Question

If $n$ is an odd number, prove that $n^6+3n^4+7n^2-11$ is a multiple of $128$.

Answer 1

Let any odd number be written as $2n+1$.

$(n^6+3n^4+7n^2-11)=(2n+1{)}^6+3(2n+1{)}^4+7(2n+1{)}^2-11=64n(n+1\left)\right(n+2n^2+4n^3+5n^4+3n^5+n^6)$

$n$ and $n+1$ are two consecutive terms.

Hence, one of them would be divisible by $2$.

Hence, the total number is divisible by $128$.