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 Diophantine Pair Decision (Posted on 2016-08-13)
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?

 No Solution Yet Submitted by K Sengupta No Rating

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 Possible solution Comment 3 of 3 |

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

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