Any solutions anyone ?

utkarshgupta wrote: Consider the polynomial $P_n(x) = \binom {n}{2}+\binom {n}{5}x+\binom {n}{8}x^2 + \cdots + \binom {n}{3k+2}x^{3k}$ where $n \ge 2$ is a natural number and $k = \left\lfloor \frac{n-2}{3} \right \rfloor$ (a) Prove that $P_{n+3}(x)=3P_{n+2}(x)-3P_{n+1}(x)+(x+1)P_n(x)$ (b) Find all integer numbers $a$ such that $P_n(a^3)$ is divisible by $3^{ \lfloor \frac{n-1}{2} \rfloor}$ for all $n \ge 2$ I assume that you meant to write $P_n(x) = \binom {n}{2}+\binom {n}{5}x+\binom {n}{8}x^2 + \cdots + \binom {n}{3k+2}x^{k}$. Then for part (a) it suffices to prove that for any $m$ both sides have the same coefficient of $x^m$ i.e. we shall prove that $\binom{n+3}{3m+2}-3\binom{n+2}{3m+2}+3\binom{n+1}{3m+2}-\binom{n}{3m+2}-\binom{n}{3m-1}=0$. Using $\binom{n+2}{3m+2}-\binom{n+1}{3m+2}=\binom{n+1}{3m+1}$ this is equivalent to $\binom{n+3}{3m+2}-3\binom{n+1}{3m+1}-\binom{n}{3m+2}-\binom{n}{3m-1}=0$. But we have $LHS=\left(\binom{n+1}{3m+2}+2\binom{n+1}{3m+1}+\binom{n+1}{3m}\right)-3\binom{n+1}{3m+1}-\binom{n}{3m+2}-\binom{n}{3m-1}$ $LHS=\binom{n+1}{3m+2}-\binom{n+1}{3m+1}+\binom{n+1}{3m}-\binom{n}{3m+2}-\binom{n}{3m-1}$ $LHS=\left(\binom{n}{3m+2}+\binom{n}{3m+1}\right)-\left( \binom{n}{3m+1}+\binom{n}{3m}\right)+\left(\binom{n}{3m}+\binom{n}{3m-1}\right)-\binom{n}{3m+2}-\binom{n}{3m-1}=0$ For part (b) just note that this implies $9|P_5(a^3)=a^3+10$ i.e. $9|2a^3+20$ and $27|P_6(a^3)=6a^3+15$ i.e. $9|2a^3+5$ and in comparison we obtain $9|15$. Absurd! Hence no such $a$ can exist.