Question 1209383
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I assume that n is an integer.
With many integer proofs, we break things up into two cases. 
We see what happens when n is even and when n is odd.


If n is even then we'd have n = 2k, where k is some integer.
Then we could say,
n^2-n
= (2k)^2-2k
= 4k^2-2k
= 2*(2k^2-k)
= 2*(some integer)
= even integer
We have shown that n^2-n is even when n is even.


If n is odd then it means n = 2k+1
Which leads to...
n^2-n
= (2k+1)^2-(2k+1)
= (4k^2+4k+1)-(2k+1)
= 4k^2+4k+1-2k-1
= 4k^2+2k
= 2*(2k^2+k)
= 2*(some integer)
= even integer
This proves that n^2-n is even despite n being odd.


Regardless if n is even or odd, n^2-n is always even.
Therefore n^2-n is always even for any integer n.
The proof is complete.


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Here is a faster alternative proof method.


n^2-n = n(n-1)


Either n is even or n-1 is even
By "or", I refer to "exclusive or".
So either n = 2k or n-1 = 2k
2 is buried somewhere in the factorization of n(n-1)
This proves n^2-n is even regardless of any integer you pick for n.


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Let's generate a few examples trying n = 1 through n = 5.
n^2-n = 1^2-1 = 1-1 = 0
n^2-n = 2^2-2 = 4-2 = 2
n^2-n = 3^2-3 = 9-3 = 6
n^2-n = 4^2-4 = 16-4 = 12
n^2-n = 5^2-5 = 25-5 = 20
I encourage you to try other examples.
Recall that a number is even (i.e. a multiple of 2) when the number ends with 0, 2, 4, 6 or 8.
The examples on their own do not constitute a formal proof since we'd need to check infinitely many examples. However, they are useful to help solidify understanding.
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