INTRODUCTION
11, 111, 1111 in base ten are not pure powers, we can easily verify it, but
is it true for a number of n digits equal 1 in base ten, for all n. This study
is an answer to this question. A number with n digits equal to 1 (111
1)
in base ten is equal to a number with n digits equal to 9 (999
9) in base
ten divided by 9, which means
,
the theorem stipulates that
,
.
This research proposes an original and analytic proof of this theorem called
Bugeaud and Mignotte theorem (Bugeaud and Mignotte, 1999).
THE PROOF
Let Bugeaud-Mignotte equation:
 |
Then with j2 = -1 |
10n-9Yq = 1 = 10n+j(9jYq) |
We pose: |
10n = x, 9jYq = y, u = 1, z = 9jYq10n |
Lemma 1
Proof of Lemma 1
 |
And: |
 |
We pose: |
 |
until infinty. |
Lemma 2
The expression of the sequences is |
 |
Proof of Lemma 2
For I = 2, ,
it is verified. We suppose the expressions true for I |
 |
But: |
 |
Lemma 3
xi-jyi = x-jy
Lemma 4
The lemmas 1, 2, 3 imply that jy-x = 0
Proof of Lemma 4
 |
We pose: |
 |
We pose: |
 |
 |
And: |
 |
And: |
 |
If we make the hypothesis that  |
 |
And: |
 |
 |
But: |
 |
The hypothesis
is false and |
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In all cases |
 |
 |
is convergent, then the general term of the series tends to zero |
 |
And it is impossible: the existence of Y and q is impossible! |
CONCLUSION
Bugeaud-Mignotte equation has effectively no solution, and an analytic proof
exists. The generalization is
is it possible? It seems that there are only three solutions: