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.



Can repunits be perfect cubes?
Rather to my surprise I find an entirely elementary proof that
$R\_1=1$ is the only (decimal) repunit cube, using nothing beyond
quadratic reciprocity (namely the formula for the Legendre symbol $(-5/p)$).


Let us first dispose of the case that $n$ is even, say $n=2k$.
This is routine: write the equation $10^n - 9m^3 = 1$ as
$$
9m^3 = 10^n - 1 = 10^{2k} - 1 = (10^k-1) (10^k+1).
$$
The two factors are relatively prime, and $10^k-1$ is a multiple of $9$.
Thus once $k>0$ it follows that $10^k-1 = 9m\_1^3$ and $10^k+1 = m\_2^3$
for some $m\_1^{\phantom.},m\_2^{\phantom.}$
with $m = m\_1^{\phantom.} m\_2^{\phantom.}$.
But then $m\_2^3 \equiv 2 \bmod 9$, which is impossible.
(The solution $(m,n) = (0,0)$ escapes because in that one case
the factor $10^k-1$ is zero so there's no condition on $10^k+1$.
We could also have used descent, since $10^k-1 = 9m\_1^3$
would be a smaller solution of the same Diophantine equation;
the solution $(m,n) = (0,0)$ escapes this argument because $k=0=n$
so the new solution is no smaller.)


The hard case is $n$ odd. For $n=1$ we obtain the
known solution $(m,n)=(1,1)$; and there is no solution with $n=3$
because $R\_3 = 111$ is not a cube. We may thus suppose $n \geq 5$,
and write $n = 2k+3$ with $k$ a positive integer. Now we take
the strange step [**see comment at bottom**] of
adding $9$ to both sides of the equation
$9m^3 = 10^n - 1$, and writing the result as
$$
9(m+1)(m^2-m+1) = 9m^3 + 9 = 10^n + 8 = 10^{2k+3} + 8
= 8 \left(125(10^k)^2 + 1 \right).
$$
Thus if $p$ is any odd prime factor of $m+1$ then $p \neq 5$ and
$25 \cdot 10^k$ is a square root of $-5 \bmod p$, so $(-5/p) = +1$
and by quadratic reciprocity $p$ is one of $1,3,7,9 \bmod 20$.


As usual, this set of residues $\bmod 20$ is closed under multiplication,
so we conclude that **any** odd factor of $m+1$, prime or not, is one of
$1,3,7,9 \bmod 20$. In particular this is true of $(m+1)/2^f$,
where $2^f$ is the largest power of $2$ dividing $m+1$.


But since $n > 3$ (*this is how the solution $(m,n)=(1,1)$ escapes
the coming contradiction*) we have $f=3$ because $125(10^k)^2+1$ is odd
(as is the complementary factor $9(m^2-m+1)$ on the left-hand side).
Moreover, once $n \geq 5$ we have $R\_n \equiv 11111 \equiv 7 \bmod 2^5$,
so the putative cube root $m$ of $R\_n$ would be $23 \bmod 2^5$, making
$(m+1)/8 \equiv 3 \bmod 4$. Since also $m \equiv 1 \bmod 5$ we'd conclude
that $(m+1)/8 \equiv 4 \bmod 5$ and thus $(m+1)/8 \equiv 19 \bmod 20$.
Since this is not among the four allowed residues we are done. **QED**


The same approach deals with some other cases of the repunit-power problem,
but does not settle it completely. For example, in the decimal case
$9m^q+1 = 10^n-1$ has no solution for any $q \equiv 3 \bmod 4$
(using descent to reduce to the case of odd $n$, and then factoring
$9(m^q+1) = 10^n+8$ as before). There are no nontrivial solutions for
$2|q$ or $5|q$ (by reduction mod $2^2$ and $5^2$ respectively), and
$q=9$ is a special case of $q=3$, so $q=13$ is the first exponent
(other than $q=1$...) that we cannot exclude this way.


**about the "strange step"**: this feels very artificial, though
I'm not familiar enough with the literature on this Nagell-Ljunggren equation
or related problems to tell if it's a standard technique.
The only other time I remember such a thing working is for a problem
that I concocted for the purpose 30+ years ago:
Prove that the Diophantine equation
$$
y^2 = 7x^2+8x-3
$$
has no *positive* integer solutions. There are infinitely many
solutions with $x<0$, such as $(x,y) = (-2,3)$, so there's no easy
congruence argument (though the problem is routine using the
theory of Pell's equation). However, adding $9x^2$ to both sides
yields $(3x)^2 + y^2 = (4x-1)(4x+3)$, at which point the two-square theorem
soon produces a contradiction
(NB if $x<0$ then $\left|4x-1\right|$ and $\left|4x+3\right|$ are
$+1 \bmod 4$ !). 


