Proof by induction $6\mid n^3+5n$

Question

$6\mid n^3+5n$

$6\mid (n+1)^3+5(n + 1)= 6k+ 3(n^2+n+2)= 6k + 3(n(n+1)+2) =6m$

I don't understand why this term $n(n+1)+2$ is always divisible by $2$. Can someone please explain it to me?

Answer 1

$n(n+1)$ is always divisible by $2$ it is a product of an odd and an even integer.

Answer 2

$(n+1)^3+5(n+1)=n^3+5n+3n(n+1)+6$, $3n(n+1)$ is divisible by 6 since $n(n+1)$ is even.

Answer 3

Here is a different take, without induction: $$ n^3+5n = n^3-n+6n = 6\binom{n+1}{3}+6n $$

Answer 4

Imagine that $n$ is an odd number. So $n + 1$ is an even number.
Now imagine that $n$ is an even number. So $n + 1$ is an odd number.
So in both cases we get an odd number and an even number.
The product of an odd and an even number is always even.
So $n(n + 1)$ is even.
Thus $n(n + 1) + 2$ is also even and thus is always divisible by $2$.

Answer 5

$n(n+1)+2$ is divible by $2$ because $n(n+1)$ is. The latter is divisible by $2$ because $$n \text{ is }\begin{cases}\text{even}\\\text{odd}\end{cases}\iff n+1\text{ is }\begin{cases}\text{odd}\\\text{even}\end{cases}.$$

Answer 6

without induction in one line $$n^3+5n=(n-1)n(n+1)+6n$$ and this is a short proof

Answer 7

$$n^3+5n \equiv 0 \mod 6$$ $$n^3+5n -6n \equiv 0 \mod 6$$ $$n^3-n \equiv 0 \mod 6$$ $$n(n^2-1) \equiv 0 \mod 6$$ $$(n-1)n(n+1) \equiv 0 \mod 6$$

One factor is a mutliple of 2, another one a mutltiple of 3 ..