Let $x=(\sqrt{1+\frac{4}{a}}-1)(\sqrt{1+\frac{4}{b}}-1)>0$ is rational Let $y=(\sqrt{1+\frac{4}{a}}+1)(\sqrt{1+\frac{4}{b}}+1)$ $xy = \frac{16}{ab}$ is rational, so $y$ is rational too But then $\frac{y-x}{2}=\sqrt{1+\frac{4}{a}}+\sqrt{1+\frac{4}{b}}$ is rational But $\sqrt{1+\frac{4}{a}}-\sqrt{1+\frac{4}{b}}$ is also rational, so $\sqrt{1+\frac{4}{a}}$ is rational and so $\sqrt{a^2+4a}$ is rational too But as $a$ is integer then $a^2+4a$ is square of integer $a^2+4a=m^2 \to (a+2)^2-m^2=4$ but this equation don't have positive integer solutions. So there are no such $a,b$

RagvaloD wrote: But $\sqrt{1+\frac{4}{a}}-\sqrt{1+\frac{4}{b}}$ is also rational Can you explain why is that true?

Same logic as above, because $(\sqrt{1+\frac{4}{a}}-\sqrt{1+\frac{4}{b}})(\sqrt{1+\frac{4}{a}}+\sqrt{1+\frac{4}{b}})=\frac{4}{a}-\frac{4}{b}$ is rational

Thank you!