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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14$\begingroup$ like $\sqrt{9} + \sqrt{16} = 7$? $\endgroup$– Jonathan ChristensenJan 14, 2013 at 23:07
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$\begingroup$ If you choose only integers that are squares of other integers this will always work, since their square roots are integers and the sum of a finite number of integers is an integer. $\endgroup$– FredJan 14, 2013 at 23:10
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1$\begingroup$ I think this link is a pretty good answer to your question. However, it might be at a level which is too advanced for you, since this is a pretty natural question to ask relatively early on in life, but it takes some significantly more difficult mathematics to prove. The direct, yes/no answer to the question is "Yes, but only if the numbers inside the square roots were already perfect squares," or equivalently "If you've already done all the simplifying that you can do, then no." $\endgroup$– Eric StuckyJan 14, 2013 at 23:13
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1$\begingroup$ I don't know if this blog is viewable without an Art of Problem Solving account, but a generalized version of this problem with a solution can be found on this blog post : artofproblemsolving.com/blog/80001 $\endgroup$– dinoboyJan 15, 2013 at 0:13
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$\begingroup$ See also this answer $\endgroup$– Bart MichelsApr 6, 2015 at 13:55
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4 Answers
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At least there's an elementary way to see that if $\sqrt{a} + \sqrt{b}$ is an integer, then $a$ and $b$ are perfect squares.
Suppose $\sqrt{a} + \sqrt{b} = c\in\mathbb{Z}.$ If $c=0$ the result is trivial. Otherwise, squaring both sides we get that $$a + b + 2\sqrt{ab} = c^2$$ and therefore $ab$ must be a perfect square. Let's say $ab = d^2$. Then $a=\frac{d^2}{b}$ and \begin{align*}\frac{d}{\sqrt{b}} + \sqrt{b} &= c\\ d + b &= c\sqrt{b}, \end{align*} so $b$ is a perfect square, and $a$ must be as well.
answered Jan 14, 2013 at 23:18
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3$\begingroup$ Sadly this argument doesn't generalize in an obvious way, since squaring $\sqrt{a}+\sqrt{b}+\sqrt{c}$ doesn't help. See Eric's answer for the fully general proof. $\endgroup$– user7530Jan 14, 2013 at 23:29
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2$\begingroup$ @user7530 Squaring will work for up to 4 square roots. For example, take squares on both sides of $\sqrt{a} + \sqrt{b} = n - \sqrt{c}$. $\endgroup$ Jan 15, 2013 at 0:25
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$\begingroup$ b is a perfect square... or c is zero. $\endgroup$– user14972Jan 15, 2013 at 1:11
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Suppose that $a,b,\sqrt a+\sqrt b\in\mathbb Z$.
$(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=a-b\in\mathbb Z$. Since $\sqrt a-\sqrt b=\frac{a-b}{\sqrt a+\sqrt b}\in\mathbb Q$. Therefore, $\sqrt a-\sqrt b$ is an algebraic integer and rational; thus, $\sqrt a-\sqrt b\in\mathbb Z$.
Next, $(\sqrt a+\sqrt b)+(\sqrt a-\sqrt b)=2\sqrt a\in\mathbb Z$ and $(\sqrt a+\sqrt b)-(\sqrt a-\sqrt b)=2\sqrt b\in\mathbb Z$. Thus, $\sqrt a$ and $\sqrt b$ are algebraic integers and rational, therefore $\sqrt a,\sqrt b\in\mathbb Z$.
Thus, $a,b,\sqrt a+\sqrt b\in\mathbb Z\Rightarrow\sqrt a,\sqrt b\in\mathbb Z$
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$\begingroup$ What do you mean by "algebraic integer"? I don't understand how you went from saying that $\sqrt{a}-\sqrt{b}$ is a rational to saying it is an integer. $\endgroup$– PlatoDec 24, 2016 at 9:26
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$\begingroup$ An Algebraic Number is a root to a polynomial with integer coefficients. An Algebraic Integer is a root to a polynomial with integer coefficients with $1$ as the leading coefficient. An algebraic integer that is also rational is an integer (see this answer). $\endgroup$– robjohn ♦Dec 24, 2016 at 11:23
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2$\begingroup$ Ah ok thank you! I was also thinking that rather than saying $\sqrt{a}-\sqrt{b}$ is an integer, you can keep saying it is rational. Since $\sqrt{a}$ is either rational (integer) or irrational, you get that it must be a whole number. $\endgroup$– PlatoDec 24, 2016 at 12:47
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$\begingroup$ Why the downvote? When this answer was written, part of the question (including the title) was concerned with the sum of two integers. Since that is a much simpler part of the question, that is what I answered. In full generality, this question is much harder. $\endgroup$– robjohn ♦Jan 14, 2019 at 22:24
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Yes. For instance, 8 has two distinct square roots: $\sqrt 8$ and $-\sqrt 8$. These add to zero, which is an integer.
The same thing happens with higher order roots in the complex plane. When we add the roots of a number together, we get zero. This is because they form equally distributed points on the unit circle in the complex plane, and so, if we regard them as vectors, we can readily see that they cancel out under addition.
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3$\begingroup$ I didn't downvote, but your answer does not address the question. The OP asked whether the statement "$a_1, \dots , a_n \in \mathbb{Z} \implies \sum_{i=1}^n \sqrt{a_i} \notin \mathbb{Z}$" is true or not. Since $-\sqrt{8}$ is never the square root of an integer, your answer sheds no light on the situation. $\endgroup$– JavaManJan 15, 2013 at 23:23
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2$\begingroup$ If one is to be pedantic, then I suppose you are correct. Then again, if one was to be pedantic, the equation $\sqrt{0} + \sqrt{0} = 0$ would also seem relevant here. $\endgroup$– JavaManJan 15, 2013 at 23:29
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2$\begingroup$ @JavaMan That is incorrect. The question is worded in such a way that it rules out $\sqrt 0$, $\sqrt 1$ and $\sqrt 4$. It plainly says "The square roots need to be irrational." Mathematics demands that we be pedantic. $\endgroup$– KazJan 15, 2013 at 23:30
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3$\begingroup$ When a question is tagged with
algebra-precalculus, it is perfectly reasonable to assume that "square root" refers to the principal square root. $\endgroup$– user7530Jul 20, 2013 at 18:29
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For two summands:
$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b}) = a- b$.
So either both factors are irrational or both are rational. The second case can only happen if $a$ and $b$ are perfect squares.
answered Jan 28, 2019 at 15:06