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### Can a finite sum of square roots be an integer? [duplicate]

Can a sum of a finite number of square roots of integers be an integer? If yes can a sum of two square roots of integers be an integer? The square roots need to be irrational.
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### Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.

Question: Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational. Following from the question, I tried: ...
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I have to find a basis for $\Bbb{Q}(\sqrt{2}+\sqrt{3})$ over $\Bbb{Q}$. I determined that $\sqrt{2}+\sqrt{3}$ satisfies the equation $(x^2-5)^2-24$ in $\Bbb{Q}$. Hence, the basis should be $1,(\... 3answers 456 views ### Showing that$\sqrt5$is not in$\mathbb{Q}(\sqrt7)$How can I prove that$\sqrt5$is not in$\mathbb{Q}(\sqrt7)$? I can only think of trying to write$\sqrt5 = a+b\sqrt7$(where$a,b$are in$\mathbb{Q}$), but I can't think of a good reason that ... 2answers 1k views ### Linear independence of roots over Q Let$p_1,\ldots,p_k$be$k$distinct primes (in$\mathbb{N}$) and$n>1$. Is it true that$[\mathbb{Q}(\sqrt[n]{p_1},\ldots,\sqrt[n]{p_k}):\mathbb{Q}]=n^k$? (all the roots are in$\mathbb{R}^+$) ... 2answers 842 views ### Show that$ a，b，c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q \implies \sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q $Assume that$a，b，c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q$are rational，prove$\sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q$,are rational. I know that can be proved, would like to know that there is ... 1answer 515 views ### Elementary proof for$\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})$seems to be quite ... 2answers 133 views ### Is$(5+\sqrt[3]2)^n$ever an integer for$n \in \Bbb Z \setminus \{0\}$? In general, I would like to prove that if$m>2$is an integer, then$(5+\sqrt[m]2)^n$is never an integer (unless for$n=0$). First, I'm interested in the simple case$m=3\$ (I already solved it ...

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