Let it be $$(1+x+x^2+x^3+x^4)^{496}=a_0+a_1x+a_2x^2+...+a_{1984}x^{1984}.$$ Find $$\gcd(a_3,a_8,...,a_{1983})$$ I know the Leibniz Theorem, so all coefficients are of the form $$\frac{496!}{a!b!c!d!e!}1^a\cdot x^b\cdot x^{2c}\cdot x^{3d}\cdot x^{4e}$$ and $a+b+c+d+e=496$, $a,b,c,d,e\in\mathbb N$ So, if $d$ is the $\gcd$ of these numbers, $$d\mid\sum_{i=3}^{1983}a_i,$$ but I couldn't found the value of $d$.
Can someone help me?
Thanks for attention!