Question

Prove that $n^7-7n^5+14n^3-8n$ is divisible by $840$ for all $n\in N$

Answer 1

$n^7-7n^5+14n^3-8^n$

$=n(n-1)(n+1)(n^4-6n^2+8)$

$=n(n-1)(n+1)(n^2-2)(n^2-4)$

$=(n-2)(n-1)\left(n\right)(n+1)(n+2)(n^2-2)$

This expression is a product of $5$ consecutive terms. Hence the given expression is divisible by $5!=120$

We need further to prove that the given expression is divisible by $7$.

Let $n=7a+k$ where $k=0,1,2,3,4,5,6$

$n=7a+0$

$⟹n$ is divisible by $7$

$n=7a+1$

$⟹n-1$ is divisible by $7$

$n=7a+2$

$⟹n-2$ is divisible by $7$

$n=7a+5$

$⟹n+2$ is divisible by $7$

$n=7a+6$

$⟹n+1$ is divisible by $7$

$n=7a+3$

$n^2-2=(7a+3{)}^2-2$

$=7p+q-2=7q$

$n=7a+4$

$n^2-2=(7a+4{)}^2-2$

$=7r+16-2=7r$

Hence, the given expression is divisible by $120\times 7=840$.