Originally posted by David113Weird!! Just tried a simple brute force sampling, but the pairings I get are:
For which positive integers x,y does the following equality hold?
(x^2 + xy - y^2)^2 = 1
(1,1)
(2,3)
(5,8)
(13,21)
(34,55)
which is just the Fibonacci numbers taken in pairs. Interesting! I'll have to take a look at the Fibonacci identities to see where this comes from.
Originally posted by PBE6Here's part of the explanation.
Weird!! Just tried a simple brute force sampling, but the pairings I get are:
(1,1)
(2,3)
(5,8)
(13,21)
(34,55)
which is just the Fibonacci numbers taken in pairs. Interesting! I'll have to take a look at the Fibonacci identities to see where this comes from.
If F(x, y) = x^2 + xy - y^2
Then F(y, x + y) = -x^2 - xy + y^2 = -F(x, y)
(just try the substitution)
So once you have (1, 1) as a solution, you can construct the rest of the Fibonacci series from it: (1, 2), (2, 3), (3, 5), (5, 8) etc...
So all you need to show is that there aren't any other solutions 🙂
1 edit
Originally posted by PBE6Very interesting. I'm thinking the geometric representation (with the squares) of the Fibonacci sequence might be an easy way to find out why, but I can't pin it down just yet.
Weird!! Just tried a simple brute force sampling, but the pairings I get are:
(1,1)
(2,3)
(5,8)
(13,21)
(34,55)
which is just the Fibonacci numbers taken in pairs. Interesting! I'll have to take a look at the Fibonacci identities to see where this comes from.
Edit - Nice one, mtthw.