# How can I prove

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!

• 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. Jan 13, 2019 at 1:09
• 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. Jan 13, 2019 at 1:14

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

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

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 $$(3,4)=1$$

• I really like that trick, using a product of three consecutive numbers. I'll remember that. +1 Jan 13, 2019 at 11:48
• Jan 13, 2019 at 12:04
• 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. Jan 13, 2019 at 12:34