Recursion in 2018

We will prove the following proposition first:

Proposition : Given a sequence of real numbers $\{a_n\}$ that obeys the recursive formula $a_n = c_1a_{n - 1} + c_2a_{n - 2}$ with initial conditions $a_0 = 0, a_1 = 1,$ we have

$a_w a_x - a_y a_z = (-c_2)^r(a_{w - r} a_{x - r} - a_{y - r} a_{z - r}),$

where $w, x, y, z, r$ are positive integers such that $w + x = y + z.$

Proof : Consider the matrix

$M = \begin{pmatrix} c_1 & c_2 \\ 1 & 0 \end{pmatrix}.$

It can be shown through induction that

$M^k = \begin{pmatrix} a_{k + 1} & c_2a_k \\ a_k & c_2a_{k - 1} \end{pmatrix},$

for all positive integers $k.$ Since $w + x = y + z,$ we can write $M^wM^x = M^yM^z,$ which can be expanded into

$\begin{aligned} \begin{pmatrix} a_{w + 1} & c_2a_w \\ a_w & c_2a_{w - 1} \end{pmatrix} \begin{pmatrix} a_{x + 1} & c_2a_x \\ a_x & c_2a_{x - 1} \end{pmatrix} &= \begin{pmatrix} a_{y + 1} & c_2a_y \\ a_y & c_2a_{y - 1} \end{pmatrix} \begin{pmatrix} a_{z + 1} & c_2a_z \\ a_z & c_2a_{z - 1} \end{pmatrix} \\ \begin{pmatrix} a_{w + 1} a_{x + 1} + c_2a_wa_x & \cdots \\ \cdots & \cdots \end{pmatrix} &= \begin{pmatrix} a_{y + 1}a_{z + 1} + c_2a_ya_z & \cdots \\ \cdots & \cdots \end{pmatrix} \\ a_{w + 1} a_{x + 1} + c_2a_wa_x &= a_{y + 1}a_{z + 1} + c_2a_ya_z \\ a_{w + 1} a_{x + 1} - a_{y + 1}a_{z + 1} &= -c_2(a_wa_x - a_ya_z). \end{aligned}$

Iterating this formula $r$ times and decreasing the initial indices by one yields $a_w a_x - a_y a_z = (-c_2)^r(a_{w - r} a_{x - r} - a_{y - r} a_{z - r}). \, \blacksquare$

We can apply this result directly to solve the problem. From the problem statement, we have $c_2 = 1$ and $a_{2018}a_{2015} - a_{2017}a_{2016} = 2018.$ Since, 2018 + 2015 = 2017 + 2016,

$a_{2018}a_{2015} - a_{2017}a_{2016} = (-1)^{2015}(a_3a_0 - a_2a_1) = -(a_3(0) - a_2(1)) = a_2,$

and $a_2 = 2018.$ However, we also have $a_2 = ca_1 + a_0 = c(1) + 0 = c,$ giving us $c = \boxed{2018}$ as the only possible value of $c$ that will satisfy the condition.

(Apologies for anyone who was trying the earlier versions of this problem before I deleted them. That's what I get for staying up until 3 in the morning writing math problems.)