Can the sum of irrational square roots of two different rational numbers be another irrational square root of a rational number?

Question

Prove or refute that $\exists x, y, z \in$ with $\mathbb {Q}$ and $x\neq y$ such that $\sqrt{x}+\sqrt{y}=\sqrt{z}$ and $\sqrt{x}, \sqrt{y}, \sqrt{z}\in \mathbb{R}\setminus\mathbb{Q}.$

If $\sqrt{x}, \sqrt{y}, \sqrt{z}\in \mathbb{R}$, we find that $\sqrt{1}+\sqrt{4}=\sqrt{9}$ or even $\sqrt{0}+\sqrt{1}=\sqrt{1}$ holds. Thus we are interested in the case where the roots are irrational numbers but the radicands are rational numbers.

After squaring both members of the equality we get $x+2\sqrt{xy}+y=z\implies\sqrt{xy}=\frac{z-x-y}{2}$. This means that the product of $\sqrt{x}$ by $\sqrt{y}$ is a rational number. We can only conclude that both $\sqrt{x}$ and $\sqrt{y}$ are rational or irrational.

I can't find any good aproach to this problem and I don't even know if this result is relevant. Thank you for advance.

Answer 1

Letting $x = 2$, $y = 8$, and $z=18$ works: $$\sqrt 2 + \sqrt 8 = \sqrt 2 + 2 \sqrt 2 = 3 \sqrt 2 = \sqrt 18.$$ More generally, for any $a, b, c \in \mathbb Q$ we have $$ \sqrt{ab^2} + \sqrt{ac^2} = b\sqrt a + c \sqrt a = \sqrt{a(b+c)^2}.$$ This is an example whenever $b \ne \pm c$ and $\sqrt a$ is irrational.

Answer 2

If $a,b,c$ are positive rationals but not squares of rationals and $\sqrt c=\sqrt a +\sqrt b$ then $$c=(\sqrt a +\sqrt b)^2=a+b+2\sqrt {ab}\in \Bbb Q,$$ which requires $$\sqrt {ab}\in \Bbb Q,$$ which requires $ab=d^2$ where $0<d\in \Bbb Q,$ which requires $b=d^2/a,$ and hence $$c=a+b+2\sqrt {ab}=a+d^2/a+2d=(a+d)^2/a.$$ And so $\sqrt c=(a+d)/\sqrt a\not \in \Bbb Q.$

This is also sufficient. That is, if $a,d$ are positive rationals and $a$ is not the square of a rational then $\sqrt a$ and $\sqrt {d^2/a}=d/\sqrt a$ are irrational and so is $\sqrt a +\sqrt {d^2/a}=\sqrt {(a+d)^2/a}=(a+d)/\sqrt a.$

And to ensure $b\ne a$ it is necessary and sufficient that $b=d^2/a\ne a.$ That is, $a\ne d. $

For example $a=2, d=1.$ That is $(\sqrt 2+1/\sqrt 2)^2=9/2$ so $\sqrt 2 +\sqrt {1/2}=\sqrt {9/2}=3/\sqrt 2.$