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I'm in high school so I don't have a lot of proof techniques.I also don't know modular arithmetic so I can't use that either. Any help is appreciated, thank you!

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    $\begingroup$ as far as 2, you can calculate the thing for cases $n=2k$ and $n=2k+1.$ In both cases the product is even. Similar for 3, cases are $n = 3w,$ $n=3w+1,$ finally $n=3w+2.$ Just calculation, but you do need to do that much. $\endgroup$
    – Will Jagy
    Jan 13, 2019 at 1:09
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    $\begingroup$ If you know the sum of squares formula, you'll know that $(2n + 1)(n + 1)n$ is $6$ times the sum of the first $n$ squares. $\endgroup$ Jan 13, 2019 at 1:14

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First note that either $n$ or $n+1$ is a multiple of 2. Then, there are three cases to see that the number is a multiple of 3 as well:

  1. If $n$ is a multiple of 3, no much more to add.
  2. If $n$ is a multiple of 3 plus 1, then $2n+1$ is a multiple of 3, as $2(3k+1)+1 = 6k + 3 = 3 (2k+1)$
  3. If $n$ is a multiple of 3 minus 1, then $n+1$ is a multiple of 3.

In any case, $(2n+1)(n+1)n$ is multiple of 6

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  • $\begingroup$ Right idea but not quite right. The first sentence is wrong when $n=6$. Settle divisibility by $2$ by looking at $n(n+1)$. $\endgroup$ Jan 13, 2019 at 1:18
  • $\begingroup$ Corrected. Thx! $\endgroup$
    – pendermath
    Jan 13, 2019 at 1:23
  • $\begingroup$ You don't need anything about $n$ mod $3$ to know $n(n+1)$ is even. Do that first. Then just worry about $3$. $\endgroup$ Jan 13, 2019 at 1:42
  • $\begingroup$ Good point. My solution is effective, it could be more efficient. I changed it following your suggestions $\endgroup$
    – pendermath
    Jan 13, 2019 at 1:43
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As $n(n+1)$ is even, it is sufficient to show $3$ divides $(2n+1)n(n+1)$

Now $2n(2n+1)(2n+2)$ is divisible by $3,$ being product of three consecutive integers for any integer $n$

$3$ will divide $\dfrac{2n(2n+1)(2n+2)}{2\cdot2}$ as $(3,4)=1$

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  • $\begingroup$ I really like that trick, using a product of three consecutive numbers. I'll remember that. +1 $\endgroup$ Jan 13, 2019 at 11:48
  • $\begingroup$ @Theo, math.stackexchange.com/questions/12065/… $\endgroup$ Jan 13, 2019 at 12:04
  • $\begingroup$ I was more referring to the trick of relating $(2n + 1)n(n + 1)$ to $(2n)(2n + 1)(2n + 2)$. I never thought of explaining the divisibility by $3$ in this way. $\endgroup$ Jan 13, 2019 at 12:34

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