I was playing around on Maple with some nested radicals and I notices that $$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\cdots}}}}=2$$ I thought my mind was playing tricks on me so I did the algebra; I let $$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}$$ Then $x^2=2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}$, and thus $$x^2=2+x$$ Solving for $x$, $x=-1,$ or $2$, so $x=2$. Then I did it for the multiplied nested radical. Again, I let $$x=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\cdots}}}}$$ Then $x^2=2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\cdots}}}}$, and thus $$x^2=2x$$ Solving for $x$, $x=0,2$, so $x=2$ So naturally I wondered if this could happen with any other number other than 2. It did not work with 3, or 4 (i figured that was the case) so again, I let algebra take over. I replaced $2$ with $a$ and I said that a solution would be where $$a+x=ax$$ But this is not the correct approach. Then I thought that I needed to solve the summed radical first using the quadratic formula, so $$x=\frac{1\pm\sqrt{1+4a}}{2} $$ And then the multiplied radical $$x=0, x=a$$ So plugging in $a$ for $x$ in the first equation gives $a=2$. Thus the only number that satisfies this equality is $a=2$.
Now (A) is this correct? (B) are there any more interesting results such as this perhaps with multiple variables or other operations involved?
