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Prove or disprove that $n^3-n$ is divisible by $6$, without using induction

I have no idea how to go about this.

I should add that n is an integer. I started by looking for some integer that was a single case disproving it, but I couldn't find one.

Any help/suggestions are welcome!

dvanaria
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3 Answers3

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We have $$n^3 - n = n(n^2-1) = n(n+1)(n-1)$$

One of these $(n, n+1, n-1)$ must be even (why?) and one must be divisible by 3 (why?).

twnly
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  • I can prove one of those must be even, but I'm not sure how to prove that one of them must be divisible by 3... – dvanaria Jan 17 '19 at 23:49
  • $n-1, n, n+1$ are three consecutive numbers. Whenever you have three consecutive numbers and take it $\pmod 3$, then one has to be $0 \pmod 3$. – twnly Jan 18 '19 at 00:01
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By the division theorem we have $n=6k+r$ for some $0 \le r < 6$. Then $$(n^3-n) = (6k+r)^3 - (6k+r) = 6(x-k) + (r^3-r)$$ where $x = \sum_{i=1}^3 \binom{3}{i} 6^{i-1}k^ir^{3-i} \in \mathbb{Z}$ by the binomial theorem.

But $r \in \{0,1,2,3,4,5\}$, so $r^3-r \in \{0,0,6,24,60,120\}$, each of which is a multiple of $6$.

Disclaimer: I would argue that both this answer and twnly's answer 'use induction', because both depend on results that themselves require induction in some form to be proved.

Clive Newstead
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  • That disclaimer is so, so true. You really can't do much with regards to the natural numbers without induction. In some definitions of natural numbers, induction is actually part of them. – Git Gud Jan 17 '19 at 23:37
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Hint:

Fermat's theorem and Chinese remainder theorem.

Bernard
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