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Find all positive integers n

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anemone

MHB POTW Director
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Feb 14, 2012
3,682
Find all positive integers \(\displaystyle n\) for which \(\displaystyle \sqrt{n+\sqrt{1996}}\) exceeds \(\displaystyle \sqrt{n-1}\) by an integer.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Find all positive integers \(\displaystyle n\) for which \(\displaystyle \sqrt{n+\sqrt{1996}}\) exceeds \(\displaystyle \sqrt{n-1}\) by an integer.
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

$\chi$ $\sigma$
 

topsquark

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MHB Math Helper
Aug 30, 2012
1,123
The only solution of (2) for n and k integers seems to be n=500 and k=1...
Thank Heavens! My head was insisting there be more solutions, otherwise it seemed too easy.

(Wait) Of course now I just jinxed myself!

-Dan
 

Jester

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MHB Math Helper
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.
 
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anemone

MHB POTW Director
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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.
 

topsquark

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MHB Math Helper
Aug 30, 2012
1,123
Nice trick with the radicands. (Bow)

-Dan
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
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

$\chi$ $\sigma$
nice deduction by making radicals same on both sides