We can search for sides being integers,
that is x² + y² = z².
The smallest is 3² + 4² = 5² giving the "rope square" in the problem.
Such "Pythagorean triples" were known by Babylonians, thousand years before Pythagora !
If x and y are both odd, x = 2m + 1, y = 2n + 1,
z² = x² + y² =4(m² + n² + m + n) + 2
can't be a square, as a square must be a multiple of 4, or a multiple of 4 plus 1.
Hence at least one of x or y is even : let x = 2m and then y and z have same parity.
x² = 4m² = z² - y² = (z + y)(z - y), z + y and z - y being each even, let z + y = 2p and z - y = 2q, y = p - q, z = p + q, m² = pq. Let p = kr², q = k's² with k and k' square free. kk' should be a square, hence k = k' and m = krs.
Then we must have :
x = 2krs
y = k(r² - s²)
z = k(r² + s²)
The solutions are said to be primitive if x,y and z have no commun divisor.
This requires k = 1, r and s of opposite parity (else r² + s² and r² - s² would be even).
If p is a prime dividing both r and s, p divides r² and s² hence x, y and z. So r and s must then be coprime.
Reciprocally (if r and s coprime) let p a prime dividing rs, If it divides r then it doesn't divide s, hence neither r² + s² nor r² - s²
Then GCD(x,y,z) = 1 if and only if :
|k = 1, GCD(r,s) = 1, r and s of opposite parity|
If we vary r and s as said above, with r > s > 0,
every triple is given only once (prove that).
We can then easily get tables of Pythagorean triples.
Studying ternary quadratic forms gives another method to get all Pythagorean triples
Starting from the "trivial" solution X0 = (1,0,1), and multiplying by each of the matrices
( 2, 1, 2)
( 2, 2, 3)
|W =||( 1, 2, 2)|
( 2, 2, 3)
gives 3 new solutions. Multiplying each of these by U, V and W give 9
new solutions etc...
The first step gives just (1,0,1)×U = (3,4,5) (X0×V = X0, X0×W = X0×U)
The next steps give :
(3,4,5)×U = (21,20,29) (3,4,5)×V = (15,8,17) (3,4,5)×W = (5,12,13)
then each one multiplied by U, V, W give :
(55,48,73), (45,28,53), (7,24,25)
(119,120,169), (77,36,85), (39,80,89)
(65,72,98), (35,12,37), (33,56,65) and so on...
Roberts proved in 1977 that all primitive solutions are given by (x,y,z) = (1,0,1)×L
where L is some product of U,V,W matrices.
We can prove (Fermat-Girard theorem) that the only numbers which are sum of two squares are :
As we want r > s > 0, 2 = 1² + 1² is not suitable. A square of a prime a = 4k + 3 is sum of square only as a² + 0², and is not suitable either. Finally :
|A number is the hypothenuse of a right triangle with integer sides if and only if it has at least one prime factor of the form 4k + 1.|
To count how many triples have this hypothenuse, we deal with the problem of decomposition into a sum of squares.
|If hypothenuse is not multiple of 5, area is multiple of 30|