 + 3x^2 + 4x - 8\) are \(\color{red}{a}\), \(\color{blue}{b}\) and \(\color{purple}{c}\), what is the value of

\[\color{red}{a}^2 \big(1 + \color{red}{a}^2\big) + \color{blue}{b}^2 \big(1 + \color{blue}{b}^2\big) + \color{purple}{c}^2 \big(1 + \color{purple}{c}^2\big)?\]

Consider all pairs of non-zero integers \( (a,b) \) such that the equation

\[ ( ax-b)^2 + (bx-a)^2 = x \]

has at least one integer solution.

The sum of all (distinct) values of \(x\) which satisfy the above condition can be written as \( \frac{ m}{n} \), where \(m\) and \(n\) are coprime positive integers. What is the value of \(m + n \)?

Let \(x_1,x_2,\ldots,x_{2015} \) be the roots of the equation \[ x^{2015} + x^{2014} + x^{2013} + \ldots + x^2 + x + 1 =0.\] Evaluate

\[ \frac1{1-x_1} + \frac1{1-x_2} + \ldots + \frac1{1-x_{2015}}. \]

What is the largest integer \( n \leq 1000 \), such that there exist 2 non-negative integers \((a, b)\) satisfying

\[ n = \frac{ a^2 + b^2 } { ab - 1 } ? \]

**Hint**:

\( (a,b) = (0,0) \) gives us \( \frac{ 0^2 + 0^2 } { 0 \times 0 - 1 } = 0\), so the answer is at least \( 0 .\)

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