Odd divisibility induction proof

Question

Prove that for odd $n>3$

$$64\ | \ n^4-18n^2+17$$

I checked that for $n=5$ it works. I think I need to assume that for $2n+1$ it holds and show that $2n+3$ also holds. Any ideas?

Answer 1

we have: $$n^4-18n^2+17+64=(n^2-9)^2 $$

and because $(n^2-9)=(n-3)(n+3)$ is divisible by $8$ for odd numbers we can conclude.

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By induction:

Assume that $n^4-18n^2+17=(n^2-9)^2-64$ is divisible by $64$ for an odd number $n$ we want to prove that this still true for $n+2$, we have $$\begin{align}(n+2)^4-18(n+2)^2+17+64&= ((n+2)^2-9)^2\\ &=(n^2-9+4(1+n))^2 \\ &=(n^2-9)^2+2.4(1+n)(n^2-9)+2.16.(1+n)^2\end{align}$$

and because $n$ is odd we have $2$ divides $(1+n)$, and by induction hypothesis $8$ divides $(n^2-9)$ and $64$ divides $(n^2-9)^2$ which implies the result.

Answer 2

First, we note that,

$$n^4-18n^2+17=(n-1)(n+1)(n^2-17)$$

Now, since $n\gt 3$ is an odd value, use $n=2k+1~,~k\geq 2~,k\in\Bbb{Z}$. This makes our expression,

$$2k(2k+2)((2k+1)^2-17)=4k(k+1)(4k^2+4k-16)=16k(k+1)(k^2+k-4)$$

For any $k\geq 2$, one of $k$ or $k+1$ is even.

Case 1 (k=Even): If $k$ is even, we have $k^2+k-4$ also even since $k,k^2,4$ all are even and we get the remaining $4$ to go with $16$ and hence $64|n^4-18n^2+17$ where $n=2k+1~,~k\geq 2\textrm { is even.}$

Case 2 (k=Odd): If $k$ is odd, we have $k+1$ even and since $k^2,k$ both are odd and $4$ is even, we also have $k^2+k-4$ even. Hence, this case also works and we have $64|n^4-18n^2+17$ where $n=2k+1~,~k\geq 2\textrm { is odd.}$

All cases are completed and we have in general,

$$64|(n^4-18n^2+17)~,~\forall~n\gt 3~\land~n\textrm{ is odd}.$$

$$\Bbb{Q.E.D}$$

Answer 3

$\ f(n) = n^4\!-18n^2\!+17\, =\, \overbrace{(n^2\!-\!1)^2}^{\large (\color{#c00}8k)^{\color{#c00}2}}\! - \overbrace{16(n^2\!-\!1)}^{\large 2\cdot \color{#c00}8\,(\color{#c00}8k)\quad }.\ $ odd $\,n\,\Rightarrow\, \!\overbrace{8\mid n^2\!-\!1}^{\large 8k\,=\,n^2-1\,\ }\Rightarrow\, \color{#c00}{8^2}\mid f(n)$

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Remark $\,\ 8\mid n^2\!-\!1\,$ is easy with congruences: $\,{\rm mod}\ 8\!:\ {\rm odd}\ n\equiv \pm1,\pm3\,\Rightarrow\ n^2\equiv 1$

If you must use induction: $\ 8\mid 1^2-1,\ $ and $\,\ (n\!+\!2)^2-1\, =\, \color{#0a0}{n^2-1} + \color{#90f}{4(n\!+\!1)}$

$\,8\mid \color{#90f}{4(n\!+\!1)}\,$ by $\,n\,$ odd, so $\,\color{#0a0}{8\mid n^2-1}\,\Rightarrow\, 8\mid (n\!+\!2)^2-1,\,$ the inductive step.

Answer 4

If $n=2k+1$, we have $$ \begin{align} n^4-18n^2+17 &=(2k+1)^4-18(2k+1)^2+17\\[6pt] &=16k^4+32k^3-48k^2-64k\\ &=384\binom{k}{4}+768\binom{k}{3}+320\binom{k}{2}-64\binom{k}{1}\\ &=64\left[6\binom{k}{4}+12\binom{k}{3}+5\binom{k}{2}-\binom{k}{1}\right] \end{align} $$ This indicates that $64$ divides $n^4-18n^2+17$ for all odd $n$.