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I've been trying to prove the following statement but have not succeeded with any proof method:

"Show that for every integer $n$, if $n^3 + n$ is divisible by $3$, then $2n^3 + 1$ is not divisible by $3$."

I've attempted various approaches, including direct proof and contraposition, but to no avail. I also tried proof by cases trying to show that $3$ has to divide $n$ if $n^3+n$ is divisible by $3$ but I also wasn't able to Could anyone provide a solution?

Bill Dubuque
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user20194358
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    If $3\mid n(n^2+1),$ then $3\mid n$ or $3\mid n^2+1$, but $3\nmid n^2+1,$ so $3\mid n$ – J. W. Tanner Feb 25 '24 at 16:25
  • Generally $\ (n,2n!+!1)!=!1,$ so by Euclid's Lemma and the Euclidean algorithm we have $$(n(n^2!+!1),2n^3!+!1) = (!!\overbrace{!!n^2!+!1}^{\large \color{#c00}{n^2\ \equiv\ -1}}!!!,2:!\color{#c00}{n^2}n!+!1) = (n^2!+!1,2(-n)!+!1)=(5,2n!-!1)\qquad$$ where the final equality follows similarly as here. Therefore if they are not coprime then they must have gcd $= 5,$ ($!!!\iff! 5\mid 2n!-!1!\iff! n\equiv 3\pmod{!5})\ \ $ – Bill Dubuque Feb 25 '24 at 18:50

2 Answers2

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Consider $$n^3+n=n\cdot (n^2+1)$$

Squares are either $0$ or $1$ modulo $3$. So $n^2+1$ is either $1$ or $2$. It means that $n^2+1$ is never divisible by $3$. So $n$ must be.

Then $2n^3$ is also divisible by $3$. If we add $1$, it is no longer so.

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Another nice reasoning came to my mind. Consider the difference of the two numbers in question:

$$(2n^3+1)-(n^3+n)=n^3-n+1=(n-1)\cdot n\cdot (n+1)+1.$$

One of the three consecutive numbers is always divisible by $3$, so $(n-1)\cdot n\cdot (n+1)+1$ is not. Now $2n^3+1$ is the sum of $n^3+n$, divisible by 3, and $(n-1)\cdot n\cdot (n+1)+1$, not divisible by $3$. Hence it is not divisible by $3$.

Aig
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By Fermat's little theorem, it follows that $n^3 \equiv n \pmod{3}$. Hence, $$2n^3 + 1 = n^3+n^3 + 1 \equiv n^3 +n + 1 \equiv 1 \pmod{3}.$$

xen
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