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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.
Find all positive integers \(\displaystyle n\) for which \(\displaystyle \sqrt{n+\sqrt{1996}}\) exceeds \(\displaystyle \sqrt{n-1}\) by an integer.
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...
nice deduction by making radicals same on both sidesThe 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
$\chi$ $\sigma$