อ้างอิง:
ข้อความเดิมเขียนโดยคุณ Euler-Fermat
$P(x,y) แทน f(xy+f(y)) = f(f(x))f(y)+y $
$P(0,0) : f(f(0)) = f(f(0))f(0)$
$f(0) = 1$ หรือ $f(f(0)) = 0 $
ถ้า $f(0) = 1 $
$P(x,0) : 1 = f(f(x)) $ได้$ f(x) = c = 1$
แทนค่ากลับ แล้ว ไม่จริง
ได้ $f(f(0)) = 0 $
$P(0,y) : f(f(y)) = y $ได้ f เป็น bijection
$f(xy+f(y)) = xf(y)+y$
แทน $y = 1 $ได้ $f(x+f(1)) = xf(1)+1 $
ได้ $f(x) = x+c $
แทน ค่ากลับได้$ c = 0 $
ดังนั้น $f(x) = x $
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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.
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.