Prove by induction that $n^4+2n^3+n^2$ is divisible by 4

Question

I'm trying to prove by induction that $n^4+2n^3+n^2$ is divisible by 4.

I know that P(1) it's true. Then $ n=k, P(k):k^4+2k^3+k^2=4w$ it's true by the hypothesis of induction. When I tried to prove $n=k+1, P(k+1):(k+1)^4+(k+1)^3+(k+1)^2 = 4t$,

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

Can I replace $(k^4+2k^3+k^2)$ of the previous expression by $4w$ and then continue the proof? Is this correct? Thanks.

Answer 1

That method will work fine, but you can simplify it by noting that $ f(n) = n^2(n+1)^2$ therefore $\,n,$ or $n+1$ is even, so $4$ divides its square, so also $f(n)$ (provable directly or by induction if need be).

Answer 2

As an alternative to proving via induction, consider the following hints:

Hint #1: $n^4 + 2n^3 + n^2 = n^2(n^2 + 2n + 1) = \bigg(n(n+1)\bigg)^2$

Hint #2: The parity of $n$ is different from that of $n+1$.

Answer 3

You are not doing it wrongly but it's easier, with $f(n)=n^4+2n^3+n^2$ , to observe that $f(n)=(n+1)^2n^2,$ so $$f(n+1)-f(n)=(n+2)^2(n+1)^2-(n+1)^2n^2=$$ $$=(n+1)^2 ((n+2)^2-n^2)=$$ $$=(n+1)^2(4n+4)=4(n+1)^3$$ which is a multiple of $4$.

Answer 4

You don't need induction to prove this.

$f(n)=n^4+2n^3+n^2 = n^2\cdot(n^2+2n+1) = n^2\cdot(n+1)^2$

Case 1: If $n$ is even, $n=2k, n^2=2k\cdot2k=4k^2$, now $4k^2\cdot(n+1)^2$, which is obvious that is divisible by 4

Case 2: If $n$ is odd then $n+1$ is even, let $m=n+1$, $m=2k, \ m^2=2k\cdot2k = 4k^2$, if we change in our first formula $n^2\cdot4k^2$, which is obvious that is divisible by 4.