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.

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

1 Answer 1


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 $f(1)=2$.


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