Mathematical induction: $n^4 + n$

Question

Question: $n^4 + n$ is even for all Natural number.

I got all the way to where Basis step is $n=1$ and Inductive step is $k^4 + k$ (evens)

If my goal is to find $(k+1)^4 + (k+1)$, how would I get there?

Answer 1

$(k+1)^4 + (k+1) = k^4 + 4k^3 + 6k^2 + 4k + 1 + k + 1 = k^4 + k + 2(2k^3 + 3k^2 + 2k + 1)$ what is even, on condition that $k^4 + k$ is even (assumption in induction proof)

Answer 2

HINT

Note that

$$(k+1)^4+(k+1)=k^4+4k^3+6k^2+5k+2=(k^4+k)+4k^3+6k^2+4k+2$$

Answer 3

Just expand it out \begin{eqnarray*} (k+1)^4+(k+1)= \underbrace{k^4+k}_{\text{Even by the I.H.}}+\color{red}{4}k^3+\color{red}{6}k^2+\color{red}{4}k+\color{red}{2}. \end{eqnarray*}

Answer 4

Maybe it is an exercise, but induction is overkill here. Just remark that $n$ and $n^k$ have same parity for any integer $k$.

Thus their sum $n^k+n$ is even.

Answer 5

Indeed... what's the problem? If $n$ is even, then $n^4$ is even so $n^4 + n$ is even. If $n$ is odd, then $n^4$ is odd, so $n^4 + n$ is even.

$\square$