Each of X and Y is a distinct positive integer, such that:
 X and Y are relatively prime, and:
 Each of (X^{2} – 12)/Y and (Y^{2 }– 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:
(K2^212)/K1 = K3, (K3^212)/K2 = K4; such that
K3 = 10(K2)(K1), and K4=(99(K2)10(K1)).
Now ((10(K2)(K1))^212)/K2 =(99(K2)10(K1)); using more conventional algebra: ((10yx)^212)/y= (99y10x),
simplifying to x^2+y^210xy12=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 20160814 05:01:27 