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Can $6111\cdots111$ be a perfect cube?

I think it can be since a perfect cube can end in $2$ or more odd digits (like $456533$) and $611\cdots111$ is $7 \pmod 8$, which is one of the cubic residues of $8$. Also, $611\dots1$ is sometimes $0,1,8\pmod{9}$.

Can $611\cdots111$ be a perfect cube? If not, what is the proof?

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    $\begingroup$ This is a duplicate of your prior question which was closed for lack of context. $\endgroup$
    – lulu
    Jan 15, 2023 at 18:06
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    $\begingroup$ If $n^3$ is $6111...111$, last digit for $n$ is $1$. Last 2 digits are $71$, last 3 digits are $471$, last 4 digits are $8471$ etc ; step by step, you can deduce the last unknown digit. But each time you guess a digit, you add new constraints. $\endgroup$
    – Lourrran
    Jan 15, 2023 at 18:11
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    $\begingroup$ Your numbers are $\frac{1}{9}(55\cdot 10^n-1)$. $\endgroup$ Jan 15, 2023 at 19:55
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    $\begingroup$ I have manually checked all the way to $611...111 \ (307 \ 1$'s$)$, and there are no cubes there. $\endgroup$ Jan 15, 2023 at 20:46
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    $\begingroup$ Expanding the binomial $(1+9)^k = 1+9k \mod 27$ in $(1+2*27)(1+9)^k-1=9n^3\mod 27$ turns into $9k=9n^3\mod 27$ which can be rewritten as $k=n \mod 3$. $\endgroup$
    – Merosity
    Jan 15, 2023 at 22:41

3 Answers 3

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There are no solutions. As noted by others, the amounts to solving the equation $$ 55 \cdot 10^n-9y^3=1. $$ Writing $n=3n_0+k$, for $k \in \{ 0, 1, 2 \}$, this leads, crudely, to $3$ Thue equations of the shape $$ Ax^3-9y^3=1, \; \; A \in \{ 55, 550, 44 \}. $$ In each case, we can use, for example, Pari GP or Magma to solve the equation and show that there are no solutions (for cubic equations of this shape, it's somewhat easier than usual due to old work of Nagell which shows that any possible solutions correspond in a very precise sense to fundamental units in the associated cubic fields).

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Assume that there is an integer $x$ such that $6111\dots1$ is $x^3$, equivalently $$ 55\cdot 10^N=9x^3 +1 $$ for the corresponding power $N$. There are two cases, $N=2n$ ($N$ even) or $N=2n+1$ ($N$ odd).

  • In the case $N=2n$ we multiply with $3^4\cdot 55^3$ and obtain $$ (3^2\cdot 55^2\cdot 10^n)^2 = (3^2\cdot 55\cdot x)^3 + 3^4\cdot 55^3\ , $$ and thus an integral point on the elliptic curve of equation $$ Y^2 =X^3 + 3^4\cdot 55^3\ . $$ It turns out that we do not have any integral point on this curve. (Its rank is one, however the generator involves denominators.)
  • In the case $N=2n+1$ we multiply with $3^4\cdot 22^3$ and obtain $$ (3^2\cdot 22^2\cdot 5\cdot 10^n)^2 = (3^2\cdot 22\cdot x)^3 + 3^4\cdot 22^3\ , $$ and thus an integral point on the elliptic curve of equation $$ Y^2 =X^3 + 3^4\cdot 22^3\ . $$ It turns out that we have only the integral points $(X,Y)=(126,\pm 1692)$ on this curve. (Its rank is one.) But $X=126$ is not of the shape $(3^2\cdot 22\cdot x)$ with an integral $x$, so we do not have any solution.

Sage code supporting the above arguments:

  • For the first curve:

      sage: E = EllipticCurve(QQ, [0, 3^4*55^3])
      sage: E
      Elliptic Curve defined by y^2 = x^3 + 13476375 over Rational Field
      sage: E.rank()
      1
      sage: E.gens()
      [(-495/4 : 27225/8 : 1)]
      sage: E.integral_points()
      []
    
  • For the second curve:

      sage: E = EllipticCurve(QQ, [0, 3^4*22^3])
      sage: E
      Elliptic Curve defined by y^2 = x^3 + 862488 over Rational Field
      sage: E.rank()
      1
      sage: E.gens()
      [(126 : 1692 : 1)]
      sage: E.integral_points(both_signs=True)
      [(126 : -1692 : 1), (126 : 1692 : 1)]
    
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Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did a quick search using Mathematica.

Well, I did a quick Mathematica search and found that for $13587$ number of $1$'s there is no solution to your problem.

Notice that I mean by the number of $1$'s:

  • 1 one: $$61$$
  • 2 ones: $$611$$
  • 3 ones: $$6111$$
  • 4 ones: $$61111$$

etc.

And notice that your numbers can be written as:

$$\text{a}_\text{n}=\frac{55\cdot10^\text{n}-1}{9}$$

With $\text{n}\in\mathbb{N}$.

The code I used is:

Clear["Global`*"];
n = 1;
Monitor[Parallelize[
  While[True, 
   If[IntegerQ[FullSimplify[((55*10^n - 1)/9)^(1/3)]], Break[]]; n++];
   n], n]
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