1. Select positive integers a and b such that a^3-(a-1)^3=(b^2)+(b-1)^2 starting with {a,b}={1,1}, numbering successive solutions {a₁,b₁}...{a_n,b_n}

2. Select positive integers x and y such that 3x^2-2y^2=1 starting with {x,y}={1,1}, again numbering successive solutions {x₁,y₁}...{x_n,y_n}

3. Let n exceed 1. Let {x_n-1, y_n-1}={X,Y}; let {a_n, b_n}={A,B}.

Prove that X+Y=2(B-A).

Part 2:
The integer solutions, (x_n, y_n), of   3x² - 2y² = 1     (1),  are given by

(sqrt(3) + sqrt(2))^2n-1 = x_n sqrt(3) + y_n sqrt(2)  for n = 1, 2..

Thus:    x_n = (p^2n-1 + q^2n-1)/(2*sqrt(3))  and   y_n = (p^2n-1 - q^2n-1)/(2*sqrt(2))

where  p = sqrt(3) + sqrt(2),    q = sqrt(3) - sqrt(2)   and  pq = 1.

This gives values of (x, y) as  (1, 1), (9, 11), (89, 109), (881, 1079),...

with recurrence relations:           x_n = 10x_n-1 - x_n-2   and   y_n = 10y_n-1 - y_n-2

Part 1:
The equation  a³ - (a - 1)³ = b² + (b - 1)² can be rearranged to give

3(2a - 1)² - 2(2b - 1)² = 1,  which is (1) with a = (x + 1)/2 and b = (y + 1)/2.

Therefore the solutions for (a, b) are  (1, 1), (5, 6), (45, 55), (441, 540),..

Part 3:
            2(B - A) = 2(b_n - a_n)  = y_n - x_n

                        = (p^2n-1 - q^2n-1)/(2*sqrt(2)) - (p^2n-1 + q^2n-1)/(2*sqrt(3))  

                        = [p^2n-1(sqrt(3) - sqrt(2)) - q^2n-1(sqrt(3) + sqrt(2)]/(2sqrt(6))

                        =(p^2n-1q  -  q^2n-1p)/(2sqrt(6))

                        = (p^2n-2  -  q^2n-2)/(2sqrt(6))           since pq = 1

Also      X + Y    = x_n-1 + y_n-1

                        = (p^2n-3 + q^2n-3)/(2sqrt(3)) + (p^2n-3 - q^2n-3)/(2sqrt(2))

                        = [p^2n-3(sqrt(3) + sqrt(2)) - q^2n-3(sqrt(3) - sqrt(2))]/(2sqrt(6))

                        = (p^2n-3p  -  q^2n-3q)/(2sqrt(6))

                        = (p^2n-2  -  q^2n-2)/(2sqrt(6))

Therefore:  2(B - A) = X + Y

Edited on May 27, 2011, 9:33 pm

Posted by Harry on 2011-05-27 21:32:24