Theorem There are no perfect cube repunits greater than 1. Proof Let denote the repunit of length n. We consider six cases: one for each of the possible residues of n modulo 6. Case 1: ≡ 0 6. We have that 6 for some positive integer k. By the formula for the sum of a geometric series, we have that 6 110100⋯10 . Let us suppose that 6 is a perfect cube, say 6 for some positive integer x. Then, it follows that if we put 10, so that 10 10, this is equivalent to the existence of a nontrivial solution in positive integers to the Thue equation 9 1. But with the assistance of the Thue equation solver in PARI/GP, a number-theoretic computer algebra system, we see that the equation 9 1 indeed has only the trivial integral solution , 0, 1. This solution does admit the existence of the perfect cube “degenerate” repunit, namely 0 00, but we may conclude that there are no other perfect cube repunits of length 6n. Q. E. D. Case 1. Case 2: ≡ 1 6. Following a line of reasoning similar to that of Case 1, let us suppose that 6 1 110100⋯10 , and moreover, that 6 1 is a perfect cube, say 6 1 for some positive integers k and x. As in Case 1, put 10, so that 10 10 and thus 10 10. It follows then that there must be a nontrivial solution in positive integers to the Thue equation 10 9 1. Once again, we turn to the Thue equation solver in PARI/GP to find that this Thue equation has only the trivial integral solution , 1, 1, which admits only the existence of the trivial perfect cube repunit 1 11, and no other perfect cube repunits of length 6n + 1. Q. E. D. Case 2. We make the observation that the hardest of the work is now behind us. Cases 3-6 follow merely from elementary properties of modular arithmetic! Case 3: ≡ 2 6. Observe that perfect cubes modulo 13 must have a residue of 0, 1, 5, 8, or 12. But 6 2 919 8 ∙ 3 11 13. Hence there cannot exist a perfect cube repunit of length 6n + 2. Q. E. D. Case 3. Case 4: ≡ 3 6. Again working modulo 13, we see that 6 3 12 1 9 11 ∙ 3 7 13, which is a cubic nonresidue modulo 13. There are hence no perfect cube repunits of length 6n + 3. Q. E. D. Case 4. Case 5: ≡ 4 6. We have that 6 4 31 9 2∙36 13, which is a cubic nonresidue modulo 13. There are hence no perfect cube repunits of length 6n + 4. Q. E. D. Case 5. Case 6: ≡ 5 6. Finally, note that 6 5 41 9 3 ∙ 3 9 13, which once again is a cubic nonresidue modulo 13. There are hence no perfect cube repunits of length 6n + 5. Q. E. D. Case 6. After considering all possible cases, we conclude that the only perfect cube repunits are the “degenerate” 0 00 and the more commonly recognized 1 11. There are no other perfect cube repunits. Q. E. D. in toto. .