Using proof by contradiction prove that if (n^2 + 2n) is even, where n is an integer, then n is even. 

To establish the proposition that if \( n^2 + 2n \) is even, then \( n \) must also be even, we will employ a proof by contradiction. We begin by assuming the opposite of what we aim to prove; that is, we will assume that \( n \) is an odd integer. An odd integer can be expressed in the form \( n = 2k + 1 \), where \( k \) is some integer. 

Substituting this expression for \( n \) into the original equation, we have \( n^2 + 2n = (2k + 1)^2 + 2(2k + 1) \). Expanding this, we find that \( (2k + 1)^2 = 4k^2 + 4k + 1 \) and \( 2(2k + 1) = 4k + 2 \). Therefore, combining these results yields \( n^2 + 2n = 4k^2 + 4k + 1 + 4k + 2 = 4k^2 + 8k + 3 \). 

The expression \( 4k^2 + 8k + 3 \) can be analysed for its parity. Notably, \( 4k^2 + 8k \) is clearly even, as it is a multiple of 4. However, when we add 3 to this even number, the result becomes odd. This leads us to a contradiction, as we initially stated that \( n^2 + 2n \) is even. Consequently, our assumption that \( n \) is odd must be incorrect, thereby confirming that if \( n^2 + 2n \) is even, \( n \) must indeed be even.