Question 1081675
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Prove that {{{ n^3 + 5n }}} is divisible by 6, where n is any positive integer.
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My proof consists of two parts.


<U>Part 1</U>.  For any integer n,  {{{ n^3 + 5n }}} is divisible by 2.

<pre>
    Consider consequently the case of even "n" and the case of odd "n".
    In both cases, the statement is true, which is OBVIOUS.
</pre>

<U>Part 2</U>.  For any integer n,  {{{ n^3 + 5n }}} is divisible by 3.

<pre>
    {{{n^3 + 5n}}} = {{{(n^3-n)}}} + {{{6n}}} = {{{n*(n-1)*(n+1)}}} + {{{6n}}}.


    {{{n*(n-1)*(n+1)}}} is always multiple of 3, since it is the product of three consecutive integers.

    {{{6n}}} is always multiple of 3, by the obvious reason.

    So the statement of the <U>Part 2</U> is proven, too.
</pre>

Thus the general statement follows regarding divisibility by 6.