### Sunday, March 22, 2009

## Updated general solution to Pell's Equation

A previous posting of mine giving a general solution to Pell's Equation had a sign error. Here is a corrected, and much simpler general solution.

In rationals, given

x^2 - Dy^2 = 1

I have proven:

y = 2[f_2*v - 1]/[(D-1) + 2f_2*v - f_2^2*v^2]

and

x = [f_2*v^2 - 2f_2*v + (D+1)]/[(D-1) + 2f_2*v - f_2^2*v^2]

where f_1*f_2 = D-1, and the f's are non-zero integer factors, while v is nonzero but is otherwise a free variable.

As Pell's Equation is normally considered in integers as a Diophantine equation note that you find rational v such that x and y are integers, which gives the 'why' of Pell's Equation. For instance, for D=2, f_1*f_2 = 1, so I have:

y = 2(v - 1)/(1 + 2v - v^2)

and

x = (v^2 - 2v + 3)/(1 + 2v - v^2)

and v = 2, gives y = 2, x = 3, and, of course,

9 - 2(4) = 1.

My apologies for the prior sign error.

In rationals, given

x^2 - Dy^2 = 1

I have proven:

y = 2[f_2*v - 1]/[(D-1) + 2f_2*v - f_2^2*v^2]

and

x = [f_2*v^2 - 2f_2*v + (D+1)]/[(D-1) + 2f_2*v - f_2^2*v^2]

where f_1*f_2 = D-1, and the f's are non-zero integer factors, while v is nonzero but is otherwise a free variable.

As Pell's Equation is normally considered in integers as a Diophantine equation note that you find rational v such that x and y are integers, which gives the 'why' of Pell's Equation. For instance, for D=2, f_1*f_2 = 1, so I have:

y = 2(v - 1)/(1 + 2v - v^2)

and

x = (v^2 - 2v + 3)/(1 + 2v - v^2)

and v = 2, gives y = 2, x = 3, and, of course,

9 - 2(4) = 1.

My apologies for the prior sign error.