What is wrong with this proof? I followed the example of the answer to another one of my questions, here
Define a general recurrence relation as $$f(x)^2=A(x)+B(x)f(x+n).$$ Substitute the root $B(y)=0$ so that $$f(y)=\sqrt{A(y)}.$$ Rewrite the expression as $$f(x)=\sqrt{A(x)}+g(x)$$ where $g(y)=0$. We can go back and say that $$f(y-n)=\sqrt{A(y-n)+B(y-n)\sqrt{A(y)}}=\sqrt{A(y-n)}+g(y-n).$$ Solve for $g(y-n)$. $$g(y-n)=\sqrt{A(y-n)+B(y-n)\sqrt{A(y)}}-\sqrt{A(y-n)}.$$ Turn $g(x)$ into the identity function so $g(x)=x$. Then we'll say $$g(y-n)=y-n=\sqrt{A(y-n)+B(y-n)\sqrt{A(y)}}-\sqrt{A(y-n)}.$$ We change the domains to simplify. $$x=\sqrt{A(x)+B(x)\sqrt{A(x+n)}}-\sqrt{A(x)}.$$ Now, for any functions, $A$ and $B$, that satisfy the equation above, we have $$f(x)=\sqrt{A(x)+B(x)f(x+n)}=\sqrt{A(x)}+x.$$ We are finished. There is certainly something flawed and illogical but I can't seem to find it.