Solving for polynomial $f(x)$ satisfying $f(x)f(y)=f(x)+f(y)+f(xy)-2 \forall x\in \mathbb{R}, f(1)\ne 1, f(3)=10$

Question

I am trying to solve this functional equation in order to find the value of $f(5)$.

Find the value of $f(5)$, if $f(x)$ satisfies $f(x)f(y)=f(x)+f(y)+f(xy)-2 \ \forall x\in\mathbb{R}$, $f(1)\ne 1$ and $f(3)=10$. $f(x)$ is a polynomial function.

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My Attempt

Putting $x=y=1$ gives $f(1)=2$. Now substituting $x\mapsto -x$ to check whether $f(x)$ is even, odd or neither. Subtracting and putting $y=1$ suggests that the function is even.

$$\begin{aligned}f(x)f(y)&=f(x)+f(y)+f(xy)-2\\f(-x)f(y)&=f(-x)+f(y)+f(-xy)-2\end{aligned}$$ This reduces $f(x)$ to being $\sum_{k=0}^{n}a_{2k}x^{2k}$ where $a_i\in\mathbb{R}\forall i\in [1,n]\cap\mathbb{Z^{+}}$. Solving for $f(0)$ gives out $f(0)=1,2$.

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Edit $1$:

Clearly if we solve for $f(9)$ the value of $f(3)$, we get $f(9)=82=81+1$. So one candidate is $f(x)=x^2+1$. But how to prove rigorously that this is the only possible polynomial function with the properties as stated above?

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I don't know how to proceed. Any hints are appreciated. Thanks

Answer 1

Define $g(x):=f(x)-1$. It is equivalent with the Cauchy functional equation:

$$g(x)g(y)=g(xy)$$

There are two cases two consider:

• If it's constant, then $f(5)=f(1)=2$, but this can not be because $f(3)=10$.
• If $f$ has degree $n$, then replacing $f(y)$ gives $f(x)=x^n+1$. From $f(3)=10$, we get that $n=2$. Thus $f(5)=26$.

Answer 2

put x=y=1 $ f(1)^2-3f(1)+2=0 \Rightarrow f(1)=2$

Now put $y=\frac{1}{x}$, we get $f(x)f(\frac{1}{x})=f(x)+f(\frac{1}{x})$ which is a standard functional eqn

$(f(x)-1)(f(\frac{1}{x})-1)=1 \Rightarrow f(x)=\pm x^n+1$