Question on the coefficients of $(1+x+x^2+x^3+x^4)^{496}$ 
Consider the expansion $$(1+x+x^2+x^3+x^4)^{496} = a_0+a_1x+\cdots+a_{1984}x^{1984}.$$
  $\quad$ (a) Determine the greatest common divisor of the coefficients $a_3,a_8,a_{13},\ldots,a_{1983}$.
$\quad$ (b) Prove that $10^{340} < a_{992} < 10^{347}$.

Is there an easier way to solve this and is there a formula for the multisection for just the sum of the coefficients and not including $x^k$?
I thought of using the Multisection formula to prove (a). That is, $$\sum_{k \equiv r \pmod{m}}a_kx^k = \dfrac{1}{m} \sum_{s=0}^{m-1} \epsilon^{-rs} f(\epsilon^s x)$$ where $\epsilon$ is a primitive $m$th root of unity and $\displaystyle f(x) = (1+x+x^2+x^3+x^4)^{496}$.
Thus we have $r = 3, m = 5$ and so \begin{align*}\sum_{k \equiv 3 \pmod{5}}a_kx^k &= \frac{1}{5} \sum_{s=0}^{4} w^{-3s} (1+w^s x+(w^{s}x)^2+(w^{s} x)^3+(w^{s}x)^4)^{496}\\&=\dfrac{1}{5}\sum_{s=0}^4 w^{-3s}\left(\dfrac{(w^s x)^{5}-1}{w^s x-1}\right)^{496}\\&= \dfrac{1}{5}\sum_{s=0}^4 w^{-3s} \left(\dfrac{x^{5}-1}{w^s x-1}\right)^{496}.\end{align*}

About point $(b)$, we may notice that $992=\frac{1984}{2}$, hence $a_{992}$ is the largest coefficient (we are dealing with palyndromic polynomials) and
$$ a_{992} = \frac{1}{2\pi}\int_{-\pi}^{\pi}\left(e^{-2iz}+e^{-iz}+1+e^{iz}+e^{2iz}\right)^{496}\,dz $$
is a real integral not so difficult to approximate:
$$ a_{992} = \frac{5^{496}}{2\pi}\int_{-\pi}^{\pi}\left(\frac{1+2\cos(z)+2\cos(2z)}{5}\right)^{496}\,dz $$
gives:
$$ a_{992} \approx \frac{5^{496}}{2\pi}\int_{-\infty}^{+\infty}e^{-496z^2}\,dz = \frac{5^{496}}{8\sqrt{31\pi}}$$ 
so $a_{992}$ is between $10^{\color{red}{344}}$ and $10^{\color{red}{345}}$.

About point $(a)$, given $f(x)=\left(\frac{1-x^5}{1-x}\right)^{496}=\sum_{n\geq 0}a_n x^n$, by the discrete Fourier transform:
$$ \sum_{k\equiv 3\!\pmod{5}}\!\!\!a_k x^k = \frac{1}{5}\sum_{s=0}^{4}w^{-3s}\left(\frac{x^5-1}{w^s x-1}\right)^{496}=\frac{(x^5-1)^{496}}{5}\sum_{s=0}^{4}\frac{w^{2s}}{(w^s x-1)^{496}} $$
so:
$$ \sum_{k\equiv 3\!\pmod{5}}\!\!\!a_k x^k =(1-x^5)^{496}\sum_{j\geq 0}\binom{495+5j+3}{5j+3}x^{5j+3}.$$
If we have $g(x)=\sum_{n\geq 0}g_n x^n$, then $\frac{g(x)}{1-x}=\sum_{n\geq 0}G_n x^n$, with $G_n=\sum_{k=0}^{n}g_k$. The sequences $\{g_n\}_{n\geq 0}$ and $\{G_n\}_{n\geq 0}$ have the same greatest common divisor, hence in our case it is enough to find the greatest common divisor of the sequence
$$ \left\{\binom{495+5j+3}{495}\right\}_{j\geq 0}. $$
The $\gcd$ is preserved by the forward difference operator, and the $j$-th term of our sequence is a polynomial in $j$ with degree $495$, with leading term $C_{495}\cdot j^{495}$. The $\gcd$ of the whole sequence is so
$$ \gcd\left(C_{495},\binom{495+3}{495}\right) = \color{red}{2\cdot 31\cdot 71\cdot 83}=365366.$$
