# Adding square roots to receive Integers

Inspired by this question, I ask the following:

For any $a\in\mathbb{N}_0$, do integers $x\ne a,y\ne a$ exist such that $$y=\sqrt x+\sqrt a$$ $$\text{or}$$

$$y=\sqrt x-\sqrt a$$ And if yes, how is such a solution found?

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Do you mean for every $a$? –  lhf Dec 16 '12 at 13:39
Yes. Isnt that what "any a" (=any arbitrary a that is in the specified set) means? –  CBenni Dec 16 '12 at 14:34
any is ambiguous in mathematics. Better use some or every, according to what is meant. –  lhf Dec 16 '12 at 14:36

Suppose that $a$ is not a square, and suppose $y$ and $x$ exist so that $y = \sqrt{x} + \sqrt{a}$. Since $\sqrt{a}$ is a root of the polynomial $t^2 - a$, it follows that $\sqrt{x}$ is a root of the polynomial $$(y - t)^2 - a = t^2 - 2yt + y^2-a.$$ There are now two cases. First, if $x$ is a square, then $\sqrt{a} = y - \sqrt{x}$ is an integer, contradicting that $a$ is not a square. Second, if $x$ is not a square, then the minimal polynomial for $\sqrt{x}$ is $t^2 - x$, and hence $$t^2-x\mid t^2-2yt + y^2-a.$$ Since these polynomial have the same degree and are both monic, it follows that they are equal, i.e., $y = 0$ and $x = a$. This shows there are no $x$ and $y$ of the desired form when $a$ is not a square. A similar argument holds for $y = \sqrt{x}-\sqrt{a}$.
When $a$ is a square, you can (obviously) always find such $x$ and $y$.