, 2007

Let
$$u = u(x, y, z) = 3x - 2y - 2z,$$
$$v = v(x, y, z) = -2x + 2y + z,$$
$$w = w(x, y, z) = -2x + y + 2z.$$
First of all, if $(x, y, z) \neq (5, 4, 3)$ is a primitive Pythagorean triplet with $x > y > z > 0$, then, it is easy to check that $u^2 = v^2 + w^2$. Further, $x^2 + (2y - 2z)^2 > 0 \Rightarrow 9x^2 > 4y^2 + 4z^2 + 8yz \Rightarrow 3x > 2y + 2z \Rightarrow u(x, y, z) > 0$.
Similarly, we have $4y > 3z \Rightarrow 4yz - 3z^2 > 0 \Rightarrow 4y^2 + z^2 + 4yz > 4x^2 \Rightarrow 2y + z > 2x \Rightarrow v(x, y, z) > 0$.
Next, $(3y - 4z)^2 > 0 \Rightarrow 25x^2 + (3y - 4z)^2 > 25x^2 \Rightarrow 25x^2 > 16y^2 + 9z^2 + 24yz \Rightarrow 25x^2 > (4y + 3x)^2 \Rightarrow 5x > 4y + 3z \Rightarrow u(x, y, z) > v(x, y, z)$.

Similarly, we have $u(x, y, z) > w(x, y, z)$. Also, since $x > y$, it follows that $u(x, y, z) > -w(x, y, z)$. And, $y > z$ implies $v(x, y, z) > w(x, y, z)$. Finally, we notice that if $y + z > x$ then it follows that $u(x, y, z) < x$. Now, let
$$f_1(x, y, z) = (u, v, w),$$
$$g_1(x, y, z) = (u, v, -w),$$
$$h_1(x, y, z) = (u, -w, v).$$
Then, it is easy to see that $f_1(f(x, y, z)) = g_1(g(x, y, z)) = h_1(h(x, y, z)) = (x, y, z)$. Further, it is easy to check that $\text{gcd}(u, v, w) = \text{gcd}(x, y, z)$.

Now, we shall prove the result by contradiction. If there are primitive Pythagorean triplets that are not obtained from $(5, 4, 3)$ by repeated application of $f, g$ and $h$ in some order, then let $(a, b, c)$ be the triplet that cannot be obtained from $(5, 4, 3)$ with $a > b > c > 0$ and with least possible value for $a$. So, all the primitive Pythagorean triplets $(x, y, z)$ with $a > x > y > z > 0$ are obtained from $(5, 4, 3)$ (uniquely). Clearly, we have $a > 5$.
Now, let $f_1(a, b, c) = (k, l, m)$. Then, $g_1(a, b, c) = (k, l, -m)$ and $h_1(a, b, c) = (k, -m, l)$. As noted before, $k$ and $l$ are positive, and $k > l, k > m$ and $k > -m$. Thus, there is a unique $\Delta \in \{f_1, g_1, h_1\}$ such that $\Delta(a, b, c) = (r, s, t)$ with $r > s > t > 0$. Also, from the previous calculations, we have

that $a > r$. Thus, by the minimality of $a$, it follows that $(r, s, t)$ is obtained from $(5, 4, 3)$, in a unique way, by repeated applications of $f, g, h$ in some order. But, now $\Delta(a, b, c) = (r, s, t)$ implies that by applying one of $f, g$ and $h$ to $(r, s, t)$ we would get $(a, b, c)$. Therefore, we can obtain $(a, b, c)$ from $(5, 4, 3)$ by repeated applications of $f, g, h$ in some order. This is a contradiction. Uniqueness follows from the uniqueness of $\Delta$. This completes the solution.