Can a finite sum of square roots be an integer?

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.