04 Jun '08 15:042 edits

For which positive integers x,y does the following equality hold?

(x^2 + xy - y^2)^2 = 1

(x^2 + xy - y^2)^2 = 1

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False berry04 Jun '08 15:59

Weird!! Just tried a simple brute force sampling, but the pairings I get are:*Originally posted by David113***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.- Joined
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04 Jun '08 16:04

Here's part of the explanation.*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.

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 ðŸ™‚- Joined
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Halfway04 Jun '08 16:081 edit

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.*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.

Edit - Nice one, mtthw.