Question

show that n^4-n^3+n^2-n is divisible by 2 for all positive integers

Answer 1

Given  $f(n)=n^4-n^3+n^2-n$. Group terms as shown.

$f(n)=n^4-n-(n^3-n^2)\\</p><p>f(n)=n(n^3-1)-n^2(n-1)\\</p><p>f(n)=n(n^2+n+1)(n-1)-n^2(n-1)\\</p><p>f(n)=n(n-1)[(n^2+n+1)-n]\\</p><p>f(n)=n(n-1)[n^2+1]$

Now note that $n(n-1),n=1,2,3,4,...$ is the product of two consecutive integers. So $n(n-1)$ is divisible by 2. hence the product $f(n)=n(n-1)[n^2+1]$ is also divisible by 2.  $f(n)=n^4-n^3+n^2-n$ is divisible by 2.

The proof is complete.