The question has been [posted](https://math.stackexchange.com/questions/3952063/prove-that-m-leq-4) on math.SE but had no response.


There are positive integers $a,b,c,d\_i$, s.t. $\sqrt{a+\sqrt{b}+\sqrt{c}}=\sum\_{i=1}^m \sqrt{d\_i}$, and for any $i\ne j$, $\sqrt{d\_i/d\_j}$ is not a rational number. Prove that, $m\le 4$.


I tried some algebraic transformations, but they didn't work. Would it be related to [Galois Theory](https://en.wikipedia.org/wiki/Galois_theory)?


Can anyone help?



Proving that $m\leq 4$ when $\sqrt{a+\sqrt{b}+\sqrt{c}}=\sum_{i=1}^m \sqrt{d_i}$ with each $d_i/d_j$ non-square
I claim that $m\leqslant 2$. Taking the square we get $a+\sqrt{b}+\sqrt{c}=(\sum d\_i)+2\sum \sqrt{d\_id\_j}$.


By Besicovitch theorem, the square roots of positive integers do not admit a non-trivial linear dependence over $\mathbb{Q}$, that is, $\sum c\_i\sqrt{n\_i}\ne 0$ for non-zero rational coefficients $c\_i$ and distinct squarefree positive integers $n\_i$ (one-line proof: divide by $\sqrt{n\_1}$ and take the trace).


Define a squarefree part $s(N)$ of a positive integer $N$ as the squarefree $t=s(N)$ such that $N=tr^2$ for an integer $r$.


If $m\geqslant 3$, then $d\_1d\_2$, $d\_2d\_3$, $d\_1d\_3$ have mutually distinct squarefree parts, also they are not equal to 1. Thus at least one of them is not equal neither to $s(b)$ nor $s(c)$. So this equality would contradict to Besicovitch theorem.


