## 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`2`y`+2`z``y"` = 2`x``y`+2`z``z"` = 2`x`2`y`+3`z`

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`2`y`+2`z``x`2`z`+2`z`=`x``x"``x``y"`= 2`x``y`+2`z`2`x``y`+2`y`= 2`x`+`y``y"``y``z"`= 2`x`2`y`+3`z`2`x`2`z`+3`z`= 2`x`+`z``z"``z`That is, each new triplet generated by this transformation will be strictly increasing.

(

`x"`)^{2}=`x`^{2}+4`y`^{2}+4`z`^{2}4`xy`+4`xz`8`yz`

(`y"`)^{2}= 4`x`^{2}+`y`^{2}+4`z`^{2}4`xy`+8`xz`4`yz`

(`z"`)^{2}= 4`x`^{2}+4`y`^{2}+9`z`^{2}8`xy`+12`xz`12`yz`Therefore ( `x"`)^{2}+(`y"`)^{2}= 5 `x`^{2}+5`y`^{2}+8`z`^{2}8`xy`+12`xz`12`yz`= `x`^{2}+`y`^{2}`z`^{2}+4`x`^{2}+4`y`^{2}+9`z`^{2}8`xy`+12`xz`12`yz`= `x`^{2}+`y`^{2}`z`^{2}+(`z"`)^{2}But as it is given that (

`x`,`y`,`z`) is a Pythagorean triplet,`x`^{2}+`y`^{2}`z`^{2}= 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

3This gives the following linear equations:

`x`=`x"`+2`y"`2`z"``y`= -2`x"``y"`+2`z"``z`= -2`x"`2`y"`+3`z"`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 M_{1} produces a different primitive Pythagorean triplet. Prove that M_{2} and M_{3} also produce new primitive Pythagorean triplets.

M_{1} | = | 1 -2 2 | 2 -1 -2 | 2 -2 3 | M_{2} | = | -1 2 2 | -2 1 2 | -2 2 3 | M_{3} | = | 1 2 2 | 2 1 2 | 2 2 3 |