Question: Let xn+1 = (axn + c) mod m, where 2 < a < m, 0 < c < m, 0 < x0 < m, 0

Let xn+1 = (axn + c) mod m, where 2 < a < m, 0 < c < m, 0 < x0 < m, 0 < xn+i < m, and n > 0. Prove that
xn = (anx0 + c[(an - 1)/(a - 1)]) mod m, 0 < xn < m.

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