The continued fraction for squareroot 3 is [1, 1, 2], use it to find 3 non-trivi

Question

The continued fraction for squareroot 3 is [1, 1, 2], use it to find 3 non-trivial solutions to u^2 - 3 upsilon^2 = 1 with u, v element Z. On HW 8 you solved x^2 + 3y^2 = 1 with rational x, y and found x = 1 - 3t^2/1 + 3t^2 and y = 2t/1 + 3t^2 Verify that dividing x^2 + 3y^2 = 1 by x^2 and making the substitution u = 1/x and upsilon = y/x brings it into the form 1 + 3 upsilon^2 = u2. Compute formulas for u and upsilon in terms of t, as well as a formula for t in terms of u and upsilon. Plug in the particular values of u and upsilon found in part a to get the corresponding values of t.

Explanation / Answer

sqrt (3) = 1 + 1/1 + ½ + …..

If we exclude the last element of the cycle and calculate the convergent fraction we get 2/1. Thus (2, 1) is the first fundamental/non trivial solution to u2 - 3 v2 = 1

Once you know the first non trivial solution, one can find the rest by with the coupled recurrence formulas

Xk+1 = aXk + nbYk
Yk+1 = bXk + aYk,

Which can be simplified to Xk+2 = 2aXk+1 - Xk and Yk+2 = 2aYk+1 - Yk, where X0 = 1, X1 = a, Y0 = 0, and Y1 = b.

Using the relation a2 - 1 = nb2, the explicit formulas for Xk and Yk are

Xk = 0.5*(a + b*sqrt(n))k + 0.5*(a - b*sqrt(n))k
Yk = (1/2sqrt(n))*(a + b*sqrt(n))k - (1/2sqrt(n))*(a - b*sqrt(n))k

n = 3; a = 2; b = 1

X2 = 0.5*(2 + 1*sqrt(3))2 + 0.5*(2-1*sqrt(3))2

Y2 = (1/2sqrt(3))*(2 + 1*sqrt(3))2- (1/2sqrt(3))*(2 - 1*sqrt(n))2

Gives us second solution (7, 4) Similarly third solution will be with values n = 3; a = 7; b = 4 and k = 3

Which gives (26,15)

b)

x2 + 3y2 -1 /x2 = 0

1 + 3 y2/x2 - (1/x)2 = 0 and replace u = 1/x and v = y/x

1+ 2v2 – u2 = 0

u2 = 2v2 +1  

c)

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