Prove that for every positive integer, this polynomial is divisible by 24.

Question

Prove that: $$24\mid n^4 + 2n^3 - n^2 - 2n, \quad \forall n\in \mathbb{Z}^+$$

I tried to prove it, but had no luck.

Answer 1

Hint: $f(n)=(n-1)(n)(n+1)(n+2)$; also, $24=4!=4\cdot3\cdot2\cdot1$. Can you continue now? See first take $n^3$ common from $n^4,2n^3$ and $-n$ common from $-n^2-2n$ then you get $(n^3-n)(n-2)$ now take $n$ comon from $n^3-n$ so you get $n(n^2-1)$ thus use $a^2-b^2=(a-b)(a+b)$ to get $(n-1)(n)(n+1)(n+2)$

Answer 2

Reduce mod $3$ and mod $8$ separately, or use the fact that the given polynomial is just $24\binom{n+2}{4}$. (As a bit of an explanation for the latter, note that any integer-valued polynomial $P(n)$ is an integer combination of binomial coefficients $\binom{n}{k}$.)

Answer 3

As suggested above, factor $n^4+2n^3-n^2-2n$ into terms with the lowest power possible- in this case, power $1$. A heuristic would be to group the even powered terms together, and the odd power terms together; doing this, we get:

$$n^2(n^2-1)+2n(n^2-1)$$ $$\implies (n^2+2n)(n^2-1)$$ Breaking it down further from here, we then get $$n(n+2)(n-1)(n+1).$$ Curiously, we rearrange these terms to find $$(n-1)n(n+1)(n+2),$$ which is an abstraction for $4$ consecutive integers. Note that for any 3 consecutive integers $k$, $k+1$, and $k+2$, their product $p=k(k+1)(k+2)$ is divisible by 3. From these four consecutive integers, can you show that 8 must divide two of them? If so, your proof is complete.

Answer 4

Aside from clever answers given, the following tedious method generally works: every integer $n$ can be written as $n=24k+m$, with $0\le m\le 23$. When you expand $n^p$ with Newton formula, every cross-term with at least a power of $(24k)$ can be divided by $24$. So only the $p$-th power of $m$ remains.

Hence, it suffices to verify your equation $m^4+2m^3−m^2−2m$, only for each $0\le m\le 23$, and this will prove the result for all integers. For instance, if $m=3$, this yields: $81+2\times 27-9-6=120 = 5\times 24$.

This is tedious as announced, but intermediate calculations may help you understand that some factoring can be done. For instance, if you do it by hand, $23^4$ or $23^2$ might seem super-tedious. So you begin to see lazy spots: $m^4−m^2$ can be turned into $m^2(m^2-1)$. You only have to compute squares. And $2m^3−2m = 2m(m^2 -1)$. Oh wait, there is a common factor: $(m^2-1)$! Let us factor it further: $(m^2-1)(m^2-2m)$. And so on, you see the $m$ factor: $(m^2-1)(m^2-2m)=m(m^2-1)(m-2)=m(m-1)(m+1)(m-2)$. Then you recognize a product of four consecutive numbers. Like $\{1,2,3,4\}$, of product $24$. Like $\{2,3,4,5\}$, of product $120$. Like $\{3,4,5,6\}$, of product $360$.

In a series of four consecutive terms, $2$ terms are even, and obviously one of those two is a multiple of $4$. So the product is likely to me divided by $8$. And there is at least one multiple of $3$. And $8$ and $3$ are relative primes. You are almost done. So, the moral is: when you have trouble finding out what a series does, it is always beneficial to compute the first terms, to see what happen. And by computing them by hand, you get free training. And you might find unexpected patterns that will help you find a clever answer on the way.

Answer 5

$$ n^4 + 2n^3 - n^2 - 2n = \underbrace{(n-1)n(n+1)(n+2)}_\text{four consecutive integers} $$ The four factors are four consecutive integers. Either the first and third are even, or the second and fourth are even. There you have two consecutive even integers. One of those is a multiple of $4$, since in every sequence of four consecutive integers, exactly one is a multiple of $4$ (just as, for example, in every sequence of seven consecutive days, exactly one is a Thursday). So you have a multiple of $4$ multiplied by another even number, and that gives you a multiple of $8$.

In every sequence of three consecutive integers, exactly one is a multiple of $3$. That means you have at least one multiple of $3$ among the four factors. (If the first or the fourth is a multiple of $3$, then both of them are, since they differ by $3$; otherwise just the second or the third is a multiple of $3$.)

Since $8$ and $3$ have no factors in common (besides the trivial factor, which is $1$), if a number is divisible by both $8$ and $3$, then it is divisible by $8\times 3$.