Finding rational solutions to a function with integer inputs

Question

I'm trying to find rational solutions to the equation $$f(x) = \frac{1-x+\sqrt{5x^2-2x+1}}{2x};\{ x \in \mathbb{N} \}$$

The only operation that could produce an irrational answer is the square root, so I figure all I need to do is find rational solutions to

$$y = \sqrt{5x^2-2x+1}$$

which I think can be further adjusted to say: $$y^2= 5x^2-2x+1; \left \{x \in \mathbb{N} \ \text{and} \ y \in \mathbb{Q} \right \}$$

(excuse my notation if it's wrong) Without the $-2x$, it looks like another similar problem I've done where I've used Pell's equation to solve it, but this is a little different. How do I find the integer solutions? Can it be solved with pells equation?

Answer 1

$5x^2 - 2x + 1 -y^2 = 0 \implies \triangle' = (-1)^2 - 5(1-y^2) = 5y^2 -4 = m^2\implies 5y^2 - m^2 = 4$. This Pell equation has initial solution $(m,y) = (\pm 1, \pm 1)$. You can google the full solution set on the internet. Check out Lenstra's Pell equation notes.

Answer 2

If you let $y=\dfrac{p}{q}$, $p,\,q\in\mathbb{Z}$, and complete the square on the polynomial under the radical you get

$$ \sqrt{\frac{(5x-1)^2+4}{5}}=\frac{p}{q} $$

then you have

$$ \frac{(5x-1)^2+4}{5}=\frac{p^2}{q^2} $$

Letting $r=5x-1$ this transforms into the question solving for $p,\,q,\in\mathbb{Z}$, $r\in\mathbb{Q}$ satisfying

$$ (r^2+4)q^2=5p^2 $$