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Let $x$ be an integer. If $\left(\sqrt{x+\frac{1}{2}\sqrt{2011}}-\sqrt{x-\frac{1}{2}\sqrt{2011}}\right)$ is an integer, find $x$.

We see that both $x+\frac{1}{2}\sqrt{2011}$ and $x-\frac{1}{2}\sqrt{2011}$ can't be perfect squares, so how do we continue?

user19405892
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  • Hint: If such a difference $\alpha-\beta$ is an integer and $x$ is an integer, $(\alpha-\beta)^2 = \alpha^2+\beta^2-2\alpha\beta $ is an integer, too. – Jack D'Aurizio Oct 05 '16 at 22:27
  • @JackD'Aurizio That's not if and only if. – user19405892 Oct 05 '16 at 22:29
  • True, but it leads to a solution, since it gives you that $$\sqrt{4x^2-2011}$$ has to be an integer, hence $4x^2-2011$ has to be a square, say $4x^2-2011=q^2$, so that $(2x-q)(2x+q)=2011$, and by factoring $2011$... – Jack D'Aurizio Oct 05 '16 at 22:31
  • It turns out to be a prime, so $2x-q=1$ and $2x+q=2011$, i.e. $x=503$, or... – Jack D'Aurizio Oct 05 '16 at 22:33
  • Same as mine: http://math.stackexchange.com/questions/1954197/denesting-a-nested-radical-where-x-y-in-mathbbz – Frank Oct 06 '16 at 18:05

1 Answers1

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Let $$k=\sqrt{x+\frac{1}{2}\sqrt{2011}}-\sqrt{x-\frac{1}{2}\sqrt{2011}}$$ Then $$k\left(\sqrt{x+\frac{1}{2}\sqrt{2011}}+\sqrt{x-\frac{1}{2}\sqrt{2011}}\right)=\sqrt{2011}$$

So $$\sqrt{x+\frac{1}{2}\sqrt{2011}}+\sqrt{x-\frac{1}{2}\sqrt{2011}}=\frac{\sqrt{2011}}{k}$$

Adding the first and the last we have $$2\sqrt{x+\frac{1}{2}\sqrt{2011}}=k+\frac{\sqrt{2011}}{k}$$

Squaring we get

$$4x+2\sqrt{2011}=k^2+\frac{2011}{k^2}+2\sqrt{2011}$$

or

$$4x=k^2+\frac{2011}{k^2}$$ since $x$ is an integer $\frac{2011}{k^2}$ is also an integer and thus $k^2=1$ and so $x=503$.

sbares
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Rene Schipperus
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