Does there exist an infinite number of pairs (A,B) of positive integers such that:
 A divides B^{2} + 1 and:
 B divides A^{2} + 1?
Give valid reasons for your answer.
(In reply to
a suggestive series by xdog)
First off your series contains an error: you left out 34.
Take n=3 for an example, where A(3)=F(5) and B(3)=F(7).
what your really found is F(7)^2+1=F(5)*F(9) and F(5)^2+1=F(3)*F(7)
Which is the oddnumbered Fibonacci terms as you pointed out.
What's left is to show F(n)^2+1 = F(n2)*F(n+2) for the odd terms.
F(n)=[(phi)^n(1)^n/(phi)^n]/√5
I'll leave out the tricky math but these are the results.
F(n)^2=[(phi)^(2n)+phi^(2n)2*(1)^n]/5
F(n2)*F(n+2)=[(phi)^(2n)+(phi)^(2n)+7]/5
As you can see when n is odd,
F(n)^2+1=[(phi)^(2n)+phi^(2n)+2*(1)]/5+1 = [(phi)^(2n)+(phi)^(2n)+7]/5
but if n is even instead the result is [(phi)^(2n)+(phi)^(2n)+3]/5

Posted by Jer
on 20150809 22:06:40 