Divisibility of $n^3 +6n^2-7n$

Question

Let $n = 2, 3, 4, ...$ be an integer. Show that $n^3 +6n^2-7n$ is divisible by $6$.

How should one approach this? Using modular arithmetic or some other approach?

Answer 1

$$n^3+6n^2-7n=n(n-1)(n+1)+6(n^2-n)$$ The first term is product of three consecutive integers and the second term is divisible by $6$, hence we are done.

Answer 2

$n^3 + 6n^2 -7n = n(n-1)(n+7)$

If $n$ is odd, then the expression is a product of at least one odd and one even integer. Similarly, when $n$ is even.

Answer 3

Let's try induction.

Base case: $(n=2)$. Then we have $2^3 + 6(2)^2 - 7(2) = 24$, and we know $6 \mid 24$ (since $6 \cdot 4 = 24)$.

Induction Hypothesis: Assume for $n = k$. So there exists $m \in \mathbb{Z}$ such that $6m = 6k^3 + 6k^2 - 7k$.

Induction Step: Prove for $n = k + 1$. \begin{align*} 6(k+1)^3 + 6(k+1)^2 - 7k & = 6(k^3 + 3k^2 + 3k + 1) + 6(k^2 + 2k + 1) \\ & = 6(k^3 + 3k^2 + 3k + 1 + k^2 + 2k + 1) \end{align*} In a strange twist, I didn't even have to use the induction hypothesis. I thought it would be instructive to see an inductive proof, but I quite prefer one of the direct proofs from some of the previous answers for this problem.

Answer 4

It is so simple with lil' Fermat and congruences! Namely:

• It is even since for all $n$, $n^2\equiv n\bmod 2$, hence $n^3\equiv n^2\equiv n$, so that $$n^3 +6n^2-7n\equiv n+0\cdot n-1\cdot n=0.$$
• It is divisible by $3$, since $n^3\equiv n\bmod 3$, and similarly, $$n^3 +6n^2-7n\equiv n+0\cdot n-1\cdot n=0.$$

Answer 5

Here are unnecessarily complicated but fun ways (but also general).

There is a theorem (see for example Theorem $2.1$ in https://arxiv.org/pdf/1709.08026.pdf) stating that $\gcd$ of all values of polynomial $f \in \mathbb{Z}[x]$ is equal to $\gcd$ of its values on any $n+1$ consecutive integers, where $n=\deg f$. So in your case, we can for example check $f(0)=0$,$f(1)=0$,$f(2)=18$$,f(3)=60$, all divisible by $6$, and so all of the values of $f$ are divisible by $6$ by the Theorem.

Another interesting way is to express the polynomial in form $$f(x)=c_0\binom{n}{0}+c_1\binom{n}{1}+c_2\binom{n}{2}+c_3\binom{n}{3}$$ and then look at the coefficients $c_i$. In this case we can find $n^3 +6n^2-7n=18\binom{n}{2}+6\binom{n}{3}$, and the conclusion follows.