 Posers and Puzzles

1. 04 Jun '08 15:042 edits
For which positive integers x,y does the following equality hold?

(x^2 + xy - y^2)^2 = 1
2. 04 Jun '08 15:56
is 0 a positive integer?
3. 04 Jun '08 15:57
Originally posted by forkedknight
is 0 a positive integer?
No
4. 04 Jun '08 15:59
Originally posted by David113
For which positive integers x,y does the following equality hold?

(x^2 + xy - y^2)^2 = 1
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.
5. 04 Jun '08 16:01
Originally posted by David113
For which positive integers x,y does the following equality hold?

(x^2 + xy - y^2)^2 = 1
Nice! It's a Fibonacci series.
6. 04 Jun '08 16:04
Originally posted by PBE6
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.
Here's part of the explanation.

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 🙂
7. 04 Jun '08 16:081 edit
Originally posted by PBE6
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.
Very 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.

Edit - Nice one, mtthw.