Pythagorean Generator Matrix

Using matrix multiplication, we have $a' = ax + by + cy$ , $b' = ay + bx + cy$ , and $c' = ay + by + cz$ . As Pythagorean triples, $a^2 + b^2 = c^2$ and $a'^2 + b'^2 = c'^2$ . Therefore, $(ax + by + cy)^2 + (ay + bx + cy)^2 = (ay + by + cz)^2$ , which simplifies to $(x^2 + 2y^2 - z^2)c^2$ $+$ $2(xy + xy - y^2)ab$ $+$ $2(xy + y^2 - yz)ac$ $+$ $2(y^2 + xy - yz)bc$ $=$ $0$ , which is true if $x^2 + 2y^2 - z^2 = 0$ , $2xy - y^2 = 0$ , and $xy + y^2 - yz = 0$ .

From $2xy - y^2 = 0$ we have $y(2x - y) = 0$ , and since $y$ is a positive integer, $y = 2x$ .

From $xy + y^2 - yz = 0$ we have $y(x + y - z) = 0$ , and since y is a positive integer, $z = x + y$ , and since $y = 2x$ , $z = 3x$ .

Since $y = 2x$ and $z = 3x$ , $x^2 + 2y^2 - z^2 = x^2 + 2(2x)^2 - (3x)^2 = 0$ .

So $(x, y, z) = (x, 2x, 3x)$ , and for $x$ , $y$ , and $z$ to be coprime we must have $x = 1$ , $y = 2$ , and $z = 3$ , and $1 + 2 + 3 = \boxed{6}$ .

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Alternatively, let $a^2 + b^2 = c^2$ . Then $(n - a)^2 + (n - b)^2 = (n + c)^2$ for $n = 0$ and, as a quadratic, possibly a second value of $n$ . Solving this equation for the other value gives $n = 2a + 2b + 2c$ . Therefore, $a' = n - a = a + 2b + 2c$ $b' = n - b = 2a + b + 2c$ $c' = n + c = 2a + 2b + 3c$ (We also note that $a'^2 + b'^2 = c'^2$ , so it is not an extraneous solution.) This means $\begin{pmatrix}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{pmatrix}\begin{pmatrix}a \\ b \\ c \end{pmatrix} = \begin{pmatrix}a' \\ b' \\ c' \end{pmatrix}$ Therefore, $x = 1$ , $y = 2$ , and $z = 3$ , and $1 + 2 + 3 = \boxed{6}$ .

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This and some other Pythagorean Generator Matrices can be found here .

Since David gave us matrix notation, we'll solve using matrix techniques!

$(a,b,c)$ is a Pythagorean triple iff $a^2 + b^2 - c^2 = 0$ , or $\begin{pmatrix} a & b & c \end{pmatrix} \begin{pmatrix} 1 & & \\ & 1 & \\ & & -1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = 0.$ Likewise, for $(a',b',c')$ to be a Pythagorean triple we must have $\begin{pmatrix} a & b & c \end{pmatrix} \begin{pmatrix} x & y & y \\ y & x & y \\ y & y & z \end{pmatrix} \begin{pmatrix} 1 & & \\ & 1 & \\ & & -1 \end{pmatrix} \begin{pmatrix} x & y & y \\ y & x & y \\ y & y & z \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = 0.$ The former equation must imply the latter for a large number of linearly independent Pythagorean triples $(a,b,c)$ . This is only possible if the following matrix equation holds: $\begin{pmatrix} x & y & y \\ y & x & y \\ y & y & z \end{pmatrix} \begin{pmatrix} 1 & & \\ & 1 & \\ & & -1 \end{pmatrix} \begin{pmatrix} x & y & y \\ y & x & y \\ y & y & z \end{pmatrix} = \begin{pmatrix} 1 & & \\ & 1 & \\ & & -1 \end{pmatrix} N,$ for a constant $N$ . We will start taking $N = 1$ , and afterward multiply by any common denominator or divide by any common factor if necessary.

Performing the matrix multiplication, we find four independent equations: $\begin{cases} x^2 = 1 \\ (2x-y)y = 0 \\ (x+y-z)y = 0 \\ 2y^2-z^2 = -1 \end{cases}$ Working with positive numbers only, we derive from the first three equations $\boxed{x = 1;\ \ y = 2;\ \ z = 3},$ and the fourth equation checks out (remarkably!): $2\cdot 2^2 - 3^2 = -1$ .

Since these three values are integer and coprime, there is no need to adjust the constant $N$ . Thus the answer to the problem is $1 + 2 + 3 = \boxed{6}$ .