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How would I show this?
I tried letting $k=2n+1$ for some $n\in\mathbb{Z}$, then

$$(k-1)k(k+1) = 4n(2n + 1)(n+1).$$

I have shown that $4n(n+1)$ is divisible by $8$... but am unsure of how to get the $3$ factor there.

Natash1
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3 Answers3

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HINT:

$$4n(2n+1)(n+1)=2n(2n+1)(2n+2)$$ is the product of three consecutive integers, hence exactly one of them must be divisible by $3$

lab bhattacharjee
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  • If I just consider $2n(2n+1)(2n+2)$, I can conclude it's divisible by $2$ (since there's a factor of $2$ there) and $3$, since one of them is divisible by $3$. How do I get the factor of $4$ there? (I thought I can look at the coefficient on the LHS... but wouldn't I be "double counting" the factors since 2*2 = 4)? – Natash1 Sep 15 '17 at 08:32
  • I seem to only be able to conclude it's divisible by $12$ – Natash1 Sep 15 '17 at 08:34
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    @Natash1, consider two cases: if $n$ itself is divisible by 2, and when it is not. – Mees de Vries Sep 15 '17 at 08:35
  • @Natash1 If you have two consecutive even numbers (like 10 and 12; or 104 and 106) then one of them must be divisible by $4$, not just $2$. So your outer two numbers are collectively divisible by $8$ (or maybe even a higher power of $2$). – 2'5 9'2 Sep 15 '17 at 08:41
  • @Natash1, See https://math.stackexchange.com/questions/126766/prove-that-one-of-n-consecutive-integers-must-be-divisible-by-n https://math.stackexchange.com/questions/12065/the-product-of-n-consecutive-integers-is-divisible-by-n-factorial/12070 https://math.stackexchange.com/questions/12067/the-product-of-n-consecutive-integers-is-divisible-by-n-without-using-the-prop – lab bhattacharjee Sep 15 '17 at 08:45
  • If there are three consecutive numbers such that the first is even, one will be divisible by 2, one by 3 and one by 4, and that's it. @Natash1 – MysteryGuy Sep 15 '17 at 08:53
  • @Natash1 It is much clearer to view this in terms of $k,,$ viz. one of $,k!-!1,,k,,k!+!1,$ must be divisible by $3.\ $ – Bill Dubuque Sep 15 '17 at 14:50
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An alternative.

Mod $3$, $k^3\equiv k$ by Fermat's Little Theorem.

It's a useful fact that $1^2\equiv3^2\equiv5^2\equiv7^2\equiv1$ mod $8$. In other words, if $k$ is odd, then $k^2\equiv1$ mod $8$. So mod $8$, $k^3\equiv k$.

Together this gives that $k^3\equiv k$ mod $24$ when $k$ is odd. So $24$ divides $k^3-k=k(k-1)(k+1)$ when $k$ is odd.

2'5 9'2
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This is true for $k=1$ because the product is zero.

Assume it is true for $k=2r-1$, then $$(2r+2)(2r+1)2r-2r(2r-1)(2r-2)=2r(4r^2+6r+2-4r^2+6r-2)=24r$$ so it is true also for $2r+1=2(r+1)-1$ and you are done by induction.

Easier, perhaps, to observe that of three consecutive integers one will be divisible by $3$ and of two consecutive even numbers one will be divisible by $4$.

Mark Bennet
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