Question
For any positive integer n, prove that $n^3 – n$ is divisible by 6.
✓
Answer
Let $a = n^3 - n$
$\Rightarrow a = n(n^2 - 1)$
$= n (n - 1)(n + 1)$
$(n - 1), n, (n + 1$) are consecutive integers so out of three consecutive numbers at least one will be even. So, a is divisible by 2.
Sum of numbers $= (n - 1) + n + (n + 1)$
$= n - 1 + n + n + 1$
$= 3n$
Clearly, the sum of three consective numbers is divisible by 3, so any one of them must be divisible by 3.
So, out of n, (n - 1), (n + 1), one is divisible by 2 and one is divisible by 3 and
a = (n - 1) × n × (n + 1)
Hence, out of three factors of a, one is divisible by 2 and one is divisible by 3. So, a is divisible by 6 or $n^3 - n$ is divisible by 6.
$\Rightarrow a = n(n^2 - 1)$
$= n (n - 1)(n + 1)$
$(n - 1), n, (n + 1$) are consecutive integers so out of three consecutive numbers at least one will be even. So, a is divisible by 2.
Sum of numbers $= (n - 1) + n + (n + 1)$
$= n - 1 + n + n + 1$
$= 3n$
Clearly, the sum of three consective numbers is divisible by 3, so any one of them must be divisible by 3.
So, out of n, (n - 1), (n + 1), one is divisible by 2 and one is divisible by 3 and
a = (n - 1) × n × (n + 1)
Hence, out of three factors of a, one is divisible by 2 and one is divisible by 3. So, a is divisible by 6 or $n^3 - n$ is divisible by 6.
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