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- Feb 14, 2012
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Find all positive integers \(\displaystyle n\) for which \(\displaystyle \sqrt{n+\sqrt{1996}}\) exceeds \(\displaystyle \sqrt{n-1}\) by an integer.
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Find all positive integers n
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chisigma
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- Feb 13, 2012
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The condition leads to the equation...
$$\sqrt{n + \sqrt{1996}} = \sqrt{n-1}+ k\ (1)$$
... 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
$\chi$ $\sigma$
$$\sqrt{n + \sqrt{1996}} = \sqrt{n-1}+ k\ (1)$$
... 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.
Of course now I just jinxed myself!-Dan
- Jan 26, 2012
- 183
First
$\displaystyle \lim_{n \to \infty} \sqrt{n+\sqrt{1996}}-\sqrt{n-1} = 0$.
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.
$\displaystyle \lim_{n \to \infty} \sqrt{n+\sqrt{1996}}-\sqrt{n-1} = 0$.
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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Thanks to those for participating in this challenge problem.
My solution:
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.
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
kaliprasad
Well-known member
- Mar 31, 2013
- 1,341
nice deduction by making radicals same on both sides