Let $f : Z_{\ge 0} \to Z_{\ge 0}$ be a function which satisfies for all integer $n \ge 0$: (a) $f(2n + 1)^2 - f(2n)^2 = 6f(n) + 1$, (b) $f(2n) \ge f(n)$ where $Z_{\ge 0}$ is the set of nonnegative integers. Solve the equation $f(n) = 1000$
Source: 2013 Saudi Arabia BMO TST III p4
Tags: algebra, recurrence relation, function
Let $f : Z_{\ge 0} \to Z_{\ge 0}$ be a function which satisfies for all integer $n \ge 0$: (a) $f(2n + 1)^2 - f(2n)^2 = 6f(n) + 1$, (b) $f(2n) \ge f(n)$ where $Z_{\ge 0}$ is the set of nonnegative integers. Solve the equation $f(n) = 1000$