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Take $p_1, p_2, \ldots, p_n, p_{n+1}$ be $n+1$ prime numbers in $\mathbb{P} \subseteq \mathbb{N}$. $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ seems to be quite clear, but still need a proof. I know some proofs are involved with Galois theory, which is not I want.

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marked as duplicate by Jyrki Lahtonen May 9 '15 at 12:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ See this answer for a simple inductive proof that works more generally. $\endgroup$ – Bill Dubuque Apr 11 '15 at 20:43
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    $\begingroup$ How many times has this been asked before on math.SE? ... $\endgroup$ – Martin Brandenburg Apr 11 '15 at 23:50
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(I will prove the following more general statement)

Theorem Let $P(n)$ be the following statement $$\forall m\in \Bbb N^+\ \ \sqrt{q_1\cdots q_m} \notin \mathbb{Q}(\sqrt{p_1}, \ldots, \sqrt{p_n}) \text{ for any }\text{distinct primes } p_1,\ldots,p_{n},q_1,\ldots,q_m$$ we claim that $P(n)$ is true for any integer $n\in \Bbb N$

Proof by induction

  • Basis step: Given a positive integer $m$ and $q_1,\cdots ,q_m$ distinct prime numbers assume that: $$\sqrt{q_1\cdots q_m}\in \Bbb Q $$ hence there exists $a$ and $b$ integers such that $q_1\cdots q_m=\frac{a^2}{b^2}$ thus $$1=v_{q_1}(q_1\cdots q_m)=2(v_{q_1}(a)-v_{q_1}(b))$$ the first equality holds because $q_1,\cdots q_m$ are distinct, It follows that $1$ is even which is absurd, finally $P(0)$ is true.

  • Induction step : Assume that $P(n-1)$ is true we will prove $P(n)$ by contradiction, assume that $P(n)$ is false then there exists an integer $m\geq 1$ and distinct primes $p_1,\cdots,p_n,q_1,\cdots,q_m$ such that: $$\sqrt{q_1\cdots q_m} \in \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$$ hence there exists $a,b\in \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_{n-1}})$ such that $\sqrt{q_{1}\cdots q_m}=a+b\sqrt{p_n}$ by squaring either:

    • one have $b=0$ then $\sqrt{q_{1}\cdots q_m}\in \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_{n-1}})$ and $p_1,\cdots,p_{n-1},q_1,\cdots,q_m$ are distinct
    • or one have $a=0$ in which case $bp_n=\sqrt{q_{1}\cdots q_mp_n}\in \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_{n-1}})$and $p_1,\cdots,p_{n-1},q_1,\cdots,q_m,q_{m+1}=p_n$ are distinct
    • or one have : $$\sqrt{p_n}=\frac{q_{1}\cdots q_m-a^2-b^2p_n}{2ab}\in \mathbb{Q}(\sqrt{p_1}, \ldots, \sqrt{p_{n-1}}) $$ and $p_1,\cdots,p_{n-1},q_1=p_n$ are distinct here the new positive integer $m$ is $1$

In all cases there is a contradiction with $P(n-1)$, finally $P(n)$ is true.

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  • $\begingroup$ can you explain about the existence of $a,b$ ? why does the linear combination does not include combinations of the other roots of $p_1,...,p_{n-1}$ ? $\endgroup$ – Belgi Apr 11 '15 at 17:18
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    $\begingroup$ because $a,b$ are elements on $\Bbb Q(\sqrt{p_1},\cdots,\sqrt{p_{n-1}})$ $\endgroup$ – Elaqqad Apr 11 '15 at 17:19
  • $\begingroup$ Something is fishy since the proof doesn't appear to use the hypotheses that the $\,p_i\,$ are distinct primes. $\endgroup$ – Bill Dubuque Apr 11 '15 at 21:02
  • $\begingroup$ I see no such argument above, so the "proof" is either incomplete or incorrect as it currently stands. $\endgroup$ – Bill Dubuque Apr 11 '15 at 21:42
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    $\begingroup$ Alas, there are still errors, e.g. you have not used anywhere that the $\,q_i\,$ are distinct from the $\,p_i,\,$ which is necessary for correctness. Also there are many typos, e.g. mixing p's and q's. $\endgroup$ – Bill Dubuque Apr 11 '15 at 23:32

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