SOLUTION: Prove that n^2-n is always even

Question 1209383: Prove that n^2-n is always even
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

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.

----------

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.

----------

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.

Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

The statement of the problem is deficient; n^2-n is NOT always even. It is always even IF n IS AN INTEGER.

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

If n is an integer, then that expression is the product of two consecutive integers. In any two consecutive integers, one of them is odd and the other is even; and the product of two integers is even whenever at least one of them is even.

So the product of two consecutive integers is always even.

TRUE: n^2-n is always even if n is an integer

RELATED QUESTIONS
prove that n^3-n is always divided by... (answered by htmentor)
Prove that (z^n) + (z*)^n is always real for integer... (answered by ikleyn)
Prove: If n is an integer, then n^2 + n^3 is an even... (answered by jim_thompson5910)
Prove that if 'n' is a positive integer, and 'n^2' is even, then n is also even. Thank (answered by edjones)
3. Prove that if n is odd, then gcd(3n; 3n + 2) = 1. What if n is... (answered by Edwin McCravy)
using deductive methods prove that if n is an interger, n is greater than or equal to 2 , (answered by venugopalramana)
prove that : n! (n+2) = n! +... (answered by tommyt3rd)
prove that p(n,n)=... (answered by Edwin McCravy)
Prove that for every positive integer n, n3 + n is... (answered by jim_thompson5910)