$5\nmid n(n+1)\implies 5\mid (n^3-6n^2+n-1)$

Question

Prove $5\nmid n(n+1)\implies 5\mid (n^3-6n^2+n-1)$. My try:

$5\nmid n(n+1)\implies 5\mid (n-2)(n-1)(n+2)=n^3-n^2-4n+4\implies$
$5\mid (n^3-n^2-4n+4)-(5n^2+ 5(n-1))=n^3-6n^2+n-1$.

This time a simple trick worked, but how to solve this (kind of) problem more methodically, maybe using modular arithmetic?

Answer 1

$$ n^3 - 6 n^2 + n - 1 \equiv n^3 - n^2 + n - 1 \equiv (n^2 + 1)(n-1) \pmod 5 $$

The hypothesis says that $n \neq 0,4 \pmod 5,$ so that $n \equiv 1,2,3 \pmod 5.$ Both $2^2 + 1 \equiv 3^2 + 1 \equiv 0 \pmod 5,$ so the $n^2 + 1$ factor takes care of 2,3. The $n-1$ factor takes care of $1 \pmod 5$

Answer 2

Not really answering the question, I believe, but ... here is another way $$5 \mid n(n+1)(n+2)(n+3)(n+4)$$ since $5$ is prime and using Euclid's lemma $$5 \nmid n(n+1) \Rightarrow \\ 5 \mid (n+2)(n+3)(n+4)=24+26n+9n^2+n^3= (25-1)+(25+1)n+(15-6)n^2+n^3 \Rightarrow \\ 5 \mid -1+n-6n^2+n^2$$

Answer 3

$\!\bmod 5\!:\,\ f(n)\equiv n^2(\overbrace{n\!-\!1}^{\Large n - 6})+n\!-\!1\, =\, (\color{#c00}{n\!-\!1})\ (\overbrace{n^2+1}^{\Large\color{#0a0}{n^2\ -\ 4}})$

therefore we conclude $\ \underbrace{ n\not\equiv 0,-1}_{\Large 5\ \nmid\ n(n+1)}\Rightarrow\ \color{#c00}{n\equiv 1}\,$ or $\!\!\!\!\underbrace{n\equiv \pm2}_{\Large \Rightarrow\ \color{#0a0}{n^2\ \equiv\ 4}\ \ \ \ }\!\!\!\!$ so $\,f(n)\equiv 0$

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Or: $\,\ (\color{#c00}{n+1})\,f(n)\equiv {n^4-1\equiv 0\ \ {\rm by}\ \ n\not\equiv 0}\,$ and little Fermat

so $\,\color{#c00}{n\not\equiv -1}\Rightarrow f(n)\equiv 0$

Answer 4

$n^3 - 6n^2 + n-1 \equiv n^3 - n^2 + n-1 \pmod 5\equiv$

$(n^2 + 1)(n-1) \equiv (n^2 -4)(n-1) \pmod 5$

$(n-2)(n+2)(n-1) \pmod(5)$.

If $n \equiv 2,-2, 1 \pmod 5$ then $n^3 - 5n^2 + n-1 \equiv 0 \pmod 5$ and $5|n^3 - 5n^2 + n-1$.

As $5\not \mid n(n+1)$ and $5$ is prie $5\not \mid n$ and $5\not \mid n+1$ and $n \not \equiv 0\pmod 5$ and $n\not \equiv -1 \pmod 5$.

$\{-2, -1, 0, 1, 2\}$ is a complete residue class of $5$ so $n\equiv -2,-1, 0, 1, $ or $2 \pmod 5$ and since $n\not \equiv 0,-1$ then $n \equiv -2, 2$ or $1 \pmod 5$. So we are done.

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Alternatively (now that we know where we want to go)

$n\not \mid n(n+1) \implies$

$n \not \mid 0$ or $-1\pmod 5\implies$

$n \mid 1,2,$ or $-2\pmod 5\implies$

One of $n-1, n-2, n+2\equiv 0 \pmod 5\implies$

$(n-1)(n-2)(n+2)\equiv 0 \pmod 5 \implies$

$ n^3 -n^2 - 4n + 4\equiv 0 \pmod 5\implies$

$n^3 - 6n^2 + n - 1\equiv 0 \pmod 5 \implies $

$5|n^3 - 6n^2 + n-1$

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Or a third way. $5$ must divide exactly on of $n,n+1, n+2, n+3, n=4$ so i$5$ doesn't divide $n$ or $n+1$ it divides $n+2, n+3, n+4$ which means $5|(n+2)(n+3)(n+4) = n^3 + 9n^2 + 26n + 24$.

$n^3 + 9n^2 +26n + 24 -(n^3 - 6n^2 +n -1) =15n^2 + 25n +25 = 5(3n^2 + 5n + 5)$