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Find the smallest solution in positive integers of { x }^{ 2 }

Question

Find the smallest solution in positive integers of $x^{2} = 41 y^{2} - 1$ .

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Elijah Discussed

Find the smallest solution in positive integers of

x^{2} = 41 y^{2} - 1

10 mins ago

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Text solutionVerified

We can write

41 = 6 + 41 - 6

= 6 + \frac{5}{41 + 6}

\frac{41 + 6}{5} = 2 + \frac{41 - 4}{5}

= 2 + \frac{5}{41 + 4}

\frac{41 + 4}{5} = 2 + \frac{41 - 6}{5} = 2 + \frac{1}{41 + 6}

Therefore,

41 = 6 + \frac{1}{2 +} \frac{1}{2 +} \frac{1}{12 +} ....

Here penultimate is

\frac{32}{5}

, and since the no. of quotients in the period is odd,

x = 32, y = 5

is a solution.

Penultimate convergent

= 6 + \frac{1}{2 + \frac{1}{2}}

= 6 + \frac{2}{5} = \frac{32}{5}

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Question Text	Find the smallest solution in positive integers of $x^{2} = 41 y^{2} - 1$ .
Answer Type	Text solution:1
Upvotes	150