[Mr.Com]

ข้อความเดิมเขียนโดยคุณ SixGoldsForThailand

Dear Euler-Fermat,
I would like to ask you to check the argument in red. From there, you missed a solution $f(x)=-x$. Other than that small point, your solution looks great!

I will provide an alternative solution, which is almost the same as Euler-Fermat's solution above.

Solution

First, substitution $(x,y) \mapsto (0,y)$ gives $f(f(y))=f(f(0))f(y)+y$___(1) which shows that $f$ is injective.

Substitution $(x,y) \mapsto (x,0)$ gives $f(f(0))=f(f(x))f(0)$. Since $f$ is injective, $f(f(x))$ cannot be constant, and therefore $f(0)=0$. Using this in (1) above gives $f(f(x))=x, \forall x$.

Substitution $(x,y) \mapsto (x-f(1),1)$ shows that $\forall x, f(x)= f(1) \cdot x + 1-f(1)^2$. A fortiori, $f$ is linear. Substituting linear $f$ back in the main equation shows that $f$ can be only $f(x) \equiv x, \forall x$ or $f(x) \equiv -x,\forall x$, both of which can be easily checked to satisfy the equation.

Can you explain further why we can conclude the injectivity of $f$ from the first substitution?