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 Quadratic and Divisibility Puzzle (Posted on 2015-08-09)
Does there exist an infinite number of pairs (A,B) of positive integers such that:
• A divides B2 + 1 and:
• B divides A2 + 1?

 No Solution Yet Submitted by K Sengupta No Rating

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 re: a suggestive series / explained Comment 2 of 2 |
(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 odd-numbered Fibonacci terms as you pointed out.

What's left is to show F(n)^2+1 = F(n-2)*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(n-2)*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 2015-08-09 22:06:40

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