Question 924485
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It is easy to prove more general statement:


<pre>
    For any integer n the number  {{{n^2 -n}}}  is even.


    Indeed,  {{{n^2-n}}} = n*(n-1)  is the product of two consecutive integers.

    Of the two, one integer inevitably is even.


    Therefore, the product,  n*(n-1)  is even.


    Hence, the original  {{{n^2-n}}}  is even.
</pre>