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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 $x$??

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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

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Yes it is possible and I think the value of $x= 0$.

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