
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" = x
2y+2z
y" = 2x
y+2z
z" = 2x
2y+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.
From each of the equations above:
x" = x
2y+2z
x
2z+2z = x
x"
x
y" = 2x
y+2z
2x
y+2y = 2x+y
y"
y
z" = 2x
2y+3z
2x
2z+3z = 2x+z
z"
zThat is, each new triplet generated by this transformation will be strictly increasing.
(x")2 = x2+4y2+4z2
4xy+4xz
8yz
(y")2 = 4x2+y2+4z2
4xy+8xz
4yz
(z")2 = 4x2+4y2+9z2
8xy+12xz
12yzTherefore (x")2+(y")2 = 5x2+5y2+8z2
8xy+12xz
12yz= x2+y2
z2+4x2+4y2+9z2
8xy+12xz
12yz= x2+y2
z2+(z")2But as it is given that (x,y,z) is a Pythagorean triplet, x2+y2
z2 = 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).
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 |
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