Show that $ a，b, \sqrt{a}+ \sqrt{b} \in\mathbb Q \implies \sqrt{a},\sqrt{b} \in\mathbb Q $

Inspired by this, I was wondering if there is a simple logical argument to

Show that $ a,b, \sqrt{a}+ \sqrt{b} \in\mathbb Q \implies \sqrt{a},\sqrt{b} \in\mathbb Q $

Note that the original link is using a computational method, where as I am looking for a simple logical argument.

I tried (unjutifiably) to argue that if some of two square roots is rational then each one is rational, this is a different than the (incorrect) argument that if sum of two algebraic numbers is rational then each one is rational ( counter example $a=1-\sqrt{2},b= 1+\sqrt{2} $)

Solution 1:

Hint $\rm\ \sqrt{a}-\sqrt{b}\: = \dfrac{a-b}{\sqrt{a}+\sqrt{b}}\ $ so $\rm\ \sqrt{a}+\sqrt{b}\in\mathbb Q\:\Rightarrow\:\sqrt{a}-\sqrt{b}\in\mathbb Q\:\Rightarrow\:$ sum/2 $\rm = \sqrt{a}\in \mathbb Q$

Remark $ $ This generalizes to a positive sum of any number of square roots over an ordered field.

Solution 2:

$$\sqrt a + \sqrt b \in \mathbb{Q} \Rightarrow \sqrt a + \sqrt b = \dfrac{p}{q}$$

$$\sqrt a = \dfrac{p}{q} - \sqrt b$$ $$a = \dfrac{p^2}{q^2} - 2 \cdot \dfrac{p}{q} \sqrt b + b$$

So if $a,b$ are rational, this forces their square roots to be also.