## Primitive Pythagorean Triplets

#### Problem

Given that (x, y, z) is a primitive Pythagorean triplet, prove that the following transformation will produce another primitive Pythagorean triplet.

 (x", y", z") = (x, y, z) 1-2 2 2-1 2 2-2 3

#### Solution

By multiplying, we get the following three linear equations:

x" = x2y+2z
y" = 2xy+2z
z" = 2x2y+3z

Without loss of generality we shall say that x y z.

The proof shall be done in three parts. Firstly we shall show that (x", y", z") is different to (x, y, z), then we shall prove that it is a Pythagorean triplet, finally showing that it will be primitive.

1. From each of the equations above:

x" = x2y+2z x2z+2z = x x" x
y" = 2xy+2z 2xy+2y = 2x+y y" y
z" = 2x2y+3z 2x2z+3z = 2x+z z" z

That is, each new triplet generated by this transformation will be strictly increasing.

2. (x")2 = x2+4y2+4z24xy+4xz8yz
(y")2 = 4x2+y2+4z24xy+8xz4yz
(z")2 = 4x2+4y2+9z28xy+12xz12yz

 Therefore (x")2+(y")2 = 5x2+5y2+8z28xy+12xz12yz = x2+y2z2+4x2+4y2+9z28xy+12xz12yz = x2+y2z2+(z")2

But as it is given that (x,y,z) is a Pythagorean triplet, x2+y2z2 = 0.

Hence (x")2+(y")2 = (z")2, and we prove the first two parts: (x",y",z") is a different Pythagorean triplet to (x, y, z).

3. By using the inverse matrix we are able to transform (x" y", z") back to (x, y, z):

 (x, y, z) = (x", y", z") 1 2-2 -2-1 2 -2-2 3

This gives the following linear equations:

x = x"+2y"2z"
y = -2x"y"+2z"
z = -2x"2y"+3z"

If HCF(x", y", z") = h, then each of x, y, and z will share the same common factor. But as we are given that (x, y, z) is primitive, HCF(x, y, z) = 1, hence h = 1, and we prove that (x", y", z") is also primitve; our proof is complete.

We have just proved that a primitive Pythagorean triplet transformed by the matrix M1 produces a different primitive Pythagorean triplet. Prove that M2 and M3 also produce new primitive Pythagorean triplets.

 M1 = 1-2 2 2-1-2 2-2 3 M2 = -1 2 2 -2 1 2 -2 2 3 M3 = 1 2 2 2 1 2 2 2 3
Problem ID: 205 (24 Jan 2005)     Difficulty: 4 Star

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