How can $\sqrt{100+\sqrt{6156}}+\sqrt{100-\sqrt{6156}}$ be equal to $18$?

Question

According to this answer, when $\sqrt{m}$+$\sqrt{n}$ is rational, then $\sqrt{m}$ and $\sqrt{n}$ are rational.

But $\sqrt{100+\sqrt{6156}}+\sqrt{100-\sqrt{6156}} = 18 \in \mathbb{Q}$.

So that would imply $\sqrt{100+\sqrt{6156}}$ and $\sqrt{100-\sqrt{6156}}$ both rational. However, the former is $9 + \sqrt{19}$ and the latter is $9 - \sqrt{19}$, both irrational.

Where is the mistake here?

Thanks.

Answer 1

The problem is that the proof of what you linked is actually equivalent to $m-n$ being rational which isn't correct here

Answer 2

Your interpretation of "when $\sqrt{m}+\sqrt{n}$ is rational, then $\sqrt{m}$ and $\sqrt{n}$ are rational" is too broad. For example:

$$\sqrt{2}+\sqrt{(3-\sqrt{2})^2}$$

is rational, because it equals $3$. But of course $\sqrt{2}$ is not rational.

The statement "when $\sqrt{m}+\sqrt{n}$ is rational, then $\sqrt{m}$ and $\sqrt{n}$ are rational" might be true when $m,n$ are integers. But the example here has $m=100+\sqrt{6156}$, not an integer.