Show that 16 divides $ n^4 -1 $ whenever n is an odd number

Question

I want to show that 16 divides $ n^4 -1$ whenever n is odd

I started using the definition of odd with $ n=2k+1$ and got $ n^4=(2k+1)^4=16 k^4+32 k^3+24 k^2+8 k+1$ which means that $n^4-1 = 16 k^4+32 k^3+24 k^2+8 k$

But I am not sure how to go from here or if this is even the correct approach.

Answer 1

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

Since $n$ is odd, then $n=2k+1$, then you can see it as:

$$(n^2+1)(n+1)(n-1)=((2k+1)^2+1)(2k+2)(2k)=(4k^2+4k+2)(2k+2)(2k)=8(k^2+k+1)(k+1)k$$

But see that $k$ and $k+1$ are consecutive, so one of them has the fourth factor $2$ and that proves that $16\mid n^4-1$.

Answer 2

Use congruences: modulo $16$, an odd integer is congruent to $\pm1, \pm 3,\pm 5,\pm 7$, hence its square is congruent to $1^2=1,\; 3^2\equiv -7, 5^2\equiv 9\equiv -7,\;7^2\equiv 1$, so the 4-th powers are congruent to $1,\;7^2\equiv 1,\;9^2\equiv 1,\;1^2=1$.

Answer 3

Hint: we can $\rm\color{#c00}{double}$ an $\rm\color{#0a0}{even\ modulus}$ when $\rm\color{#c00}{squaring}$ a congruence, by the Lemma below.

Thus odd $\,n\equiv \pm1\pmod{\!4}\Rightarrow n^2\equiv 1^2\pmod{\!8}\Rightarrow n^4\equiv 1^4\pmod{\!16}. \ \ \small\bf QED$

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Lemma $\ \ a\equiv b\pmod{\!\color{#0a0}{2n}}\,\Rightarrow\, a^{\large\color{#c00}2}\equiv b^{\large \color{#c00}2}\pmod{\!\color{#c00}2\cdot\color{#0a0}{2n}}\ $

Proof $\:\ \color{#0a0}{2\mid 2n\mid a\!-\!b}\,\Rightarrow\,\color{#90f}{2\mid a\!+\!b}\,\Rightarrow\,\color{#90f}2\cdot\color{#0a0}{2n}\mid(\color{#0a0}{a\!-\!b})(\color{#90f}{a\!+\!b})=a^{\large 2}-b^2$

Remark $ $ More generally the Lemma is a special case of LTE = Lifting The Exponent.