# Two different values for equation $f(x)\cdot f(y)=f(x)+f(y)+f(xy)-2$

Let's say I have this functional equation

$$f(x)\cdot f(y)=f(x)+f(y)+f(xy)-2$$

For all $$x,y \in R$$

And I have given with $$f(2)=5$$

If I proceed to find $$f(1)$$

By substituting $$x=1$$and $$y=2$$

Then I will be getting $$f(1)=2$$ by using the given relation [$$f(2)=5$$]

But if I substitute $$x=1$$and $$y=1$$

And on solving for $$f(1)$$.

I am getting two values for $$f(1)$$ i.e $$2$$ and $$1$$

Now what is wrong in second approach.

• Nothing wrong. Just because you got $f(1)=1$ or $f(1)=2$ cannot conclude that $1$ sand $2$ are both solutions. May 13 at 4:32

Your method is correct. You get $$f(1)=1$$ or $$f(1)=2$$ for candidates. But you need to plug into the original equation to verify these two candicates.

Assume $$f(1)=1$$

$$f(1)\cdot f(2)=f(1)+f(2)+f(1\cdot2)-2\Longrightarrow f(2)=1$$

But this contradicts with the given condition $$f(2)=5$$, hence drop this candidate.

Next, you verify the other candidate $$f(1)=2$$

$$f(1)\cdot f(2)=f(1)+f(2)+f(1\cdot2)-2\Longrightarrow f(2)=5$$

This agrees with the given condition, so the correct value is $$f(1)=2$$.