Find all natural numbers $$x, y, z$$ such that $$\sqrt{\frac{2026}{x+y}} + \sqrt{\frac{2026}{y+z}} + \sqrt{\frac{2026}{z+x}}$$ is also a natural number.

A sum of square roots of positive rationals can be an integer only if each square root is rational, and since $$2026 = 2 \times 1013$$ is square-free, each root must be the reciprocal of a natural number. Put $$\sqrt{\frac{2026}{x+y}} = \frac{1}{a}$$, $$\sqrt{\frac{2026}{y+z}} = \frac{1}{b}$$, $$\sqrt{\frac{2026}{z+x}} = \frac{1}{c}$$, so $$x + y = 2026a^2$$, $$y + z = 2026b^2$$, $$z + x = 2026c^2$$, which give $$x = 1013(a^2 - b^2 + c^2)$$, $$y = 1013(a^2 + b^2 - c^2)$$, $$z = 1013(b^2 + c^2 - a^2)$$; positivity needs each of $$a^2, b^2, c^2$$ to be less than the sum of the other two. Now $$N = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \le 3$$ is a natural number, and up to order the only solutions are $$(1, 1, 1)$$ for $$N = 3$$, $$(1, 2, 2)$$ for $$N = 2$$, and $$(2, 3, 6)$$, $$(2, 4, 4)$$, $$(3, 3, 3)$$ for $$N = 1$$; the triple $$(2, 3, 6)$$ fails because $$4 + 9 < 36$$. Translating back, $$(1, 1, 1)$$ gives $$(1013, 1013, 1013)$$, the arrangements of $$(1, 2, 2)$$ give $$(1013, 1013, 7091)$$, $$(1013, 7091, 1013)$$, $$(7091, 1013, 1013)$$, the arrangements of $$(2, 4, 4)$$ give $$(4052, 4052, 28364)$$, $$(4052, 28364, 4052)$$, $$(28364, 4052, 4052)$$, and $$(3, 3, 3)$$ gives $$(9117, 9117, 9117)$$. So there are 8 triplets $$(x, y, z)$$ in all.