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because 58 is the order of G, the multiplicative group of nonzero integers modulo 59 (a prime). If
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then m must be a multiple of r, the order of 10 in the group G. Now r must be a divisor of the group order |G| = 58, i.e. its possible values are 1, 2, 29, 58. In my posts above, I eliminated 1, 2, 29. Therefore r = 58 (i.e. 10 is generator of the cyclic group G).
]]>So there you have it:
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is also a solution to
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i.e. to
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and vice versa. By Fermat’s little theorem, as 59 is prime,
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Therefore the smallest positive value of n+1 must divide 58. Clearly 10 ≢ 1 (mod 59) and 10² = 100 ≢ 1 (mod 59). If you can show that
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then n+1 will be a multiple of 58 and the general solution will be n = 58k − 1, k = 1, 2, 3, ….
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