If such a function F(x)=x+ 1, now let x=a be brought into this equation to get F(a)=a+ 1, right? Let x=a+ 1 and bring it directly into the original equation to get f (a+1) = (a+1)+1. Can you understand this? I now set b=a+ 1, then the equation f (a+1) = (a+1)+1is F(b)=b+ 1, and this is the substitution method! Replace the equation about A with the equation about B, and the two equations are equivalent!

Then the square of f (root number x+1) = (root number X+ 1)-1, so f (x) = square of x-1, which is easy to understand. Is to use method of substitution to change the whole number (root number X+ 1) into X, of course, it can also be changed into other unknowns such as A, B, C, etc. Because we are used to writing the equation as an expression about X, we finally changed it into the form of X, that is, F(x)=x+ 1 and F (A) = A+6544.

Matching method is to transform the original equation into the required form, and does not involve the substitution of unknowns. F (root number X+ 1)=X+2 root number X= (root number X+ 1) squared-1, that is, matching, in order to appear (root number X+ 1), and then exchange elements.

This problem can be directly replaced. Let t= radical X+ 1, then X = (T- 1) 2, so f (radical X+ 1) becomes F(t), which is the form we need. Then replace all x in the equation with (T- 1) 2 to get the expression of t, and just write t as x after sorting.