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 $x$??
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asked Nov 4, 2014 at 6:10
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2 Answers
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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
answered Nov 4, 2014 at 6:12
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1$\begingroup$ I think it should be $n=\dfrac{1\pm\sqrt{1+4x}}2$ $\endgroup$ Commented Nov 4, 2014 at 6:28
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1$\begingroup$ @user109899, Thanks for your observation. I was just thinking on my way $\endgroup$ Commented Nov 4, 2014 at 6:39
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1$\begingroup$ 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? $\endgroup$ Commented Nov 4, 2014 at 6:50
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2$\begingroup$ @JyrkiLahtonen, Good Observation. Till I've found math.stackexchange.com/questions/61048/… and math.stackexchange.com/questions/410413/… also mathworld.wolfram.com/NestedRadical.html $\endgroup$ Commented Nov 4, 2014 at 6:55
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1$\begingroup$ That settles my doubts, thanks! $\endgroup$ Commented Nov 4, 2014 at 7:37
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Yes it is possible and I think the value of $x= 0$.
