to Radical twins that are rational and not irrational

Question

So we have two similar expressions in terms of $y$, $(y^2+5)^{1/2}$ and its "twin" $(y^2-5)^{1/2}$, we are to find $y$ so that both are Rational. I though somehow $y^2-25$ could be used here since it factors into terms under the radical. I still am lost on the first step but think if someone helped with that I could finish the rest. Sorry i cant write in math jax so If someone edits this to make it math stack legible i greatly appreciate it.

Answer 1

What we're really looking for here is a rational number $\frac{a}{b}$ such that $\left(\frac{a}{b}\right)^2 + 10$ is also the square of a rational number (here, $\frac{1}{b}$ takes the place of $(y^2 - 5)^{1/2}$). So, we want $\frac{a^2 + 10b^2}{b^2}$ to be the square of a rational number; this would follow if $a^2 + 10b^2$ is also a square. There may be a more rigorous way to find such an example, but observe that $a = 3, b = 2$ gives $49$, which is a perfect square. Then $\frac{a}{b} = \frac{3}{2} = (y^2 - 5)^{1/2}$, and you can solve to get $y = \frac{\sqrt{29}}{2}$.

Answer 2

If we don't need $y$ to be rational, replace $y^2=z\ge0$ and let $$z-5=(p-q)^2,z+5=(p+q)^2,z\ge0$$ where $p,q$ are rational

$$z=p^2+q^2$$

$$10=(p+q)^2-(p-q)^2=4pq\iff q=\dfrac5{2p}$$

$$\implies z=p^2+\dfrac{5^2}{4p^2}=\dfrac{4p^4+25}{4p^2}$$

Trivially $p=1$