Find a + b where the continued fraction has recurring 2-0-1-8 pattern as shown below: b a = 4 + 2 + 0 + 1 + 8 + 2 + 0 + 1 + 8 + ⋱ 1 1 1 1 1 1 1 1
a and b are coprime positive integers.
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We write b a = 4 + x , so that
x x = 2 + 0 + 1 + 8 + 2 + 0 + 1 + 8 + . . . . 1 1 1 1 1 1 1 1 = 2 + 0 + 1 + 8 + x 1 1 1 1 = 2 + 0 + x + 9 x + 8 1 1 = 2 + x + 8 x + 9 1 = 3 x + 2 5 x + 8
This leads to 3 x 2 + 2 4 x − 8 = 0 , which yields 3 ( x + 4 ) 2 − 5 6 = 0 and x = - 4 ± 3 5 6 . We discard the negative root as the continued fraction is clearly positive.
Thus b a = x + 4 = 3 5 6 , so a + b = 5 9 .