Question
Find all the rational values of $x$ at which $y=\sqrt{x^2+x+3}$
My attempt
Since we only have to find the rational values of $x$ and $y$, we can assume that $$ x \in Q$$ $$ y \in Q$$ $$ y-x \in Q $$ Let$$ d = y-x$$ $$d=\sqrt{x^2+x+3}-x$$ $$d+x=\sqrt{x^2+x+3}$$ $$(d+x)^2=(\sqrt{x^2+x+3})^2$$ $$d^2 + x^2 + 2dx =x^2+x+3$$ $$d^2 +2dx = x +3$$ $$x = \frac{3-d^2}{2d-1}$$
$$d \neq \frac{1}{2}$$
So $x$ will be rational as long as $d \neq \frac{1}{2}$.
Now $$ y = \sqrt{x^2+x+3}$$ $$ y = \sqrt{(\frac{3-d^2}{2d-1})^2 + \frac{3-d^2}{2d-1} + 3}$$ $$ y = \sqrt{\frac{(3-d^2)^2}{(2d-1)^2} + \frac{(3-d^2)(2d-1)}{(2d-1)^2} + 3\frac{(2d-1)^2}{(2d-1)^2}}$$ $$ y = \sqrt{\frac{(3-d^2)^2 + (3-d^2)(2d-1) + 3(2d-1)^2}{(2d-1)^2}} $$ $$ y = \frac{\sqrt{(3-d^2)^2 + (3-d^2)(2d-1) + 3(2d-1)^2}}{(2d-1)}$$ $$ y = \frac{\sqrt{d^4-2d^3+7d^2-6d+9}}{(2d-1)}$$
I know that again $d \neq \frac{1}{2}$ but I don't know what to do with the numerator. Help
Hint:
The expression is of the following form $$(a+b-c)^2 = a^2+b^2+c^2+2ab-2bc-2ac$$