Field containing sum of square roots also contains individual square roots 
Let $F$ be a field of characteristic $\neq 2$. Let $a\neq b $ be in $F$. Suppose $\sqrt{a}+\sqrt{b}\in F$. Prove that $\sqrt{a}\in F$.

We have $(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}\in F$, so $\sqrt{ab}\in F$. Then $\sqrt{ab}(\sqrt{a}+\sqrt{b})=a\sqrt{b}+b\sqrt{a}\in F$. What then?

Subtract $a(\sqrt{b}+\sqrt{a})$ for the next step.
