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I am trying to prove that $n(n-1)(n+1)$ is divisible by $6$ for all $n$ in $\mathbb{N}$.

My attempt:

The result certainly holds for $n=0$. Suppose now that $n > 0$. Assume that $P(k)$ is true for all $k<n$. In particular $P(n-1)$ is true. Thus $(n-1)(n-2)n$ is divisible by 6. But

$n(n-1)(n+1) = (n-1)(n-2)n + 3n(n-1)$

Now I don't know how to proceed from here. It is not immediately apparent to me that the right hand side of the expression is divisible by 6.

bawse
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  • The first term on the left is a multiple of 6 by hypothesis. The second term is obviously a multiple of 3, and one of $n$ and $n-1$ must be even, because they're consecutive, so the second term is a multiple of 6, too. – symplectomorphic Nov 06 '15 at 02:42
  • In fact, no need to use induction here. You have the product of three consecutive integers; one of them must be even and one of them must be a multiple of 3, so the product must be a multiple of 6. – symplectomorphic Nov 06 '15 at 02:43
  • If you want to get some more practice with induction, you could show that $3n (n-1)$ is divisible by $6$ for all $n$ using induction. (Which would then complete your current proof) There are easier ways to do it than induction, of course, but there are easier ways of solving the original problem too. – Dylan Nov 06 '15 at 02:43

5 Answers5

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Does it have to be by induction? The solutions suggested by the other answers are more straightforward.

Anyway...

Basis: $(1)(0)(2) = 0$ is divisible by $6$ (when $n = 1$).

Induction: Suppose $k = n(n-1)(n+1)$ is divisible by $6$. Then

\begin{align} (n+1)n(n+2) & = (n+1)n[(n-1)+3] \\ & = k+3n(n+1) \end{align}

and provided $n(n+1)$ is always even, this quantity is always also divisible by $6$. We show this secondary claim, also by induction (!).

Basis: $(1)(2) = 2$ is even (when $n = 1$).

Induction: Suppose that $j = n(n+1)$ is even. Then

$$ (n+1)(n+2) = j+2(n+1) $$

is also even.

Brian Tung
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Hint : either $n$ or $n-1$ is even.

corindo
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Hint:

  1. if you multiply 2 consecutive integers together , one of them is going to be even (contains a factor of two)

  2. if you multiply 3 consecutive integers together, one of them is going to contain a factor of 2, and one of them is going to contain a factor of 3

More water plz
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$n(n+2)(n+2)=$ 6$n \choose 3$, $n \choose 3$ is always a natural number.

Black-horse
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Since the OP already has an answer, here is one more approach for "fun" (in the spirit of killing a fly with a sledgehammer).

The product of the first two terms is $n^2 - n$ which is divisible by $2$, and the product of all three terms is $n^3 - n$ which is divisible by $3$, as follows in each case from Fermat's Little Theorem.

Since $\gcd(2,3) = 1$, divisibility by each implies the expression is divisible by $2 \cdot 3 = 6$. QED.

Benjamin Dickman
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