Prove by mathematical induction that $n^4 − n^3 + n^2 − n$ is divisible by $2$ for all positive integers $n$.

Question

So I've done the base case with $n = 1$ and the general case of $k^4 − k^3 + k^2 − k = 2m$ but I can't seem to find a way to manipulate $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)$ in a manner such that I can substitute in the $2m$. Any help?

Answer 1

Hidden induction (almost directly):

$$n^4-n^3+n^2-n=n^3(n-1)+n(n-1)=n(n^2+1)(n-1)$$

and since either $\;n\;$ or $\;n-1\;$ is even...

Answer 2

Mod 2, $n$ takes only the two values 0 and 1 and so $n^k\equiv n\;[2]$. Hence

$$n^4-n^3+n^2-n\equiv n-n+n-n=0\;[2]$$

Answer 3

Claim

$n^4 − n^3 + n^2 − n$ is even for $n=1,2,\cdots.$

An Inductive Proof

Let $n=1$. Then $n^4 − n^3 + n^2 − n=0$,which is even. Thus, the claim holds for $n=1.$ Assume that the claim holds for $n=k$,namely, $k^4 − k^3 + k^2 − k$ is even. Then \begin{align*} (k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)=(k^4 − k^3 + k^2 − k)+4k^3+3k(k+1).\end{align*}

Here is a sum of three even numbers. Therefore, it is even as well, which implies that the claim also holds for $n=k+1$. As a result, by the inductive principle, the claim holds for all $n$. The proof is completed.

Answer 4

Express $(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)$ in terms of $k^4 - k^3 + k - k$.

$(k+1)^4 − (k+1)^3 + (k+1)^2 − (k+1)=$

$k^4 + (4k^3 + 6k^2 + 4k+1) - k^3 -(3k^2 + 3k + 1) + k^2 + (2k + 1) - k - (1) =$

$(k^4 - k^3 +k^2 - k) + (4k^3 + 6k^2 + 4k) -(3k^2 + 3k) + (1-1+1-1) =$

$(k^4 - k^3 + k^2 - k) +(4k^3 + 6k^2 + 4k) - (3k^2 + 3k)$.

As $k^4 - k^3 + k^2 -k$ is even and $4k^3 + 6k^2 + 4k$ is even then this is even if and only if $3k^2 + 3k$ is even.

And we prove $3k^2 + 3k$ is even the exact same process.

Base case: $k = 0$ then $3k^2 + 3k = 0$ is even.

Induction: Assume $3k^2 + 3k$ is even.

Express $3(k+1)^2 + 3(k+1)$ in terms of $3k^2 + 3k$.

$3(k+1)^2 + 3(k+1) = 3k^2 +(6k +3) + 3k + (3) = (3k^2 + 3k) + (6k + 6)$

And as both $(3k^2 + 3k)$ and $(6k+6)$ are both even, so is $(3k^2 + 3k) + (6k + 6)$