Question

Show that n square minus n is divisible by 2 for every positive integer n.

Answer 1

Suppose the positive integer is n.

∴ n = 2q or n = 2q + 1         where q∈Z.

CASE 1:-  

n = 2q

∴ n² - n = (2q)² - 2q

             = 4q² - 2q

             = 2(2q² - q)

CASE 2:-

n = 2q + 1

∴n² - n = (2q + 1)² - (2q + 1)

           = 4q² + 4q + 1 - 2q - 1

           = 4q² + 2q

           = 2(2q² + q)

Thus, in any case, n² - n is divisible by 2.

Thus, n² - n is divisible by 2 for every positive integer

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Answer 2

There can be two possibilities of the positive integer n such that it is either an odd number or an even number.
Now,
Let n = 2m that is an even number where m is a positive integer.

So,
n^2 - n = (2m)^2 - 2m = 4m^2 - 2m = 2(2m^2 - m) = 2k
Where k = 2m^2 - m
So, since here when n is even and n^2 - n = 2k , this means that n^2 - n is divisible by 2.

Now, if n = 2m+1 that is n is odd and also m is a positive integer.

So,
n^2 - n = (2m+1)^2 - (2m+1)
= 4m^2 + 4m + 1 - 2m - 1 = 4m^2 + 2m = 2(2m^2 + m) = 2k
where k = 2m^2 + m
So, this also similarly shows that if n is odd, then n^2 - n is divisible by 2.

Hope this helps you !