So here is the question

If $f(x)$ is a polynomial satisfying $$f(x)f(y) = f(x) + f(y) + f(xy) - 2$$ for all real $x$ and $y$ and $f(3) = 10$, then $f(4)$ is equal to ?

here what i have tried

Putting $x=y=1$ in the given solution,$$(f(1)^2) = 3f(1) - 2$$ on solving it we get $$f(1) = 2$$ or$$f(1)= 1$$ so putting $y=1$in the eqation $$f(x)f(y) = f(x) + f(y) + f(xy) - 2$$we get $$f(x) = 1$$ but it not true as $$f(3) = 10$$ so $f(1) = 2$

i am stuck here i don't know what to do next please help me



  • $\begingroup$ What is the source of this question ? $\endgroup$ – user2369284 Dec 25 '13 at 16:33
  • $\begingroup$ @user2369284 Tata McGraw-Hill's $\endgroup$ – Dimensionless Dec 25 '13 at 16:55


$$f(x)f(y)=f(x)+f(y)+f(xy)-2$$ $$f(x)(y)-f(x)-f(y)+1=f(xy)-2+1$$ $$(f(x)-1)(f(y)-1)=f(xy)-1$$ Define $g(a)=f(a)-1$ for every $a$, now we have: $$g(x)g(y)=g(xy)$$

We know that $f$ is a polynomial, therefore $g$ is continuous. Hence ....

  • $\begingroup$ sorry but i have still no idea what to after $g(x)g(y)=g(xy)$can you help me a little bit more $\endgroup$ – Dimensionless Dec 25 '13 at 14:46
  • $\begingroup$ see Beni's answer. $\endgroup$ – Suraj M S Dec 25 '13 at 16:21

Your equation translates to $(f(x)-1)(f(y)-1)=f(xy)-1$. Define $g(x)=f(x)-1$ and you obtain that $g(x)g(y)=g(xy)$. This can be treated as a Cauchy functional equation.

Define $h(x)=\log(g(e^x))$. Then $h(x+y)=\log(g(e^x e^y))=\log(g(e^x))+\log(g(e^y))=h(x)+h(y)$. Since $h$ is continuous the only solution has the form $h(x)=ax$ so $$ \log(g(e^x))=ax \Leftrightarrow g(e^x)=e^{ax} \Leftrightarrow g(x)=x^a (x>0). $$


Another hint: Using your functional equation, prove that $f(x)=a+bx^n$.


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