<s>[INST] Prove that $\max\{a_i \mid i \in \{0,1,\ldots,1984\}\} = a_{992}$ 
Consider the expansion
  $$\left(1 + x + x^{2} + x^{3} + x^{4}\right)^{496} =
a_{0} + a_{1}x + \cdots + a_{1984}\,\,x^{1984}
$$
  Prove that
  $\max\left\{a_{i} \mid i \in \left\{0,1,\ldots,1984\right\}\right\} =
a_{992}\ $.

Since $1 + x + x^{2} + x^{3} + x^{4}$ is a palindromic polynomial, so is $\left(1 + x + x^{2} + x^{3} + x^{4}\right)^{496}$. Now we want to show that $a_{1},a_{2},\ldots,a_{1984}$ has exactly one maximum and we are done.
 [/INST] As you have noticed, for $(1 + x + x^2 + x^3 + x^4)^n = a_0 + a_1x + \cdots + a_{4n}x^{4n}$, we have
$$
a_i = a_{4n-i}\quad\text{for } 0 \leq i \leq 4n
$$
So it is sufficient to prove that $a_0 < a_1 < \cdots < a_{2n}$. We prove by induction below.

Basis. For $n = 2$, we have
$$
(1 + x + \cdots + x^4)^2 = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + 4x^5 + 3x^6 + 2x^7 + x^8
$$
Thus $a_0 < a_1 < \cdots < a_4$.

Induction. Suppose for $n \geq 2$ we have
$$
(1 + x + \cdots + x^4)^n = a_0 + a_1x + \cdots + a_{4n}x^{4n}
$$
with $a_0 < a_1 < a_2 < \cdots < a_{2n}$ and
$$
(1 + x + \cdots + x^4)^{n+1} = b_0 + b_1x + \cdots + b_{4n+4}x^{4n+4}
$$
We aim to prove that $b_0 < b_1 < \cdots < b_{2n+2}$. Note that
$$
(1 + x + \cdots + x^4)^{n+1} = (1+x+\cdots + x^4)^n\cdot(1+x+\cdots +x^4)
$$
Therefore,
\begin{align}
b_i &= a_i + a_{i-1} + a_{i-2} + a_{i-3} + a_{i-4}\quad \text{for}\quad 4\leq i \leq 2n + 2\\
b_3 &= a_3 + a_2 + a_1 + a_0 \\
b_2 &= a_2 + a_1 + a_0 \\
b_1 &= a_1 + a_0 \\
b_0 &= a_0
\end{align}
Easy to observe that $b_0 < b_1 < b_2 < b_3 < b_4$. For $4 < i \leq 2n$, we have
$$
b_{i} - b_{i-1} = a_i - a_{i-5} > 0
$$
by our induction assumption. Moreover, we have
$$
b_{2n + 1} - b_{2n} = a_{2n+1} - a_{2n-4} = a_{2n-1} - a_{2n-4} > 0
$$
and
$$
b_{2n + 2} - b_{2n+1} = a_{2n + 2} - a_{2n - 3} = a_{2n - 2} - a_{2n - 3} > 0
$$
Therefore,
$$
b_0 < b_1 < \cdots < b_{2n + 2}
$$

When $n = 496$, by our proof, $a_{992}$ is maximum.
</s>