Given the equation $(1+x+x^2+x^3+x^4)^{496}=a_0+a_1x+a_2x^2+...+a_{1984}x^{1984}$
a) What is the gcd of $(a_1,a_2,a_3,... a_{1983})$?
b) Show that $10^{340}< a_{992}< 10^{347}$.
By the Leibinz theorem, I know that the general term of this polynomial is $\frac{496!}{a!b!c!d!e!}\cdot x^{b+2c+3d+4e}$.
And $a+b+c+d+e=496$.
So, the first answer is 496, I think, but how can I solve the question b?
For the first part:
Taking $x=1$ gives us $$5^{496}=\sum_{i=0}^{1984}a_i=2+\sum_{i=1}^{1983}a_i$$
Hence, if $d=\gcd (a_1, \cdots, a_{1983})$ then $d\,|\,(5^{496}-2)$. Since $a_1=496$ we see that $d\,|\,\gcd(5^{496}-2, 496)$.
Now $496=2^4\times 31$ and, since $5^{496}-2$ is clearly odd we see that $\gcd(5^{496}-2, 496)\in \{1,31\}$. It is easy to compute that $5^{496}-2\equiv 3 \pmod {31}$ so we must have $$\gcd(5^{496}-2, 496)=1$$ hence $d=1$.