Question 924485
Suppose it's an odd number instead. That is, 


{{{(2n+1)^2-(2n+1)}}}={{{2k+1}}}, for some k∈Z


On the other hand,


{{{4n^2+4n+1-2n-1=4n^2+2n=2(2n^2+n)}}}, which is an even number. This contradicts the original hypothesis.