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The following expression seems to result in an integer for all $n$: $$\frac13n\left(n(n-3)+8\right)$$ So, how do I prove that, if $n$ is not a multiple of $3$, $n(n-3)+8$ is.

Andrei
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ordptt
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    Note that if $n$ isn't a multiple of $3$, then it is either $1$ or $2$ mod $3$. Either way $n\cdot(n-3)$ is $1$ mod 3 since $n\cdot(n-3)$ mod $3=n^2\mod 3=1$. And, $1+8=9$. – Rushabh Mehta Dec 28 '20 at 02:13
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    Huh???!? Try $n=3$. – David G. Stork Dec 28 '20 at 02:17
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    The question body has the expression multiplied by $n$, which the title does not – J. W. Tanner Dec 28 '20 at 02:22
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    $$n(n-3)+8=n^2-3n+9-1=n^2-1+9-3n=(n+1)(n-1)+3(3-n)$$ If $n$ is not divisible by $3$ then either $n+1$ or $n-1$ is divisible by $3$. – bof Dec 28 '20 at 02:34
  • It $,= (n^3-n)/3 + 3n-n^2,$ so it suffices to prove $,3\mid n^3-n,$ a much more well-known form, for which see the dupe for many methods (which subsume the answers below). – Bill Dubuque Dec 28 '20 at 15:13

2 Answers2

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$$\frac13n\left(n(n-3)+8\right)=\frac13n\left(n^2-3n+8\right)\\=\frac13n\left(n^2-3n+2+6\right)=\frac13n\left((n-1)(n-2)+6\right)\\=\frac13n(n-1)(n-2)+2$$ One of $n,\ n-1, \ n-2$ is divisible by $3$.

Andrei
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You can also use the Mathematical Induction to prove this.

CASE 1 $(n=1)$: So your equation turns out to be $$1(1(1-3)+8)=6=3(2)$$ Thus, when $n=1$, this equation is divisible by $3$.

CASE 2 $(n=m)$: We assume that $m(m(m-3)+8)$ is divisible by $3$.

CASE 3 $(n=m+1)$: The equation in this case becomes: $$\begin{align}&(m+1)\{(m+1)(m+1-3)+8\}\\ =&(m+1)\{(m+1)(m-2)+8\}\\ =&(m+1)\left(m^2-m+6\right)\\ =&m^3+5m+6\\ =&\underbrace{m(m(m-3)+8)}_{\text{divisible by 3 from CASE 2}}+3\left(m^2-m+2\right)\\ \end{align}$$ Thus, the equation in CASE 3 is also divisible by $3$.

Thus, by the Principle of Mathematical Induction, we can conclude that the equation $n(n(n-3)+8)$ is divisible by $3$ for all $n\in\mathbb{N}$

DeBARtha
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