# can a sum of square roots be an integer?

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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like $\sqrt{9} + \sqrt{16} = 7$? – Jonathan Christensen Jan 14 at 23:07
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. – Fred Jan 14 at 23:10
Ok, I forgot to say I wanted both square roots to be irrational, i know the sum of integers is an integer. – Jorge Fernández Jan 14 at 23:18
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 – dinoboy Jan 15 at 0:13

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."

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 Ding! 1000 points (exactly)! – Eric Stucky Jan 15 at 2:18 Shouldn't that read "were already perfect squares?" – BlueRaja - Danny Pflughoeft Jan 15 at 7:52 Yeap, it should :S – Eric Stucky Jan 15 at 8:17

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.

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nice, this proves that it cant be an integer. But what about the second problem? – Jorge Fernández Jan 14 at 23:20
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. – user7530 Jan 14 at 23:29
@user7530 Squaring will work for up to 4 square roots. For example, take squares on both sides of $\sqrt{a} + \sqrt{b} = n - \sqrt{c}$. – Calvin Lin Jan 15 at 0:25
b is a perfect square... or c is zero. – Hurkyl Jan 15 at 1:11
Sure. I'll add a remark. – user7530 Jan 15 at 2:10

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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 I think this is a perfect example. Why the downvote? – emory Jan 15 at 1:31 A lot of downvoting on stackexchange is just angry teenagers acting up. – Kaz Jan 15 at 1:58 If this correct answer is unsatisfying, please edit the question such that it rules out this answer and then add a helpful comment under the answer to notify me that the answer is no longer correct. I will gladly delete the answer, or edit, if possible. – Kaz Jan 15 at 2:25 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. – JavaMan Jan 15 at 23:23 @JavaMan That is false. $-\sqrt 8$ is one of the two square roots of $8$, which is an integer. $(-\sqrt 8)\times(-\sqrt 8) = 8$. The question, in its present form, does not rule out negative roots. – Kaz Jan 15 at 23:27
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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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