Prove that $8640$ divides $n^9 - 6n^7 + 9n^5 - 4n^3$.

Question

I found this problem in a book, I can't solve it unfortunately. Prove that for all integer values $n$, $n^9 - 6n^7 + 9n^5 - 4n^3$ is divisible by $8640.$ So far I've noticed that $8460 = 6! \times 12$, also I've tried to simplify that expression and I've found that it's equal to this $n^3(n^3-3n-2)(n^3-3n+2)$, but I can't move on after that.

Answer 1

We can further note that $-1$ is a root of $n^3-3n-2$, and that $1$ is a root of $n^3-3n+2$.

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You can find these roots by the Rational Root theorem:

Rational root theorem. All rational roots have the form $\frac{p}{q}$,with $p$ a divisior of the constant term and $q$ a divisior of the first coefficient.

Other techniques can also be found in this question and its answers.

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Then we simplify it to \begin{align*} n^3(n+1)(n^2-n-2)(n-1)(n^2+n-2) &= n^3(n+1)(n-2)(n+1)(n-1)(n+2)(n-1)\\ &=n^3(n+1)^2(n-1)^2(n-2)(n+2) \end{align*}

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There are 6 factors two in 8640.

• If $n$ is even, $n, n-2$ and $n+2$ are even, and furthermore at least one of them is divisible by $4$, which gives 6 factors 2.
• If $n$ is odd, $n-1$ and $n+1$ are even, and furthermore at least one of them is divisible by $4$, which gives 6 factors 2.

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There are 3 factors three in 8640.

• If $n \equiv 0 \mod 3$, then $n$ is divisible by 3, and then $27 \mid n^3$.
• If $n \equiv 1 \mod 3$, then $n-1$ and $n+2$ are divisible by 3.
• If $n \equiv 2 \mod 3$, then $n+1$ and $n-2$ are divisible by 3.

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The factor 5 can be taken care of since the product is a multiple of five consecutive integers, namely a product of $(n-2)(n-1)n(n+1)(n+2)$.

Answer 2

Hints:

• Note for example $n^3-3n+2 = (n-1)(n^2-n-2) = (n-1)(n+1)(n-2)$. You can factorise completely to a product of terms like $(n-1)$ and $(n+2)$. Do that
• How many powers of $5$ must the product have?
• How many powers of $3$? How many powers of $2$? Note for example that one of two consecutive even numbers is a multiple of $4$

Answer 3

If you do complete factorisation you get $$n^9−6n^7+9n^5−4n^3=n^3(n-2)(n-1)^2(n+1)^2(n+2)$$ and $$8640=2^6\cdot 3^3\cdot 5$$ Now just look at different cases. For $2\mid n$ we have $2^3\mid n^3$, $2\mid n-2$, $2\mid n+2$ and either $n-2$ or $n+2$ has the factor $2^2$, so we get $2^6\mid n^9−6n^7+9n^5−4n^3$. Do it for the case $n$ odd and for the other factors in a similar style.

Answer 4

$$ \begin{align} &n^9-6n^7+9n^5-4n^3\\ &\small=362880\binom{n}{9}+1451520\binom{n}{8}+2298240\binom{n}{7}+1814400\binom{n}{6}\\ &\small+734400\binom{n}{5}+138240\binom{n}{4}+8640\binom{n}{3}\\ &=\small8640\left[42\binom{n}{9}+168\binom{n}{8}+266\binom{n}{7}+210\binom{n}{6}+85\binom{n}{5}+16\binom{n}{4}+\binom{n}{3}\right] \end{align} $$