Weighted summation of irrational numbers

Question

Consider the set $\{\sqrt{p_1},\dots,\sqrt{p_n}\}$, where none of $p_i$'s are perfect squares, and ${\rm gcd}(p_i,p_j)=1$ for every $i \neq j$. Prove that $0$ cannot be expressed as $\sum\limits_{k=1}^n q_k\sqrt{p_k}$, where $q_k\in \mathbb{Q}$ for every $k$, and $(q_1,\dots,q_k)\neq (0,0,\dots,0)$.

Answer 1

Oops, I have a semi-rigorous idea. Can someone help me with checking and rigorizing it?

I will prove the result when $p_i$'s are all prime.

Lemma: There exists a large prime $p$ such that, in modulo $p$, $p_1,\dots,p_{n-1}$ are all quadratic residues, while $p_n$ is not.

Proof. Here is the construction. Let $P$ be a large prime. The strategy is to use the law of quadratic reciprocity. Select $p\equiv 1\pmod{4}$, and $p\equiv 1\pmod{p_i}$, for every $i$. Now, the law of quadratic reciprocity implies, $$ \left(\frac{P}{p_i}\right)\left(\frac{p_i}{P}\right)=\left(\frac{p_i}{P}\right)=(-1)^{\frac{p_i-1}{2}\cdot \frac{P-1}{2}}=1, $$ hence, $p_i$ is indeed a quadratic residue, in modulo $P$, for every $i =1,2,\dots,n-1$. Now, let $\alpha_i\in\{0,1,\dots,p_n-1\}$ be a number, such that $\alpha_n$ is a quadratic non-residue, in modulo $p_n$. With this, we select $P\equiv \alpha_n\pmod{p_n}$. One can immediately see, using the law of quadratic reciprocity that, $p_n$ is a quadratic non-residue, in modulo $P$. Finally, the existence of such a prime $P$ obeying the congruences $P\equiv 1\pmod{4}$, $P\equiv 1\pmod{p_i}$ for $i =1,2,\dots,n-1$, and $P\equiv \alpha_n\pmod{p_n}$, follows immediately from the Chinese Remainder Theorem and Dirichlet's theorem on primes.

Now, in this prime $P$ modulus, if $\sum_{i=1}^n q_i\sqrt{p_i}=0$, then by multiplying with a large $M$ everywhere we get $\sum_{i =1}^{n-1} M_i\sqrt{p_i} \equiv -q_i\sqrt{p_n}\pmod{P}$, from here, we deduce that $p_n$ is a quadratic-residue, a contradiction.