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Pythagorean triples and their connection to Pell's equation. It includes theorems, examples, and methods for finding solutions. primitive solutions, fundamental solutions, and the negative Pell's equation.
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An equation of the form
a 1 x 1 + a 2 x 2 + · · · + anxn = b (1)
with a 1 , a 2 ,... , an, b integers, is called a linear Diophantine equation (LDE).
The LDE a 1 x 1 + a 2 x 2 + · · · + anxn = b
has a solution x 1 , ..., xn ∈ Z if and only if gcd(a 1 , a 2 ,... , an)|b
An equation of the form
∑^ n
i,j=
aij xi xj = b (2)
with aij , b integers, is called a quadratic Diophantine equation (QDE).
The equation x^2 + y 2 = z^2
is a QDE. Any solution (x, y , z) of this equation for integers x, y , z is called a Pythagorean triple.
Any primitive solution of
x 2
is of the form
x = m 2 − n 2 , y = 2mn, z = m 2
Where m, n ≥ 1 are relatively prime positive integers.
Pell’s equation has the form
x 2 − dy 2 = 1 (5)
where d not a perfect square.
We say that (x 0 , y 0 ) is a fundamental solution of Pell’s equation if x 0 , y 0 are positive integers that are minimal amongst all solutions.
Pell’s equation has infinitely many solutions. Given the solution (x 0 , y 0 ) the solution (xn+1, yn+1) is given by
xn+1 = x 0 xn + dy 0 yn, x 1 = x 0 , n ≥ 1
yn+1 = y 0 xn + x 0 yn, y 1 = y 0 , n ≥ 1
The equation x^2 − 2 y 2 = 1, has the fund. sol. (x 0 , y 0 ) = (3,2). So
x 2 = x 2 0 +^ dy^
2 0 = 9 + 2.4 = 17,^ y^2 =^ y^0 x^0 +^ x^0 y^0 = 6 + 6 = 12
is also a solution: 17^2 − 2. 122 = 1.
Let Pell’s equation x^2 − dy 2 = 1, have the fundamental solution (x 0 , y 0 ). Then (xn, yn) is also a solution, given by
xn =
[(x 0 + y 0
d) n
d) n ]
yn =
d
[(x 0 + y 0
d) n − (x 0 − y 0
d) n ]
Solve x^2 − 2 y 2 = 1. The fund. sol. is (3,2). The general solution is:
xn =
2)n+(3− 2
2)n], yn =
2)n−(3− 2
2)n]
Let ax^2 − by 2 = 1
have an integral solution. Let (A, B) solution for least positive A, B. The general solution is
xn = Aun + bBvn
(10)
yn = Bun + aAvn
Where (un, vn) is the general solution of Pell’s resolvent u^2 − abv 2 = 1.
Solve
6 x^2 − 5 y 2 = 1 (11)
The fund. sol. is (x, y ) = (A, B) = (1, 1). The resolvent is u^2 − 30 v 2 = 1, with fund. sol. (u 0 , v 0 ) = (11, 2). The general solution of the resolvent is
un =
30)n^ + (11 − 2
30)n]
vn =
n − (11 − 2
n ]
The general solution of Eq. (11) is
xn = un + 5vn, yn = un + 6vn
Find all integers n ≥ 1 such that 2 n + 1 and 3 n + 1 are both perfect squares.
Observe that
2 n + 1 = x 2 , 3 n + 1 = y 2 =⇒ 3 x 2 − 2 y 2 = 1,
with 3.2 = 6 not a square in Z. So solving this amounts to solving the general form of Pell’s equation.
The negative Pell’s equation has the form
x 2 − dy 2 = − 1 (12)
where d not a perfect square.
Find all pairs (k, m) such that
1 + 2 + · · · + k = (k + 1) + (k + 2) + · · · + m.
Find all pairs (k, m) such that
1 + 2 + · · · + k = (k + 1) + (k + 2) + · · · + m.
Adding 1 + 2 + · · · + k to both sides of the above equality we get
2 k(k + 1) = m(m + 1) ⇐⇒ (2m + 1) 2 − 2(2k + 1) 2 = − 1.
Show that the equation x^2 + y 3 + z^3 = t^4 has infinitely many solutions x, y , z, t, ∈ Z with the greatest common divisor 1.
Show that the equation x^2 + y 3 + z^3 = t^4 has infinitely many solutions x, y , z, t, ∈ Z with the greatest common divisor 1.
Start from the equality
[1^3 + 2^3 + · · · + (n − 2)^3 ] + (n − 1)^3 + n^3 =
n(n + 1) 2
(n − 2)(n − 1) 2
n(n + 1) 2
