Prove that n2 - n is …

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Sia ? 6 years, 6 months ago
Any positive integer is of the form 2q or 2q + 1, for some integer q.
{tex}\therefore{/tex} When n = 2q
{tex}\style{font-family:Arial}{n^2\;-\;n\;=\;n(n\;-\;1)\;=\;2q(2q\;-\;1)=\;2m,}{/tex}
where m = q(2q - 1) ( m is any integer)
This is divisible by 2
When n = 2q + 1
{tex}\style{font-family:Arial}{\begin{array}{l}n^2\;-\;n\;=\;n(n\;-\;1)\;=\;(2q\;+\;1)(2q+1-1)\\=2q(2q+1)\end{array}}{/tex}
= 2m, when m = q(2q + 1) ( m is any integer)
which is divisible by 2.
Hence, n2 - n is divisible by 2 for every positive integer n.
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