Show that $n^4-6n^3+11n^2-6n$ is divisible by $4$ for every integer $n$.

Question

So I am trying to refresh my memory when it comes to modular arithmetic but I am having some difficulties. The problem as states in the title is:

Show that $n^{4}-6n^{3}+11n^{2}-6n$ is divisible by $4$ för every integer $n$.

So what I first did was reformulating the problem in terms of congruences, in the following way:

$$4\vert (n^{4}-6n^{3}+11n^{2}-6n) \iff 4\vert (n^{4}+11n^{2}-(6n^{3}+6n))\\ \hspace{6cm} \iff n^{4}+11n^{2}\equiv 6n^{3}+6n\pmod 4$$

Now, one initial question I have is if it is correct that the only cases I need to check is:

$$n^{4}+11n^{2}\equiv 0\equiv 6n^{3}+6n\pmod 4 \\ n^{4}+11n^{2}\equiv 1\equiv 6n^{3}+6n\pmod 4\\ n^{4}+11n^{2}\equiv 2\equiv 6n^{3}+6n\pmod 4\\n^{4}+11n^{2}\equiv 3\equiv 6n^{3}+6n\pmod 4,$$

since $0,1,2,3$ are the possible remainders when considering division by $4$?

If this is true, it still seams like a really tedious approach to solving the problem. So I am wondering if there is some other way of solving it?

Thanks in advance.

Answer 1

Hint: Use that $$n^4-6n^3+11n^2-6n=(n-3)(n-2)(n-1)n$$

Answer 2

$2\mid n\Rightarrow 4\mid n^2\,$ so $\!\bmod 4\!:\, f_n\equiv\overbrace{ -6n\equiv 0;}^{\large -12n(n/2)}\,$ else $n$ odd so $\,\color{#c00}{n^2\equiv 1}\,$ so $\,f_n\equiv\overbrace{1-6n+11-6n}^{\large \color{#c00}{n^4} -\,6\color{#c00}{n^3}+11\color{#c00}{n^2}-6n\ }\equiv 0$

Answer 3

Continuing your idea (which isn't quite what I would have done) you can reduce the coefficients.

$n^{4}-6n^{3}+11n^{2}-6n \equiv 0 \pmod 4 \iff$

$n^4 + 11n^2 \equiv 6n^3 + 6n \pmod 4\iff$

$n^4 - n^2 \equiv 2n^3 + 2n\pmod 4\iff$

$n^2(n-1)(n+1) \equiv 2n(n^2 + 1)\mod 4\iff$

The check if $n\equiv 0\pmod 4$ then LHS and RHS are both $\equiv 0\pmod 4$.

If $n\equiv 3\equiv -1 \pmod 4$ or $n \equiv 1\pmod 4$ then $n\equiv \pm 1\pmod 4$ and $n\mp 1 \equiv 0\pmod 4$ so LHS is $\equiv 0 \pmod 4$. And $n^2 +1\equiv 2\pmod 4$ an $2n \equiv \pm 2$ so RHS is $\equiv \pm 4 \equiv 0 \pmod 4$.

And if $n \equiv 2 \pmod 4$ then $n^2 \equiv 4 \equiv 0\pmod 4$ so LHS $\equiv 0\pmod 4$. meanwhile $2n\equiv 4\equiv 0 \pmod 4$. So RHS $\equiv 0 \pmod 4$.

So in all cases: $n^2(n+1)(n-1) \equiv 2n(n^2+1)\equiv 0\pmod 4$.

......

But I wouldn't have done it that way.

Perhaps the must informative, but most needlessly calculating way would be to factor.

$n^4 -6n^2 + 11n^2 -6n = n(n^3 - 6n^2 + 11n -6)=$

$n(n-1)(n^2 -5n^2 + 6)=n(n-1)(n-2)(n-3)$

Note: $n-3, n-2,n-1, n$ are four consecutive numbers! So ONE of them is divisible by $4$. So the product must be divisible by $4$.

That is an important lesson! And an important strategy to learn.

But ... factoring is hard.

.......

If I were to do it my way, I'd reduce

$n^4 -6n^3 +11n^2 -6n \equiv n^4 + 2n^3 -n^2+2n\pmod 4$.

For $n\equiv 0,1,2,-1\pmod n$ we have $n^2 \equiv0,1,0,1\pmod n$ and $n^3\equiv 0,1,0,-1\pmod n$ and $n^4 \equiv 0,1,0,1\equiv n^2$ we'd have

$2n \equiv 0,2,0,-2\equiv 0,2,0,2$ and $2n^3 \equiv 0,2,0,2\equiv 2n$

So we have $n^4 + 2n^3 - n^2 + 2n \equiv n^2 +2n - n^2+ 2n\equiv 4n\equiv 0*n\equiv 0\pmod 4$.

.....

Which hints at Bill Dubuque's parity argument.

If $n$ is even than $n^4, n^3, n^2, 6n$ are all divisible by $4$.

If $n$ is odd then $n\equiv \pm 1$ and $n^4-6n^3 + 11n^2 -6n\equiv 1\mp 6+11 \mp 6=12 \mp 12\equiv 0\mp 0 \equiv 0\pmod 4$