If $n^3-9n+27$ is congruent to 0 mod 81 we have that $n^3+27n$ is congruent to 54 mod 81. $n^3+27n=n(n^2+27)$ and I believe the residues start to cycle? I have not checked yet though.

Assume $81\mid n^3-9n+27$ for contr. Then $3\mid n^3-9n+27$, so $3\mid n^3\,\Rightarrow\, n=3k$. $n^3-9n+27=27(k^3-k+1)$. By Little Fermat $k^3-k+1\equiv 1\not\equiv 0\pmod{3}$, so contr.

Nice solution randomasdf97!

If you don't know Fermat's Little Theorem, you can also note that $k^3 - k + 1 = (k-1)k(k+1) + 1$. As $k-1$, $k$, $k+1$ are consecutive integers, one of them has to be a multiple of $3$.

techguy2 wrote:

Is this a valid proof?

It should be if you proved all of the cases correctly.

randomasdf97 nice clever solution!

franchester wrote: It should be if you proved all of the cases correctly. -__- I thought so. I guess I was asking if all the cases were proved correctly.

If n^3 - 9 (n-3) is a multiple of 81 then n^3 must be a multiple of 9, that is n should be a multiple of 3. Let N = 3k. So, given expression = 27 (k^3 -k+1), which will be a multiple of 81 only if k^3 - k +1 is a multiple of 3. But k^3 -k = (k-1)k (k+1)= multiple of 3. Hence, k^3 - k +1 is not a multiple of 3.

I did like the others but I put here anyways. So, we want to prove that for any integer $n$, $81\nmid n^3-9n+27$. For contradiction, let's suppose that for some $n$, $81\mid n^3-9n+27$. As $3\mid 81\mid n^3-9n+27\implies 3\mid n^3-9n+27\implies 3\mid n^3$. Then, we must have $n^3\equiv 0 (mod 3)\implies n\equiv 0(mod 3)$. Now, let's say that $n= 3k, k \in\mathbb{Z}$. Putting this on the original equation, we will have: $81\mid (3k)^3-9\cdot (3k)+27\implies 81\mid 27k^3-27k+27$. Simplifying this by $27\implies 3\mid k^3-k+1\implies$ $$k^3-k+1\equiv 0(mod 3)\implies k(k^2-1)+1\equiv 0(mod 3)\implies (k-1)k(k+1)\equiv -1 (mod 3)$$That's obviously false, because, if we have 3 integers consecutive, we must have one of these multiple by 3. So, the product would be $\equiv 0(mod 3)$, and not $\equiv -1 (mod 3)$. Therefore, there isn't any integer $n$ such that $81\mid n^3-9n+27$

take n=3k and then 3 divides $ k^3-k $