Question 1209383
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The statement of the problem is deficient; n^2-n is NOT always even.  It is always even IF n IS AN INTEGER.<br>
n^2-n = n(n-1)<br>
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.<br>
So the product of two consecutive integers is always even.<br>
TRUE: n^2-n is always even if n is an integer<br>