Calculating $f(x)$

Question

Let $f(x)$ be a non-constant polynomial satisfying the relation $$f(x)f(y)=f(x)+f(y)+f(xy)-2; \forall x,y \in \Bbb R;\\ f(0)\ne1, f(4)=65,$$

By plugging $x=y=1$, we get two values of $f(1)=1,2$. But now which is the correct value? Similar when we do it for $x=0$, $f(0)=1,2$ and following the question $f(0)=2$. But as it is a non-constant polynomial $f(1)=1$, is this argument correct? We can easily calculate all the values of $f(x)$ as we did for $x=0,1$. We can also plug $y=\frac{1}{x}$ but now how should we calculate $f(x)$. Because we have two values each for $f(1),f(2),f(3)....$so on. But which is the correct one and what's the reason?

Answer 1

Hint: Let $p(x)=f(x)-1$ then we have $$p(x)p(y)= p(xy)$$ and $p(0)\ne 0$ and $p(4)=64$.

Then $p(1)=1$ (plug $x=0$ and $y=1$) and $p(0)=1$ (plug $x=y=0$). Also $p(2)^2 = p(4) = 64 \implies p(2) = \pm 8$.

Let $p(x) = ax^n+...$ and $a\ne 0$.

• If $p(2)=8$ then we get $8p(x) = p(2x)$ for all $x$ and thus $8ax^n + ...= a2^nx^n+...\implies \boxed{n=3}$...
• If $p(2)=-8$ then we get $-8p(x) = p(2x)$ for all $x$ and thus $-8ax^n + ...= a2^nx^n+...$ and thus no solution.

Answer 2

Let $c=f(0)\ne1$ and plug in $y=0$ to arrive at $$ c f(x)=f(x)+2c-2 $$ and simplify.