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Friday, April 17, 2009

Mystery with Pell's Equation parametric solution

What I've managed to find from a result that followed from my Quadratic Diophantine Theorem is a parametric solution to Pell's Equation in rationals:

With x2 - Dy2 = 1

I have proven:

y = 2[f2v - 1]/[D - (f2v - 1)2]

and

x = [D + (f2v - 1)2]/[D - (f2v - 1)2]

where f1f2 = D-1, and the f's are non-zero integer factors, while v is nonzero but is otherwise a free variable.

Which is, of course, looks similar to the parametric solution for circles in rationals discovered in antiquity:

x = (1 - t2)/(1 + t2)

y = 2t/(1 + t2)

See: http://mathworld.wolfram.com/Circle.html eqns. 16 & 17

It looks similar as it is the parametric solution for circles when D=-1, for ellipses with D<0, and of course, hyperbolas are given by D>0.

In considering what I see as the mystery of the quietness in the mathematical literature around this solution, it seems possible that the focus on Pell's Equation as a Diophantine equation, considering integers only, can be part of it.

Research on the web indicates that the parametric solution I've found was known much earlier, as Fermat was aware of it. So now I wonder why it gets so little mention and why wasn't the relation to rational parametric solutions for circles, ellipses and hyperbolas noted? Or was it? I'm continuing research.

There are other interesting things I've found.

For example, for any positive integer D, if D+2 is a perfect square, then D+1 is the first solution to Pell's Equation for x.

e.g. 82 - 7*32 = 1, 152 - 14*42 = 1

In general, if D+2 is a perfect square, then D = n2 - 2 for some natural number n, and

(n2 - 1)2 - (n2 - 2)n2 = 1

and notice n4 - 2n2 + 1 - n4 + 2n2 = 1.

But that is probably something so well-known that it's considered trivial.

Another thing is the results with the negative Pell's Equation and two other alternates shows that in general the alternates can be solved in order to solve Pell's Equation, as they must exist for prime D, and a method can be generalized for composite D as well.

e.g. For the famous D=61 case, the solution from the negative Pell's Equation is not as spectacular:

j2 - Dk2 = -1

297182 - 61*38052 = -1

gives x = 1766319049, from x = 2j2 + 1.

17663190492 - 61*2261539802 = 1

The difference in computation is staggering though as my example above suggests, as the first x for Pell's Equation is roughly the square of the first j for the alternates in all cases, and the same techniques that work for Pell's Equation work for the alternates, for instance, continued fractions.

The issue is I'm guessing not minor, as consider a paper I found with some digging on the web from Los Alamos National Laboratory researchers:

...Zeros of weight 1 3j- and 6j-coefficients are known to be related to the solutions of classic Diophantine equations. Here it is shown how solutions of the quadratic Diophantine equation known as Pell's equation are related to weight 2 zeros of 3j- and 6j-coefficients.

Link to full citation


James Harris
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