# Elementary proof for $\sqrt{p_{n+1}} \notin \mathbb{Q}(\sqrt{p_1}, \sqrt{p_2}, \ldots, \sqrt{p_n})$ where $p_i$ are different prime numbers. [duplicate]

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.

• See this answer for a simple inductive proof that works more generally. Apr 11 '15 at 20:43
• How many times has this been asked before on math.SE? ... Apr 11 '15 at 23:50

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 has $$b=0$$ 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 has $$a=0$$ 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 has $$\sqrt{p_n}=\frac{q_{1}\cdots q_m-a^2-b^2p_n}{2ab}\in \mathbb{Q}(\sqrt{p_1}, \ldots, \sqrt{p_{n-1}})$$ 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.

• can you explain about the existence of $a,b$ ? why does the linear combination does not include combinations of the other roots of $p_1,...,p_{n-1}$ ? Apr 11 '15 at 17:18
• because $a,b$ are elements on $\Bbb Q(\sqrt{p_1},\cdots,\sqrt{p_{n-1}})$ Apr 11 '15 at 17:19
• what is $v_{q_1}?$ Sep 5 '20 at 5:34
• $v_p (n)$ is the highest power of $p$ that divides $n$ Sep 5 '20 at 9:16