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Lemma.

If integers [MATH]a, b, c[/MATH] have only one common divisor [MATH]1[/MATH], [MATH]a+b=c[/MATH], number [MATH](c^n-a^n)[/MATH] and [MATH](c^n-b^n) [/MATH] have common prime divisor [MATH]d>3c^2[/MATH], in base d numbers [MATH]a^n, b^n, c^n[/MATH] finish by digit [MATH]1[/MATH], then with [MATH]n>1[/MATH] either number [MATH](a^n-b^n)/(a+b) [/MATH] is not divided by [MATH]d[/MATH], or number [MATH]3c^2[/MATH] is divided by [MATH]d[/MATH]d.

 

Lemma is not true. New version of the lemma:

If in the simple base q the mutually-simple numbers A^p, B^p, C^p (where C=c^n, B=b^n, A=a^n, prime n>2 and p=/q-1) finish by digit 1, then either A+B and A-B or C+B and C-B, or C+A and C-A have common divisor not equal to 1.

Today I have not its proof.

 

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