How can I prove that:
$$\frac{1}{2} \left(3 n^2-2\right) \left(3 n^2-1\right) \left(3 n^3-3 n+1\right)\in\mathbb{N}$$
When $n\in\mathbb{N}$?
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7Out of two consecutive integers, one is even. – Qi Zhu Jan 11 '21 at 13:37
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@QiZhu I don't see what you mean – Baropryl Jan 11 '21 at 13:38
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Special case $n=2$ of the dupe. – Bill Dubuque Jan 11 '21 at 13:48
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Every product of two consecutive integers is even. So $(3n^2 - 2)(3n^2 - 1)$ is even, and the whole product is even (before you divide by 2).
purple
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4I don't think so. He asked for proving that a number is divisible by 2, equivalent to proving that this number divided by 2 belongs to $\mathbb{N}$ – purple Jan 11 '21 at 13:41
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1@JMoravitz I don't think so. That fraction $1/2$, multiplied by an even number, is guaranteed to be an integer. – Jan 11 '21 at 13:45
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we want $\dfrac{1}{2}(3n^2-2)(3n^2-1)(3n^3-n+1)\in \mathbb{N}$, hence proving that $(3n^2-2)(3n^2-1)$ is even is enough – NHL Jan 11 '21 at 13:47
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1@JMoravitz Please, reread the original question. Nobody wants you to prove that this is an even number. – Jan 11 '21 at 13:47
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1I see it now. My apologies. Clearly I need more coffee. I read the question title "prove this number is even" and read it as "this number" included the $\frac{1}{2}$ term, not seeing the $\in\Bbb N$ bit. Poor phrasing of the question in my opinion. – JMoravitz Jan 11 '21 at 13:50