Given that n can be any integer such that n > 1, prove that n^n - n is never an odd number. - Edexcel - GCSE Maths - Question 14 - 2019 - Paper 1

To prove that the expression $n^n - n$ is never odd, we need to consider the parity of $n$. An integer is either even or odd, so we will investigate both cases:

1. Case 1: n is even
   Let n be any even integer. By definition, even integers can be expressed as:
   $n = 2k$
   where $k$ is an integer.
   Thus, the expression becomes:
   $n^n - n = (2k)^{2k} - 2k$ Since $(2k)^{2k}$ is clearly even (as a power of an even number is even), and $2k$ is also even, we find that:
   $n^n - n ext{ is even}.$

2. Case 2: n is odd
   Let n be any odd integer. Odd integers can be expressed as:
   $n = 2k + 1$
   where $k$ is an integer.
   Hence, the expression becomes:
   $n^n - n = (2k + 1)^{(2k + 1)} - (2k + 1)$
   According to properties of odd numbers, this simplifies as follows:
   $ext{Odd} - ext{Odd} = ext{Even}.$
   Therefore, $n^n - n$ is also even.

In both cases, whether $n$ is even or odd, the result shows that $n^n - n$ is always even. Thus, we have proven that for any integer $n > 1$, the expression $n^n - n$ is never an odd number.