Prove that for any $n>2$, $n \in \Bbb N$, $\lfloor (\sqrt[3]{n}-\sqrt[3]{n+2})^3 \rfloor \equiv -1 \mod \ 8$

Question

Prove that for any $n>2$, $n \in \Bbb N$, $\lfloor (\sqrt[3]{n}-\sqrt[3]{n+2})^3 \rfloor \equiv -1 \mod \ 8$.

I went with the very general approach of trying to find the binomial expansion of the expression given inside the greatest integer function(g.i.f). but couldn't really get to anywhere after that. Trying to find some similar problems hinted at trying to work backwards and thus tried to assume that $8$ divides it which implies the expression itself is a number of the form $8k+7$.

Now by property of g.i.f and a bit of motivation from earlier I am trying to prove that $8n+7 \leq \lfloor (\sqrt[3]{n}-\sqrt[3]{n+2})^3 \rfloor <8n+8$. Is this the right approach and can anyone help me complete the solution.

@TheRevolution Your statement that you're trying to prove $8n+7\leq\lfloor(\sqrt[3]{n}-\sqrt[3]{n+2})^3\rfloor\lt 8n+8$ is true only if the $-$ is replaced by $+$ (otherwise, the lower bound should be $-1$ and the upper bound is $0$). If so, then note that an Approach0 search indicates there's [How show $8|[(\sqrt[3]{n}+\sqrt[3]{n+2})^3]+1$](/q/567745), and there's also several AoPS threads, e.g., ... — John Omielan, Oct 26 '23 at 10:22
@TheRevolution number theory, Prove divisibility by 8, ssilwa's number theory problem, Integer functions, etc. — John Omielan, Oct 26 '23 at 10:24
Thanks a lot for providing clarity regarding the question! It helped me quite a lot. — The Revolution, Oct 26 '23 at 11:30
@TheRevolution You're welcome. FYI, your question, and its answer, inspired me to consider, and confirm, a different approach to solving the problem in [How show $8|[(\sqrt[3]{n}+\sqrt[3]{n+2})^3]+1$](https://math.stackexchange.com/q/567745/602049), with my technique actually giving a somewhat stronger result than required, as shown in my answer there. — John Omielan, Oct 26 '23 at 18:22

score 2 · Accepted Answer · answered Oct 26 '23 at 10:18

Clearly, $\sqrt[3]{n}-\sqrt[3]{n+2}<0$, so $(\sqrt[3]{n}-\sqrt[3]{n+2})^3<0$

Let $k=\sqrt[3]n\Rightarrow n=k^3$. Sińce $n>2>1$, $k>1$.

If $\sqrt[3]{k^3+2}=\sqrt[3]{n+2}\ge\sqrt[3]n+1=k+1=\sqrt[3]{k^3+3k^2+3k+1}$, then $$k^3+2\ge k^3+3k^2+3k+1\Rightarrow 3k^2+3k\le1$$ Since $3k^2+3k$ increases as $k$ increases, and the minimum value of $k$ is $1$, the minimum value of $3k^2+3k$ is $6$, and so there are no solutions for this.

Therefore, $\sqrt[3]{n+2}<\sqrt[3]n+1\Rightarrow\sqrt[3]n-\sqrt[3]{n+2}>-1\Rightarrow(\sqrt[3]{n}-\sqrt[3]{n+2})^3>-1$

Now that we have found $-1<(\sqrt[3]{n}-\sqrt[3]{n+2})^3<0$, we can conclude that $\left\lfloor(\sqrt[3]{n}-\sqrt[3]{n+2})^3\right\rfloor=-1\equiv-1\pmod8$.

Thanks a lot for the solution! – The Revolution Jan 25 '24 at 11:15 — The Revolution, Jan 25 '24 at 11:15

Prove that for any $n>2$, $n \in \Bbb N$, $\lfloor (\sqrt[3]{n}-\sqrt[3]{n+2})^3 \rfloor \equiv -1 \mod \ 8$

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