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I have to show that this is divisible by both 2 and 3. For the next, part I have to show that it's divisible by 6 which from what I understand is the same as showing it's divisible by 2 and 3.

Hanul Jeon
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  • It is equal to $2n^3+3n(n+1)$. The second term it multiple of $2$ because it is multiple of two consecutive integers $n$ and $n+1$, among which one must be even. It is also equal to $3n^3+n(2n^2+1)$. This is multiple of $3$ because in the second term either $n$ is multiple of $3$ or else $2n^2+1$ is multiple of $3$. –  Feb 02 '18 at 02:33
  • Just factor. 2n^3+3n+n=n (2n+1)(n+1). One of n,n+1 must be even. One of n,n+1 or 2n+1=3n-(n-1) must be divisible by 3. – fleablood Feb 02 '18 at 03:55
  • See also https://math.stackexchange.com/q/3372218/ – lhf Sep 27 '19 at 16:54
  • @lhf fyi: You don't need to explicitly link since the dupe link will already show in the "Linked" questions list (i.e.links are bidirectional). – Bill Dubuque Sep 27 '19 at 17:08

3 Answers3

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Factorising $2n^3 + 3n^2 + 1$ we get $n(n+1)(2n+1)$ As we know that sum of squares of first n natural numbers $$1^2+2^2+...+n^2 = {n(n+1)(2n+1)\over 6}$$ And since $$1^2+2^2+ ... + n^2$$ is an positive integer we can say that $n(n+1)(2n+1)$ is divisible by 6 and hence $2n^3 + 3n^2 + 1$ is divisible by 2, 3 and 6.

dssknj
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Hint: $\;2n^3 + 3n^2 + n = 2n(n+1)(n+2) -3n(n+1)\,$, then use that the product of $\,2\,$ consecutive numbers is divisible by $\,2\,$ and the product of $\,3\,$ consecutive numbers is divisible by $\,6\,$.

dxiv
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By fermat theorem $n^2=n (mod2)$ then $2n^3+3n^2+n = 2n+3n+n = 6n =0 (mod2)$ $2n^3+3n^2+n =2n+3n^2+3n $

curieux_2014
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