Can repunits be perfect cubes? Is it true that the equation $10^{n}-9m^{3}=1$ has only one positive integer solution, namely $n=m=1$? I can't find the answer. This has an equivalent description that the repunits $R_n = 11\dots1$ are not cubic numbers.
 A: As expected $(m,n)=(1,1)$ is the only solution in positive integers 
of the exponential Diophantine equation $10^n - 9m^3 = 1$.
An entirely elementary proof of this seems unlikely because 
any finite list of congruence conditions on $n$ would have to allow 
$n=1$ (and also $n=0$, which corresponds to the empty repunit $0$), 
and thus could not exclude infinitely many other potential $n$.
A routine but non-elementary solution is to reduce to the three
cubic Thue equations $p^3-9m^3=1$, $10p^3-9m^3=1$, and $100p^3-9m^3=1$ 
(where $p = 10^{\lfloor n/3 \rfloor}$), and then use an effective and
practical algorithm for solving such equations.  gp takes only
a few milliseconds to process each of
thue(thueinit(x^3-9),1)
thue(thueinit(10*x^3-9),1)
thue(thueinit(100*x^3-9),1)

and reports that the third equation has no solutions, the second
only $(p,m)=(1,1)$, and the first only $(-2,-1)$ and $(1,0)$.
Since neither $-2$ nor $1$ is a positive power of $10$ we're done.
P.S. I see that quid linked to a page that in turn includes a link to
much the same proof except that the third case (which has no solution)
is disposed of by elementary considerations, namely reduction $\bmod 13$
(and as it happens there's a similar proof $\bmod 7$).
A: It is know by work of Bugeaud and Mignotte "Sur l'équation diophantienne $(x^n - 1)/(x-1)=y^q$, II" (see Thm 5) that a repunit in base 10 cannot even be a perfect power (so in particular not a cube). 
Another relevant reference is by the same authors "On integers with identical digits" containing among other things the result (Thm 2), which in particular gives all repunits that are perfect powers in all (nontrivial) bases up to $10$: 

Let $a$ and $b$ be integers with $2 \le b \le 10$ and $1 \le a \le b-1$. The integer $N$ with all digits equal $a$ in base $b$ is not a perfect power, except for $N=1,4,8,9$, for $N=11111$ written in base $3$, for $N=1111$ written in base $7$, for $N=4444$ written in base $7$.

There are earlier contributions to this problem by others, see the papers for references. (Both papers are freely available online on Bugeaud's webpage  http://www-irma.u-strasbg.fr/~bugeaud/publi.html see year 1999) 
It seems there was some discussion of this question on the Mersenne forum a while ago http://www.mersenneforum.org/showthread.php?t=16295 for cubes specifically and also a proof was given (scroll down a bit). I did not study it in detail, but if you want just the result for cubes it seems more accessible, yet it also involves solving Thue equations.  
Added: The more general questions to classify all repunits (in arbitrary bases) that are cubes or more generally perfect powers is AFAIK open. It is the question for solutions of the Diophantine equation (with $q=3$ for cubes)
$$\frac{x^n - 1 }{x-1} = y^q$$
mentioned in the title of the paper mentioned above, called  Nagell--Ljunggren equation.
It is conjecture that the set of all non-trivial (i.e., $n,q \gt 1$) solutions $(x,y,n,q)$ is given by $(3,11,5,2)$, $(7,20,4,2)$, and  $(18,7,3,3)$, the former two correpond to the two repunits (that are squares) mentioned in the result I recall above, and the third to the repunit in base $18$ that is a cube mentioned by Gerhard Paseman in a comment. 
Yet, as said this is open. But there are numerous partial results, for example: 


*

*no other than the mentioned ones (nontrivial) repunit is a square

*if there were another cube then it has at least $29$ digits, and the number of digits is a prime and $5$ mod $6$.
In fact in a different terminology, A geometric series equalling a power of an integer, the Nagell--Ljunggren came up a while ago on MO; there some additional informaition and pointers to literature can be found. 
A: $\frac {10^n-1} 9$ is not a cube:
Since a cube is either 0, 1 or -1 mod 9, the number of digits in the repunit is 
0, 1 or -1 mod 9.
If the number of digits is -1 mod 9, you can show it's not a cube by reducing 
mod 19.
If the number of digits is 1 mod 9, reduce mod 7 to show it's not a cube 
(except for 1).
Suppose $\frac {10^{9n}-1} 9$ is a cube.
$n =2^k m$ where $m$ is odd.
$\frac {10^{9n}-1} 9$ factors as 
$(10^{2^{k-1}9m}+1)...(10^{9m}+1)\frac {10^{9m}-1} 9$ and the factors are 
relatively prime, so each must be a cube.  
The factors that have a power of 10 plus 1 can't be cubes because a cube + 1 
isn't a cube.
Reducing the last factor mod 19, $10^9\equiv -1 (19)$ so $\frac {10^{9m}-1} 9\equiv 
\frac {-2} 9\equiv 4 (19).$  
$4^{18/3}=4^6\equiv 11 (19)\neq 1 (19)$ so this factor also is not a cube. 
So the product can't be a cube.
So no repunit base 10 is a cube.
