Some days ago i came across a question about writing $\sqrt {2001}$ as sum of two other square roots. I managed to prove that this is not possible unless one of them is zero and the other one is $2001$.
The proof was as following: $\sqrt{2001}=\sqrt a+\sqrt b$, $\sqrt{2001}-\sqrt a=\sqrt b$ so $2001+a-2\sqrt{2001a}=b$. This shows that $2\sqrt{2001a}$ is an integer so $2001*a$ is a perfect square.
We also know that $2001=3*23*29$ which is a square-free number. so $a$ must divide all of $3,23,29$ which means $a\geq2001$ so$\sqrt a\geq\sqrt{2001}$ and $\sqrt{b}\leq 0$ which means $b=0$.
With exact method we can prove that $\sqrt{s}=\sqrt a+\sqrt b$ does not have any natural solutions with $s$ being a square-free number. Then I tried to generalize the proof for $3$ or more square roots but i failed. The only thing I always get is $\sqrt {ab}+\sqrt {bc}+\sqrt {ac}$ is an integer which does not help at all.
For what numbers can we write the square root of a square-free number as sum of three or more non-zero square roots? I would appreciate any help.
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