Find all rational values of $x$, at which $ \sqrt{x^2 + x + 3}$ is rational

Question

How to find all rational values of $x$, at which $y = \sqrt{x^2 + x + 3}$ is rational?

Answer 1

This is quite easily shown to be false: The square root of $0^2 + 0 + 3 = 3$ is not rational.

For a non-zero example, just note that $\sqrt{5}$ is irrational and set $x = 1$.

Answer 2

There exist unfinitely many solutions. Putting $\frac{x}{y}$ instead of $x$ rewrite the question as

Find all integer solutions of $x^2+xy+3y^2=z^2$. The equation is equvalent the following $(2z-2x-y)(2z+2x+y)=11y^2$. Now for any $y$ choose some factorisation $uv=11y^2$ and have the system $\left\{\begin{matrix}2z-2x-y=u\\ 2z+2x+y=v\\ uv=11y^2 \end{matrix}\right.$

For which $y$ this system is solvable? Show the answer is Yes for any odd $y$. In this case $11y^2\equiv -1\pmod{4}$ and so for any factorisation we have $u+v\equiv 0\pmod {4}$ and we hawe $z=\dfrac{u+v}{4},\ \ x=\dfrac{v-y}{2}-z$.

For instance for any odd $y$ not dividing by $11$ we can choose $u=11, \ \ v=y^2$ and have $x=\dfrac{y^2-y}{2}-\dfrac{11+y^2}{4}$ providing an irreducible of the fraction $\dfrac xy$.

The even case is more complicated.

PS. \not\equiv doesn't work here