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- Feb 14, 2012

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- #1

- Feb 14, 2012

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- Feb 13, 2012

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$$\sqrt{n + \sqrt{1996}} = \sqrt{n-1}+ k\ (1)$$

... and squaring both terms of (2) we arrive to write...

$$2\ \sqrt{499} = k^{2} + 2\ k\ \sqrt{n-1} - 1\ (2)$$

The only solution of (2) for n and k integers seems to be n=500 and k=1...

Kind regards

$\chi$ $\sigma$

- Aug 30, 2012

- 1,176

Thank Heavens! My head was insisting there be more solutions, otherwise it seemed too easy.The only solution of (2) for n and k integers seems to be n=500 and k=1...

Of course now I just jinxed myself!

-Dan

- Jan 26, 2012

- 183

$\displaystyle \lim_{n \to \infty} \sqrt{n+\sqrt{1996}}-\sqrt{n-1} = 0$.

It not hard to show that the function is decreasing and when $n = 1$, $\sqrt{n+\sqrt{1996}}-\sqrt{n-1}=6.758447736$ then so you really only need to check when the difference is either $1, 2, 3, 4, 5$ or $6$ and only 1 gives that $n$ is an integer.

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- Feb 14, 2012

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Let \(\displaystyle \sqrt{n+\sqrt{1996}}-\sqrt{n-1}=k\), where \(\displaystyle k\) is an integer.

\(\displaystyle \sqrt{n+\sqrt{1996}}=k+\sqrt{n-1}\)

\(\displaystyle n+\sqrt{1996}=k^2+n-1+2k\sqrt{n-1}\)

\(\displaystyle n+\sqrt{1996}=k^2+n-1+\sqrt{4k^2(n-1)}\)

By equating the radicand from both sides, we get

\(\displaystyle 1996=4k^2(n-1)\)

\(\displaystyle 499=k^2(n-1)\)

Since 499 is a prime number, and \(\displaystyle k^2=1\), we can conclude that

\(\displaystyle n-1=499\) or \(\displaystyle n=500\) is the only solution to the problem.

- Aug 30, 2012

- 1,176

Nice trick with the radicands.

-Dan

-Dan

- Mar 31, 2013

- 1,341

nice deduction by making radicals same on both sides

$$\sqrt{n + \sqrt{1996}} = \sqrt{n-1}+ k\ (1)$$

... and squaring both terms of (2) we arrive to write...

$$2\ \sqrt{499} = k^{2} + 2\ k\ \sqrt{n-1} - 1\ (2)$$

The only solution of (2) for n and k integers seems to be n=500 and k=1...

Kind regards

$\chi$ $\sigma$