# [seqfan] Re: Embedding Knuth's Fibonacci (circle) product in a ring defined over pairs.

Don Reble djr at nk.ca
Sun Sep 11 04:33:50 CEST 2022

```> Define (j1,k1) o (j2,k2) = (j1*j2 + k1*k2, j1*k2 + k1*j2 - k1*k2),
> with (j1,k1) + (j2,k2) = (j1+j2, k1+k2) for addition as you might
> expect.

It's easy to map this ring into the real numbers.
Suppose that (a,b) = a+bx (for some real or complex x),
and that (a,b)o(c,d) = (a+bx)(c+dx) (ordinary multiplication).
Then (a,b)o(c,d) = ac + bdx^2 + (ad+bc)x.
If that equals (ac+bd) + (ad+bc-bd)x, then x^2 = 1 - x,
and x = (-1 +- sqrt(5)) / 2. Either mapping is good.

> This seems to check out (with integer components) as a commutative
> ring, possibly becoming a field if the components are rationals.

Those (any) subrings of the real numbers are commutative.

Let x,X be the two roots of that quadratic: xX = -1, x+X = -1.
The norm of (a+bx) is (a+bx)(a+bX)
= a^2 + ab(x+X) + b^2(xX)
= a^2 - ab - b^2
And 1/(a,b)
= (a+bX) / [(a+bx)(a+bX)]
= [a+b(-1-x)] / (a^2 - ab - b^2)
= (a-b,-b) / (a^2 - ab - b^2)
For nonzero a,b, if the denominator is zero, then
(a/b)^2 - (a/b) - 1 = 0, and a/b is irrational.
So with rational coefficients, it's a field.

Today's puzzle: what are the units of the integer-coefficient ring?
(For which integral (a,b) is a^2-ab-b^2 = +1 or -1?)

--
Don Reble  djr at nk.ca

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