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Let say $a_0, a_1, ..., a_k$ are a series of positive integers, with $k > 0$, if one of the number has irrational square root, $\sqrt{a_n}$, such as $\sqrt{2}$, or $\sqrt{3}$, there is not way we can find a series of $a_0, ..., a_k$ to make $$\sqrt{a_0}+ \sqrt{a_1} + \sqrt{a_2}+ ... + \sqrt{a_k}$$ is an integer.

Is this assumption correct? Is there a theorem for this? Thanks.

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  • $\begingroup$ I am not sure it is duplicate. I mean square root, not any b1, b2,,, bn. $\endgroup$
    – user325483
    Mar 29 '16 at 4:47
  • $\begingroup$ Uh Oh, how can I edit with symbol square root? It seems someone helped me but got the last equation wrong. I tried to correct and now it messed up. $\endgroup$
    – user325483
    Mar 29 '16 at 4:49
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    $\begingroup$ sure it's not an exact duplicate, but this is a special case of that question. I will fix your latex. $\endgroup$
    – vadim123
    Mar 29 '16 at 4:53
  • $\begingroup$ Thank you, @vadim123. I don't understand the other question. $\endgroup$
    – user325483
    Mar 29 '16 at 4:57
  • $\begingroup$ Rational and Integer are not synonyms. Integers are a subset of the rational numbers. Regardless, that relaxation is easy to answer with a very quick yes. Take for example $\pi + (6-\pi)$. Both $\pi$ and $6-\pi$ are irrational and positive, yet $\pi+(6-\pi)=6$ is an integer. Your original question is much harder. As for interpreting the linked question for your case, your case is the special circumstance where all of $b_1,b_2,\dots,b_k$ are all equal to two and some of your summands are possibly rational as well. $\endgroup$
    – JMoravitz
    Mar 29 '16 at 5:00

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