根式鏈

$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}=2$

$\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}=3$

$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+...}}}}}=4$

$\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20+...}}}}}=5$

....

$\sqrt{a(a-1)+\sqrt{a(a-1)+\sqrt{a(a-1)+\sqrt{a(a-1)+\sqrt{a(a-1)+...}}}}}=a$，$a$是大於1的自然數.

假設$x=\sqrt{a(a-1)+\sqrt{a(a-1)+\sqrt{a(a-1)+\sqrt{a(a-1)+\sqrt{a(a-1)+...}}}}}$，則

$x^2=a(a-1)+\sqrt{a(a-1)+\sqrt{a(a-1)+\sqrt{a(a-1)+\sqrt{a(a-1)+\sqrt{a(a-1)+...}}}}}$，

$x^2=a(a-1)+x$，

$(x-a)(x+a-1)=0$，解得$x=a$。