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If $$n= \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}......}}}$$ Is it possible that $n$ is a integer for any $x=Z( \text{zahlen number})$.If yes .What is the value of $x$??

Bart Michels
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Fazla Rabbi Mashrur
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2 Answers2

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So, $n^2=x+n\iff n^2-x-n=0\implies n=\dfrac{1\pm\sqrt{1+4x}}2$

As $n\ge0,n=\dfrac{1+\sqrt{1+4x}}2$ So, we need $1+4x$ to be Perfect Square

As $1+4x$ is odd, $1+4x=(2m+1)^2\iff x=m^2+m$ where $m$ is any integer

lab bhattacharjee
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    I think it should be $n=\dfrac{1\pm\sqrt{1+4x}}2$ – Redundant Aunt Nov 04 '14 at 06:28
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    @user109899, Thanks for your observation. I was just thinking on my way – lab bhattacharjee Nov 04 '14 at 06:39
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    Well done, as expected! My nitpicking core wants to include a proof for the fact that the sequence defined by those nested square roots converges for these choices of $x$. Otherwise the equation $n^2=x+n$ loses its footing :-) This may have been covered elsewhere on the site? – Jyrki Lahtonen Nov 04 '14 at 06:50
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    @JyrkiLahtonen, Good Observation. Till I've found http://math.stackexchange.com/questions/61048/definition-of-convergence-of-a-nested-radical-sqrta-1-sqrta-2-sqrta-3 and http://math.stackexchange.com/questions/410413/convergence-divergence-of-a-particular-infinite-nested-radical also http://mathworld.wolfram.com/NestedRadical.html – lab bhattacharjee Nov 04 '14 at 06:55
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    That settles my doubts, thanks! – Jyrki Lahtonen Nov 04 '14 at 07:37
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Yes it is possible and I think the value of $x= 0$.

thunder
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