Question

Prove that the sum of the coefficients of the odd powers of $x$ is the expansion of $(1+x+x^2+x^3+x^4{)}^{n-1}$, when $n$ is a prime number other than $5$, is divisible by $n$.

Answer 1

${(1+x+x^2+x^3+x^4)}^{n-1}=1+c_{1}x+c_2{x}^2+c_3{x}^3+........\left(A\right){(1-x+x^2-x^3+x^4)}^{n-1}=1-c_{1}x+c_2{x}^2-c_3{x}^3+..........\left(B\right)$

Putting $x=1$ and subtracting $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow 5^{n-1}-1=2(c_1+c_3+c_5.......)\Rightarrow c_1+c_3+c_5......=\frac{5^{n-1}-1}{2}$

Using fermats theorem if $N$ is prime to $p$ then $N^{p-1}-1$ is divisible by $p$

$\Rightarrow 5^{n-1}-1$ is divisible by $n$ as $n$ is prime other then $5$

Hence proved.