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For convenience, let $(f(x), g(x))$ be a solution to the problem. Now, \begin{align*} f(x) + g(x) &= f(x)g(x) \\ f(x)g(x) - f(x) - g(x) &= 0 \\ f(x)g(x) - f(x) - g(x) + 1 &= 1 \\ (f(x) - 1)(g(x) - 1) &= 1 \end{align*}

By letting $f(x) - 1 = 1$ and $g(x) - 1 = 1$, we get the solution $(2,2)$. Also, letting $f(x) - 1 = -1$ and $g(x) - 1 = -1$, we get the solution $(0,0)$.

What I am wondering now, is if there exists $f$ and $g$ where both are nonconstant polynomials?

Edit 1. The set of real numbers is the domain and range of both $f$ and $g$.

soupless
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2 Answers2

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No, because then\begin{align}\deg\bigl(f(x)g(x)\bigr)&=\deg f(x)+\deg g(x)\\&>\max\bigl\{\deg f(x),\deg g(x)\bigr\}\\&\geqslant\deg\bigl(f(x)+g(x)\bigr).\end{align}

José Carlos Santos
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    @soupless : Test your statement yourself with $f(x) = x^2$ and $g(x) = x^3$. – Eric Towers Aug 01 '21 at 15:07
  • What about $f=g=2x$ in $\mathbb{Z}/ 4\mathbb{Z}$? – rosecabbage Aug 01 '21 at 15:18
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    Sorry, the domain and range is the set of real numbers. I forgot to indicate it in the question, editing it now. – soupless Aug 01 '21 at 15:20
  • Without loss of generality, assume that $\deg f(x)=n$ and $\deg g(x) = 0$, then $\deg(f(x)g(x))=\deg(f(x)+g(x))$. Now, $g(x)=0$ implies that $f(x)g(x)=0$ and $f(x) + g(x) \neq 0$ which means that $f(x)g(x) \neq f(x) + g(x)$. Furthermore, for $g(x) = c \neq 0$, the most promising $c$ is if $c = 1$, but this means that $a_0 = a_0 + 1$ if $a_0$ is the constant term of $f$. For other $c$ not equal to zero or one, this means that $ca_{i} = a_{i}$ for all $i$ from $i = 1$ to $i =n$ which holds only if and only if $c = 1$. Therefore, no such $f$ and $g$ exists where $\deg f(x) > 0$ and $\deg g(x)=0$. – soupless Aug 01 '21 at 15:48
  • Is this a correct solution? – soupless Aug 01 '21 at 15:49
  • It looks fine to me. – José Carlos Santos Aug 01 '21 at 15:53
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You showed it yourself! The only way the product of two polynomials can be constant is if they are themselves constant: otherwise, the degree of the product polynomial would not be zero. Thus $(f(x)-1)(g(x)-1)=1$ immediately implies $f,g$ are constant.

YiFan Tey
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