2016th term of the sequence

For the sake of convenience, I'm rewriting the notation of $a_n$ as $f(n)$ .

Given: $f(m+n) + f(m-n) = \dfrac{1}{2}(f(2m)+f(2n))$

Let $m+n = X, m-n = Y$

$\therefore 2m = X+Y, 2n = X-Y$

$\therefore f(X) + f(Y) = \dfrac{1}{2}(f(X+Y) + f(X-Y))$

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Subsituting $X = Y= 0:$

$2f(0) = f(0) \implies f(0) = 0$

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Substituting $X=Y=1:$

$2f(1) = \dfrac{1}{2}(f(2)+f(0)) \implies f(2) = 4$

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Subsituting $Y=1:$

$f(X) + f(1) = \dfrac{1}{2}(f(X+1) - f(X-1))$

$\implies f(X+1) = 2f(X) - f(X-1) + 2$

Substituting $X = 2$ gives $f(3) = 9$

It looks likes the following could be true: $f(X) = X^2 \forall X \in \mathbb{N} \cup {0}$

And this can be proved easily via induction, skipping the complete formal proof and just demonstrating the inductive step:

$f(X+1) = 2f(X) - f(X-1)+2$

$\implies f(X+1) = 2X^2 - (X-1)^2 + 2$

$\implies f(X+1) = 2X^2 - X^2 + 2X -1 +2 = X^2 + 2X +1 = (X+1)^2$

$\therefore f(X) = X^2 \implies a_{2016} = f(2016) = 2016^2 = \boxed{4064256}$