Quadratic Diophantine Theorem and Pell's Equation

With the Quadratic Diophantine Theorem derived, it makes sense to try it out with a well-known equation in Diophantine theory which is Pell's Equation:

x² - Dy² = 1

with D a natural number.

Now again, the Quadratic Diophantine Theorem:

Quadratic Diophantine Theorem:

In the ring of integers, given the quadratic expression

c₁x² + c₂xy + c₃y² = c₄z² + c₅zx + c₆zy

where the c's are constants, for solutions to exist it must be true that

((c₂ - 2c₁)² + 4c₁(c₂ - c₁ - c₃))v² + (2(c₂ - 2c₁)(c₆ - c₅) + 4c₅(c₂ - c₁ - c₃))v + (c₆ - c₅)² - 4c₄(c₂ - c₁ - c₃) = n² mod p

for some n, where p is any prime coprime to z for a given solution, when

v = -(x+y)z^-1 mod p.

So with Pell's Equation I have

c₁ = 1, c₂=0, c₃ = -D, c₄ = 1, c₅ = 0, c₆ = 0, and z=1

which gives

4Dv² - 4D + 4 = n² mod p

and v = -(x+y) mod p, so I have

4D(x+y)² - 4D + 4 = n² mod p

and since that must be true for all primes p, since z=1, I have in general that the left hand side must be a perfect square so it must be true then that

D(x+y)² - D + 1 = S²

where S is some integer, and I have in general that

x+y = sqrt((S² + D - 1)/D).

Example: From a reference I have that with D=2, x=17 and y=12 are solutions.

Working backwards I found that S=41 gives that solution, verifying the result.

Notice also that S² = 1 mod D. It is of interest to consider the special case of S = 1 mod D or S = -1 mod D, and as that's tedious in what I follows I just use S = +/-1 mod D, where it's an OR, so both cases are not true.

So I can make the substitution S = jD +/- 1, to find

x+y = sqrt(Dj² +/- 2j + 1)

which is

x+y = sqrt((D-1)j² + (j +/- 1)²)

and I have the existence of solutions related to another Diophantine relation of the form

(D-1)u² + v² = w²

with the condition that u = j and v = j+/-1.

For instance with D=2, I have that I need solutions to

u² + v² = w²

with u=j, and v=j+/-1, and j=20 works as 20² + 21² = 29², and gives x+y = 29, and again x=17, y=12 is a known solution to x² - 2y² = 1.

So then x² - 2y² = 1 is related to certain Pythagorean triples, when D is prime.

Also notice that from

x+y = sqrt((S² + D - 1)/D)

I have

S² - D(x+y)² = -D + 1

which means a second Diophantine equation connected to the first!

With D=2, I get then that x² - 2y² = 1, is connected to

S² - 2(x+y)² = -1

so for every solution of the first there is a solution of the second.

So there is an immediate result with the classical Pell's Equation, with little effort at all using the theorem, which can be used against any Diophantine quadratic in 2 variables, almost as easily, and also give results in 3 variables, though not quite as generally.


James Harris