An algebra problem by Farhabi Mojib

$\begin{aligned} P(m,n): \quad a_{m+n} + a_{m-n} & = \frac 12 (a_{2m}+a_{2n}) \end{aligned}$

$\begin{aligned} P(0,0): \quad a_0 + a_0 & = \frac 12 (a_0+a_0) \\ 2a_0 & = a_0 \\ \implies a_0 & = 0 \end{aligned}$

$\begin{aligned} P(1,0): \quad a_1 + a_1 & = \frac 12 (a_2+{\color{#3D99F6}a_0}) & \small \color{#3D99F6} \text{Note that }a_0 = 0 \\ 2{\color{#3D99F6}a_1} & = \frac 12 a_2 & \small \color{#3D99F6} \text{Note that }a_1= 1 \\ \implies a_2 & = 4 \end{aligned}$

$\begin{aligned} P(2,1): \quad a_3 + a_1 & = \frac 12 ({\color{#3D99F6}a_4}+a_2) & \small \color{#3D99F6} \text{Note that }a_4 = 4a_2 = 16 \\ a_3 + 1 & = \frac 12 ({\color{#3D99F6}16}+4) \\ \implies a_3 & = 9 \end{aligned}$

We note that $a_0=0=0^2$ , $a_1=1=1^2$ , $a_2=4=2^2$ , and $a_3=9=3^2$ . It appears that we can claim $a_k = k^2$ . Let us prove by induction that the claim is true for all nonnegative integers $k$ .

Proof:

It has been shown that the claim $a_k = k^2$ is true for $k=0$ and $k=1$ .

Assuming the claim is true for $k-1$ and $k$ , then we have:

$\begin{aligned} a_{k+1} + a_{k-1} & = \frac 12 \left(a_{2k} + a_2\right) \\ a_{k+1} + a_{k-1} & = \frac 12 \left(4a_k + 4 \right) \\ a_{k+1} + (k-1)^2 & = 2a_k + 2 \\ a_{k+1} + k2-2k+1 & = 2k^2 + 2 \\ a_{k+1} & = k^2 + 2k + 1 \\ \implies a_{k+1} & = (k+1)^2 \end{aligned}$

Therefore, the claim is true for $k+1$ and hence true for all nonnegative integers $k$ . $\square$

$\implies a_{2017} = 2017^2 = \boxed{4068289}$