An algebra problem by Prem Chebrolu

If we subtract the given equations, we get

$\begin{array}{rlrrrl} p^2 - q^2 &= \ \ q - p \\ (p + q)(p - q) &= \ \ \text{-}(p - q) \\ p + q &= \ \ \text{-}1 & & \small (1) & & \small [ \ p, q \text{ are distinct, so } p - q \ne 0 \ ] \\ p^2 + 2pq + q^2 &= \ \ 1 & & \small (2) \end{array}$

Now add the original equations:

$\begin{array}{rlrrrl} p^2 + q^2 &= \ \ p + q + 4010 \\ p^2 + q^2 &= \ \ 4009 & & \small (3) & & \small \text{[ Substituting } (1) \ ] \\ 2pq &= \ \ \text{-}4008 & & & & \small \text{[ Subtracting } (2) - (3) \ ] \\ pq &= \ \ \text{-}2004 \end{array}$

so our answer is $pq + 2014 = \boxed{10}$

We can write this system as

$q^{2} - p = 2005, p^{2} - q = 2005 \Longrightarrow p^{2} - q = q^{2} - p \Longrightarrow p^{2} - q^{2} = q - p \Longrightarrow (p - q)(p + q) = -(p - q)$ .

Now since $p,q$ must be distinct we must have that $p + q = -1 \Longrightarrow q = -(1 + p)$ . Substituting this result into $2005 + p = q^{2}$ yields

$2005 + p = (-(1 + p))^{2} = 1 + 2p + p^{2} \Longrightarrow p^{2} + p - 2004 = 0$ . Similarly we can find that $q^{2} + q - 2004 = 0$ .

Thus $p,q$ are the two roots of the equation $x^{2} + x - 2004 = 0$ , and so by Vieta's we have that $pq = -2004$ , and so $2014 + pq = 2014 - 2004 = \boxed{10}$ .

Note that these roots $p,q$ are $\dfrac{-1 \pm \sqrt{8017}}{2}$ .