Each of X and Y is a distinct positive integer, such that:

1. X and Y are relatively prime, and:
2. Each of (X² – 12)/Y and (Y² – 12)/X is an integer.

Does there exist an infinite numbers of pairs (X,Y) satisfying the given conditions?

Give reasons for your answer.

Let's start from the work already done, so that we have 4 numbers in a recurrence relation: K1,K2,K3,K4, and:

(K2^2-12)/K1 = K3, (K3^2-12)/K2 = K4; such that

K3 = 10(K2)-(K1), and K4=(99(K2)-10(K1)).

Now ((10(K2)-(K1))^2-12)/K2 =(99(K2)-10(K1)); using more conventional algebra: ((10y-x)^2-12)/y= (99y-10x),

simplifying to x^2+y^2-10xy-12=0, with infinite solutions {{x == 1, y == -1}, {x == 1, y == 11}, {x == 11, y == 1}, {x == 11, y == 109}, {x == 109, y == 11}, {x == 109, y == 1079}, {x == 1079, y == 109}, {x == 1079, y == 10681},...}

From these we need to deduct those where y is negative or less than x, as we assumed earlier that x was smaller. There remain an infinite number of solutions, as was to be shown:

Xn+1 = P Xn + Q Yn 
Yn+1 = R Xn + S Yn 
with characteristics 
P = 10 
Q = -1 
R = 1 
S = 0 
and  
P = 0 
Q = 1 
R = -1 
S = 10 

Edited on August 14, 2016, 5:06 am

Posted by broll on 2016-08-14 05:01:27