# For any positive integer n, prove that n^3 – n is divisible by 6.

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For any positive integer n, prove that n3 – n is divisible by 6.

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Solution:
n3 – n = n (n2 – 1) = n (n – 1) (n + 1)
Therefore, n3 – n is product of three consecutive positive integers, where n is any positive integer.
Since one out of every two consecutive integers is divisible by 2.

Therefore, The product n3 – n is divisible by 2.
Since one out of every three consecutive integers is divisible by 3.
Therefore, The product n3 – n is divisible by 3.
Any number which is divisible by 2 and 3 is also divisible by 6.
Hence, The product n3 – n is divisible by 6.

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Let a = n^3 - n ⇒ a = n.(n^2 - 1)
⇒ a = n.(n-1).(n+1)  [∵(a^2 - b^2) = (a-b)(a+b)]
⇒ a = (n-1).n.(n+1)  ...(i)
We know that,
1. If a number is completely divisible by 2 and 3, then it is also divisible by 6.
2. If the sum of digits of any number is divisible by 3, then it is also divisible by 3.
3. If one of the factor of any number is an even number,then it is also divisible by 2.
∴ a = (n-1).n.(n+1)  [from Eq.(i)]
Now,sum of the digits = n-1+n+n+1 = 3n
= multiple of 3,where n is any positive integer,
and (n-1)-n-(n+1) will always be even, as one out of (n-1) or n or (n+1) must of even.
Since, conditions II and III is completely satify the Eq.(i).
Hence, by condition I the number n^3 - n is always divisible by 6, where n is any positive integer.
Hence proved.