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In essence {xn-1, yn-1}={X,Y}, {an, bn}={A,B} X+Y=2(B-A) implies that the two sets (‘columns’) of solutions are identical, subject to an ‘offset’ of 1.
Checking small values suggests that the expressions are indeed equivalent;
we observe
(i) that both columns appear to be of the form: 6*p^2+1=q^2; and note also
(ii) that a^3-(a-1)^3=(b^2)+(b-1)^2 simplifies to 3a(a-1) = 2b(b-1), from which expression 3x^2-2y^2=1 follows automatically, since:
I 3(2a-1)^2-3= 2(2b-1)^2-2 Multiplying by 4, completing squares.
II 3(2a-1)^2-2(2b-1)^2= 1 Simplifying
Now we know the relationship between the {a,b} and {x,y} terms, we may write WLOG, using (i) as our inital assumption : 6*(2(b-a))^2+1=c^2, 6*( (2a-1)+(2b-1))^2+1=d^2.
Using the formulation 3(2a-1)^2-2(2b-1)^2=1, we demonstrate the expected equivalence algebraically:
First column {0,2,20,198..};
I 3(2a-1)^2-2(2b-1)^2=1, 6*(2(b-a))^2+1=c^2
II 12a^2-12a-8b^2+8b+1 = 1, 24a^2-48ab+24b^2+1 = c^2
III 12a^2-12a-8b^2+8b+1+24a^2-48ab+24b^2+1 = c^2+1
IV Hence (6a-4b-1)^2=c^2; c=(6a-4b-1)
Second column {2,20,198,1960..}:
I 3(2a-1)^2-2(2b-1)^2=6*( (2a-1)+(2b-1))^2+1=d^2
II 24(a+b-1)^2+1 = d^2, or
III 24a^2+48ab-48a+24b^2-48b+25 = d^2
IV Then 12a^2-12a-8b^2+8b+1+ 24a^2+48ab-48a+24b^2-48b+25 = d^2+1
V Hence (6a+4b-5)^2 = d^2; d=(6a+4b-5)
Putting this together:
I 6*(2(b-a))^2+1=(6a-4b-1)^2, 6*( (2a-1)+(2b-1))^2+1=(6a+4b-5)^2
II Simplifying rather nicely to 3a(a-1)= 2b(b-1), 3a(a-1)= 2b(b-1).
Hence the two columns are identical, with an ‘offset’ of 1, as was to be shown.
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