Mathematical induction proof $n^4-1$ is divisible by $16$ for all odd integers $n$

Question

I'm stuck towards the end of proving this, here's my attempt:

$P(3) = 80/16 = 5$, True

$P(k) = k^4-1$

$P(k+1)= (k+1)^4-1$

Expanded $= k^4+4k^3+6k^2+4k+1-1$

This is where I am stuck at.

Sorry for the sloppy formatting im still reading how to properly write formulas on this site.

Thanks in advance for your help!

Answer 1

If $k$ is an odd integer, then next odd integer is $k+2$, not $k+1$. (why?)

You need to prove $P(k+2)=(k+2)^4-1$ is divisible by $16$: $$\begin{align}(k+2)^4-1&=k^4+8k^3+24k^2+32k+16-1 =\\ &=(k^4-1)+16(2k+1)+8k^2(k+3)\end{align}$$ Note:

1) $k^4-1$ is divisible by $16$ by inductive hypothesis.

2) $16(2k+1)$ is obviously divisible by $16$.

3) $8k^2(k+3)$ is divisible by $16$, because $k+3$ is even. (why?)

Answer 2

hint

It is true for $n=1$.

assume it is true for some odd $n\ge 1$.

let us prove it is true for $n+\color{red}{2}$.

$$(\color{green}{(n+2)^4-1})-(n^4-1)$$ $$=(n+2)^4-n^4$$ $$=((n+2)^2-n^2)((n+2)^2+n^2)$$

$$=8(n+1)(n^2+2n+2)$$

but $$n=2k+1$$ and $$n^4-1=16p$$ with $p,k$ integers. thus

$$\color{green}{(n+2)^4-1}=$$ $$16p+16(k+1)(n^2+2n+2)$$ nearly done.

Answer 3

Without loss of generality we proved the statement for positive odd integers.

Note that a positive odd integer has the form of $2k-1$ where $k\ge 1$

We proceed with induction on k.

Let $P(k)$ be $(2k-1)^4-1$ is a multiple of $16$

For $k=1$, $(2k-1)^4-1=0$ which is a multiple of $16$ so $P(1)$ is true.

Assume that $(2k-1)^4-1$ is a multiple of $16$

We need to show that $(2k+1)^4-1$ is also a multiple of $16$

Note that the binomial theorem implies $$(2k+1)^4 = [(2k-1)+2]^4 = (2k-1)^4+8(2k-1)^3 +24(2k-1)^2+32(2k-1)$$

Thus $$(2k+1)^4 - (2k-1)^4 =8(2k-1)^3 +24(2k-1)^2+32(2k-1)=$$

$$ 8(2k-1)[(2k-1)^2+3(2k-1)+4)]$$

Note that the expression in square bracket is an even integer, so the result is multiple of $16$

Answer 4

Hint $ $ a $1$-line proof shows $\ 8\mid n^2\!-\!1\,\Rightarrow\, 16\mid n^4\!-\!1.\,$ But $\,8\mid n^2\!-\!1$ for odd $\,n\,$ inducts easily

$$\qquad\ {8\mid \color{#c00}{n^2\!-\!1}}\ \Rightarrow\ {8\mid (n\!+\!2)^2-1} = \color{#c00}{n^2\!-\!1} + 4(\underbrace{n\!+\!1}_{\large\rm even})$$