# Functional problems: Find all functions such that $f(x)f(y) = f(xy + 1) + f(x - y) – 2$

Find all functions such that $$f(x)f(y) = f(xy + 1) + f(x - y) – 2$$ for all $$x, y$$ are real numbers.

I put y=0 into the equation and get $$(f(0)-1)f(x)=f(1)-2$$. If $$f(0)≠1$$, then $$f(x) = (f(1)-2)/(f(0)-1)$$ and so the function would be constant. But, since $$c^2=2c-2$$ has no real solutions, the function cannot be constant. Hence, $$f(0)=1$$, which implies that $$f(1)=2$$. Also note that f must be an even function as $$f(x-y)=f(y-x)$$.
Set $$f(x)=1+x^2+g(x)$$ for some even function of $$g$$, with $$g(0)=g(1)=0$$. Then put $$f(x)$$ into the functional equation and substitute $$y=1$$, get $$g(x+1)-g(x)=g(x)-g(x-1)$$. Define $$P(x) = {…, g(x-2), g(x-1), g(x), g(x+1), g(x+2),…}$$ , so $$P(x)$$ is an A.P. Then substitute $$y = -x , y=x$$ and using the fact that g is even, to get $$g(x^2+1) – g(x^2-1) = g(2x)$$. I’m trying to prove $$g(x)=0$$ for all $$x$$. How can I proceed?

Let $$S=\{\,x\in \Bbb R\mid f(x)=x^2+1\,\}$$. You already know: $$f(0)=1$$, $$f(1)=2$$, $$f$$ is even, i.e., $$\{0,1\}\subset S$$ and $$S=-S$$. We want to show $$S=\Bbb R$$, and this is equivalent to your endeavour to show $$g\equiv 0$$, but I find it easier to think of it this way.

One readily sees from the fact that $$x\mapsto x^2+1$$ solves the functional equation:

$$\tag0\text{If three out of }x,y,xy+1,x-y\text{ are }\in S\text{, then so is the fourth.}$$

With $$y=1\in S$$, this becomes $$\tag1\text {If two of }x-1,x,x+1\text{ are }\in S\text{, then so is the third.}$$ From this and $$0,1\in S$$, induction gives us $$\Bbb Z\subset S.$$ Likewise, with $$x=y$$ and using $$0\in S$$, $$(0)$$ becomes $$\tag2\text{If }x\in S\text{, then }x^2+1\in S.$$ More precisely, $$y=x$$ gives us $$\tag 3f(x)^2+1=f(x^2+1),$$ and so $$\tag4 f(x)\ge 1\quad\text{for }|x|\ge 1.$$ Consider $$x$$ with $$0<|x|<1$$. With $$y=2\operatorname{sgn}(x)+x$$, we get that $$|y|>1$$, $$|xy+1|>1$$, $$y-x=\pm2$$ so that by $$(4)$$, $$f(x)f(y)=f(xy+1)+3\ge 4$$, and hence $$f(x)>0$$. Thus together with $$(4)$$ and $$f(0)=1$$, $$f(x)>0\qquad\text{for all }x\in\Bbb R.$$ This and $$(3)$$ allows us to improve $$(2)$$ to $$\tag5x\in S\iff x^2+1\in S.$$ For $$0\le u_0, let $$u_n=u_{n-1}^2+1$$, $$v_n=v_{n-1}^2+1$$. Then $$u_n, $$(u_{n-1},v_{n-1})$$ is mapped bijectively to $$(u_n,v_n)$$ by $$x\mapsto x^2+1$$, and $$v_n-u_n=(u_{n-1}+v_{n-1})(v_{n-1}-u_{n-1})>2u_{n-1}(v_{n-1}-u_{n-1})\ge 2(v_{n-1}-u_{n-1})$$ for $$n\ge 2$$, hence some interval $$(u_n,v_n)$$ will have length $$>1$$ and therefore intersect $$\Bbb Z$$ and also $$S$$. We conclude from $$(5)$$ that also $$(u_0,v_0)$$ intersects $$S$$, hence $$\overline S=\Bbb R.$$ In other words,

$$f(x)=x^2+1$$ is the only continuous solution of the functional equation.

A way to show that $$g(x)=0$$ is the only continuous solution is to show that $$g(q)=0$$ for all rational $$q$$. To do this, substitute $$x=\frac mn$$ and $$y=n$$ ($$m,n\in\mathbb{Z}$$, $$n\ne0$$) into the equation and we get $$\tag1(1+n^2)\ g\left(\frac mn\right)=g\left(\frac mn-n\right)$$

Now, let $$d(x)=g(x+1)-g(x)$$. We have $$d(-x)=-d(x)$$. But $$(1)$$ implies that $$g\left(\frac mn\right)=-\frac 1n\ d\left(\frac mn\right)$$ $$m\rightarrow-m$$ then shows that $$g\left(\frac mn\right)=0$$, as desired.