Let p=sqrt(x+p) and then substitute for p in the square root expression and we get p=sqrt(x+sqrt(x+p)). If you continue the substitution, you get sqrt(x+sqrt(x+sqrt(x+sqrt(... indefinitely. Similarly, if we let p=sqrt(xp) and then substitute for p in the expression we get p=sqrt(xsqrt(xsqrt(x... indefinitely. So to solve the given equation, sqrt(x+p)=sqrt(xp), we solve p=sqrt(x+p), from which p^2=x+p and we solve p=sqrt(xp), from which p^2=xp. So dividing both sides by p, x=p. From the other equation p^2=x+p=2p, and so x=2.