# Is it possible to find a perfect cube like 111…11?

Can we find a perfect cube like $111...111$(all digits are $1$), apart from the number $1$ itself?

It's easy to prove that there can't be anything like $111...11$ that is a perfect square besides $1$, but how to do this for perfect cube? Are there some new techniques to do this?

• I've been picking away at this a little bit, but no definite answers yet. I've found that is question is equivalent to "is there an integer output for $log(9n^3 + 1)$" where n is an integer and the log is base 10. – Kaynex Feb 6 '16 at 5:26
• No solutions below $10^4$ digits. – Lucian Feb 6 '16 at 5:39
• I was just working on this exact same question yesterday ! I know the text book said that it's easy to prove there can't be a square like that ... But could you elaborate on that reason. I can't seem to get it. – user230452 Feb 7 '16 at 4:27
• I'll just leave an answer to a simpler problem like the heuristic suggests and hope there's a pattern. Every number relatively prime to 10 has a multiple consisting only of 1's. – user230452 Feb 7 '16 at 4:33
• I just wonder if the generalized statement is also valid: There exists one and the only one integer $n$ such that $$n^k=\frac{10^n-1}{9} , \forall k\in\Bbb{N}$$ which is 1. – Mythomorphic Feb 7 '16 at 6:00

(Edited version)

If $$k^3=\frac{10^n-1}9$$

then $$k^3-1=\frac{10^n-10}{9}$$ $$9(k-1)(k^2+k+1)=10(10^{n-1}-1)$$

So if such $k$ exists and both sides are non-zero (i.e. $n\neq1$ and $k\neq1$), it implies that,

First situation:

When either of the factors $k^2+k+1$ or $k-1$ is divisible by $10$,

$$k^2+k+1\equiv 0 \text{ (mod 10) or }k=10m+1$$

Notice $\forall k\ge 0$, $k^2+k+1$ must be odd, so the first condition is not satisfied.

For the second condition, $k=10m+1$, if it is true, then

$$9(10m+1)^3=10^n-1$$

Rearrange and simply and we get

$$100m^3+30m^2+3m=\frac{10^{n-1}-1}{9}.$$

So for every $n$, the premise to get this condition right is to get it correct for every $n-1$. So finally it is equivalent to prove when $n=1$, there exist a positive integer(s) $m$ such that $m(100m^2+30m+3)=0$.

But no real solution except $m=0$.

When $m=0, k=1, n=1$

Second situation:

Only $k-1$ is divisible by 5. But then $k^2+k+1$ is divisble by 2, which is impossible.

Third situation:

Only $k-1$ is divisible by 2. We write $k=2t+1$. So we investigate whether $k^2+k+1$ is divisible by 5.

Observe $$(2t+1)^2+(2t+1)+1=4t^2+6t+3$$ and $$4t^2+6t+3\equiv -t^2+t-2\pmod{5}$$

This is equivalent to state that

$$t^2-t+2\equiv 0 \pmod{5}$$

But by inspection, this is not valid. (By testing the case 5q, 5q+1, ... , 5q+4)

So this situation is impossible.

• I still don't see how you conclude $k^2+k+1\equiv0\pmod{10}\text{ or } k=10m+1$. It appears you are taking your equation $9(k-1)(k^2+k+1)=10(10^{n-1}-1)$ mod 10 to read $-(k-1)(k^2+k+1) \equiv 0 \pmod{10}$ and concluding that therefore one factor must be congruent to 0, mod 10. But that conclusion isn't valid in general: 10 is composite so the integers mod 10 are not a field; there are zero divisors. For instance, $5 \cdot 2 \equiv 0 \pmod{10}$ yet neither factor is congruent to 0. One would need another argument and I don't see what it is. – Nate Eldredge Feb 6 '16 at 7:05
• The correction propagates : you should write $10(900m^3+270m^2+27m-10^{n-1}+2)-10=0$ rather than $10(900m^3+270m^2+27m-10^{n-1}+2)-1=0$, and $10r-10=0$ rather than $10r-1=0$ – Ewan Delanoy Feb 6 '16 at 7:06
• You can indeed deduce that k=1(mod10). But 9(10m+1)^3=10n−10 is simply wrong. – Li Xinghe Feb 6 '16 at 7:17
• @NateEldredge I also don't see immediately that this alternative holds, but we can save the argument as follows: clearly $2$ doesn't divide $k^2+k+1$, and if you look at all solutions possible $k$ modulo $5$, we have that $k^2+k+1$ isn't divisible by $5$. Because RHS is divisible by $10$ and $9(k^2+k+1)$ is relatively prime to $10$, $10\mid k-1$. – Wojowu Feb 6 '16 at 7:30
• If $k=10m+1$, does this not imply that $9(10m+1)^3=10^n-1$, not $10^n-10$? – S.C.B. Feb 6 '16 at 10:22