1
$\begingroup$

The question has been posted 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?

Can anyone help?

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ (1) What is the motivation? (2) Why do you think this should hold? $\endgroup$
    – Francesco Polizzi
    Commented Dec 18, 2020 at 14:48
  • $\begingroup$ (1)The motivation of producing this problem comes from an Euler project problem. It involves considering the radical splitting in this way. (2)As we could use the arguments in abstract algebra about fields, The minimal polynomial of the left hand side has a degree less than or equal to eight, which indicates that $m$ is less than or equal to eight. However, as I tried many times, I think $m$ is smaller or equal to $4$ and all the $d_i$'s have the same pattern. I don't know exactly why this holds because I don't know the proof. If I know the proof, I may not ask this question XD. $\endgroup$
    – FFjet
    Commented Dec 18, 2020 at 14:59
  • $\begingroup$ the dimension of linear space spaned by $\{1,\sqrt{b},\sqrt{c},\sqrt{bc}\}$ over $\mathbb{Q}$ is no more than 4, so the dimension of linear space spaned by $(\sum_{i=1}^{m} \sqrt{d_{i}})^{2t}, t\in \mathbb{N}^*$ over $\mathbb{Q}$ is no more than 4 then there is a contradiction if $m\geq 5$. $\endgroup$
    – katago
    Commented Dec 18, 2020 at 15:04
  • $\begingroup$ I can conclude this fact you listed up to the eighth to the last word, but could you elaborate more about where is the contradiction when $m$ is greater or equal to $5$? It seems that the fact of an even power does not indicate something to the odd power. $\endgroup$
    – FFjet
    Commented Dec 18, 2020 at 15:13
  • $\begingroup$ $\left(\sum_{i=1}^{m} \sqrt{d_i}\right)^{2 t}=\sum_{s \in\{0.1\}^{m}} \lambda_{2t, s} \prod_{i=1}^{m}\left(\sqrt{d_{i}}\right)^{s_{i}}$ , where $s=\left(s_{1}, \cdots, s_{m}\right) \in\{0,1\}^{m}$, $v_{t}$ is the vector composed of $\lambda_{2 t,s} \quad s \in\{0.1\}^{m}$. if we can prove $\operatorname{Rank}\left(v_{1}\left|v_{2}\right| \cdots\left|v_{n}\right| \cdots\right) \geqslant m>4$ then there is a contradition. For some special case the last rank inequality related to $\left(v_{1}\left|v_{2}\right| \cdots\left|v_{n}\right| \cdots\right)$ has some Vandermonde determinant structure $\endgroup$
    – katago
    Commented Dec 18, 2020 at 15:25

1 Answer 1

Reset to default
8
$\begingroup$

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.

Improve this answer
$\endgroup$
6
  • $\begingroup$ What is this "trace"? $\endgroup$
    – abx
    Commented Dec 19, 2020 at 15:56
  • $\begingroup$ Trace of an algebraic number $\alpha\in K$, where $K$ is a number field, is is the trace of the $\mathbb{Q}$-linear operator $x\to \alpha x$ in the space $K$ (regarded as a linear space over $\mathbb{Q}$). $\endgroup$
    – Fedor Petrov
    Commented Dec 19, 2020 at 16:04
  • 1
    $\begingroup$ I know. What I was trying to point out is that you use transitivity of the trace to see that it is zero in the big field generated by all the square roots. $\endgroup$
    – abx
    Commented Dec 19, 2020 at 16:31
  • $\begingroup$ @abx well, yes, I use this. $\endgroup$
    – Fedor Petrov
    Commented Dec 19, 2020 at 16:55
  • 1
    $\begingroup$ I see, when consider the trace of $\sum_{i\in I} c_{i} \frac{\sqrt{n_{i}}}{\sqrt{n_1}}$, the trace of it over the big field generated by all the square roots is $[K: \mathbb{Q}]c_1\neq 0$, but the trace of 0 over the same field is 0. $\endgroup$
    – katago
    Commented Dec 19, 2020 at 17:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .