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Repunit Rigor (Posted on 2011-10-25) Difficulty: 4 of 5
Can any base ten repunit, other than 1, be a perfect cube?

If so, give an example. Otherwise prove that no base ten repunit (other than 1) can be a perfect cube.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

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Possible Solution Comment 5 of 5 |

No.


I x^3 mod 100 = a gives a=71, x^3 mod 1000= a gives a=471, then 8471, 88471, 288471, 8288471, 68288471... There is at most one solution for any length of a; see A153042 in Sloane for a list of the digits. Call the part of a^3 and of length n comprising 1's the 'payload', and the rest the 'booster section'.
II  The booster section is of the form y*10^n and the payload is of the form 1/9(10^n-1).
III Now we have y*10^n+1/9(10^n-1)=a^3, which is 10^n(9y+1) =9a^3+1.
IV  To get some idea of the length of y compared to n, a must be not greater than 10^n, while we can discard the small terms, so  10^n(9y+1) is around 9(10^n)^3, y is around 100^n or 10^(2n).
V So to get a payload of length n, a booster section of approximate length 2n is needed.
VI Proof by descent: Now we can throw away the zeros at the end of y and start the same process again, since the problem requires that y, too, consist just of 1s.
VII But the length for which we are now trying to solve is shorter than the length that we started with, so we already know that there is no solution for y comprising all 1's within that length.

So {n,a,y}={1,1,0} is the sole solution.
Notes:
(1) Equivalently, the sum of the first n powers of 10 cannot be a cube.
(2) Using the substitution (((5y^2+4)^(n/2)-1)/((5y^2+4)^(1/2)-1))=x^p suggests that only 121, 343 and 400 are solutions for n>2, p>1 (121=11111_3,343=111_18, 400=1111_7). But this is just a speculation.

Edited on October 26, 2011, 5:14 am
  Posted by broll on 2011-10-26 00:08:15

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