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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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    *any* is ambiguous in mathematics. Better use *some* or *every*, according to what is meant.2012-12-16

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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 $y$.