Question 984151
If k stands for an integer, then is it possible for k^2+k to stand for an odd integer? Explain.

The answer is "NO".
Case 1: k is an odd integer
For some integer p, k = 2p + 1
So, k + 1 = 2p + 2 = 2(p+1) = 2q for some integer q
k^2 + k = k(k+1) = k(2q) = 2kq = 2r for some integer r.[Integers are closed under multiplication]
Therefore, if k is an odd integer, then k^2 + k is an even integer

case 2: k is an even integer
For some integer p, k = 2p
So, k + 1 = 2p + 1 
k^2 + k = k(k+1) = 2p(2p+1) , multiple of 2
Therefore, if k is an even integer, then k^2 + k is an even integer