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Since the 2nd degree curve with equation

x + y = (y - 2x + 1)²

is a rational curve, all its rational points can be found by Euler's tangent-secant method (follow the link below for details) if we know a single rational point. Here such point is found easily:

(x₀, y₀) = (1, 3) and the parametric equations of a line through (1, 3) are

x = x₀ + t, y = y₀ + kt, where t - rational parameter.

This line intersects our curve in a rational point (of course we should take care also about x + y ≥ 0 because of the square root):

(1 + t) + (3 + kt) = (3 + kt - 2 - 2t + 1)²

(1 + k)t = (k - 2)² t² + 4(k - 2)t, now if t ≠ 0 we get 9 - 3k = (k - 2)² t, hence

(***) t = (9 - 3k)/(k - 2)², x = (k ² - 7k + 13)/(k - 2)², y = (12 - 3k)/(k - 2)²

The condition y - 2x + 1 ≥ 0 leads to the quadratic inequality

k² - 7k + 10 ≤ 0 or (k - 2)(k - 5) ≤ 0,

finally all rational solutions of your equation are given by expressions (***) where k is a rational parameter, 2 < k ≤ 5.

Examples: k = 3, t = 0, x = 1, y = 3;

k = 2.5, t = 6, x = 7, y = 18;

k = 4, t = -3/4, x = 1/4, y = 0;

k = 5, t = -2/3, x = 1/3, y = -1/3, etc.