So I've started playing around with trivial modular solutions to x2 - Dy2 = 1 as I'm bored and it's easy to do. Earlier I noted that:
x2 - Dy2 = 1 mod D-1, is solved modularly by:
2x = r + r-1 mod D-1 and 2y = r - r-1 mod D-1
Now I've thought up another easy one using some prime p for which D is a quadratic residue, so there exists n, such that:
n2 = D mod p
Then I have (x + ny)(x - ny) = 1 mod p.
And with x + ny = r mod p, and x - ny = r-1 mod p, then
2x = r + r-1 mod p and 2y = n-1(r - r-1) mod p
Giving yet another way to solve for x and y modularly.
And, of course, as it's a modular solution it is for:
x2 - Dy2 = 1 mod p
So you don't necessarily get an explicit answer to the original equation.
Of course you can generalize easily enough modulo some natural number N for which D is a quadratic residue, and continue even further to solve modularly for the equation:
x2 - Dy2 = F
Where F is some natural number.
If "mod" is unfamiliar to you, I made up a small primer covering modular arithmetic that you may find useful.
James Harris
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