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Question

Find the smallest solution in positive integers of x2=41y21.

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Solution

We can write 41=6+416
=6+541+6
41+65=2+4145
=2+541+4
41+45=2+4165=2+141+6
Therefore, 41=6+12+12+112+....
Here penultimate is 325, and since the no. of quotients in the period is odd,x=32,y=5 is a solution.
Penultimate convergent
=6+12+12
=6+25=325

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