# 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.

## marked as duplicate by YuiTo Cheng, José Carlos Santos, max_zorn, The Count, Xander HendersonJun 23 at 14:12

• like $\sqrt{9} + \sqrt{16} = 7$? – Jonathan Christensen Jan 14 '13 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 '13 at 23:10
• 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 '13 at 0:13
• See this answer. – Bill Dubuque Jul 22 '16 at 21:26

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."

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.

• 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 '13 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 '13 at 0:25
• b is a perfect square... or c is zero. – Hurkyl Jan 15 '13 at 1:11
• Sure. I'll add a remark. – user7530 Jan 15 '13 at 2:10

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, $a,b,\sqrt a+\sqrt b\in\mathbb Z\Rightarrow\sqrt a,\sqrt b\in\mathbb Z$

• 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. – Plato Dec 24 '16 at 9:26
• 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). – robjohn Dec 24 '16 at 11:23
• 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. – Plato Dec 24 '16 at 12:47
• 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. – robjohn Jan 14 at 22:24

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.

• 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 '13 at 23:23
• 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. – JavaMan Jan 15 '13 at 23:29
• @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. – Kaz Jan 15 '13 at 23:30
• Fair enough, and this is my last remark on the topic. However, mathematics does not require pedantry. In fact, mathematics is often hindered by it. Such questions have been considered here, for example. – JavaMan Jan 15 '13 at 23:37
• When a question is tagged with algebra-precalculus, it is perfectly reasonable to assume that "square root" refers to the principal square root. – user7530 Jul 20 '13 at 18:29

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.