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Is there function $f: \mathbb R^2 \rightarrow\mathbb R $ such that:

  1. $\forall a \in \mathbb R$ $\exists x, y \in \mathbb R$ such that $f(x;y)=a$

  2. $\forall x, y \in \mathbb R$ $$f(x;y)=-f(y;x)$$

  3. $\forall x, y, z \in \mathbb R$ $$f(x;f(y;z))=2021f(f(x;y);z)?$$

###My work so far:

I have proved

I. $\forall a \in \mathbb R$ $$f(a;a)=0$$

II. $\forall a \in \mathbb R$ $$f(a;0)=f(0;a)=0$$

III. I have hypothesis that $f(a;1)=a$ for any $a$ in $\mathbb R$ but I think there are not exist such function

Roman83
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1 Answers1

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This is my try:
Assume that there exists $f: \mathbb R^2 \rightarrow\mathbb R $ a function such that:

(1) $\forall a \in \mathbb R$ $\exists x, y \in \mathbb R$ such that $f(x;y)=a$

(2) $\forall x, y \in \mathbb R$ $$f(x;y)=-f(y;x)$$

(3) $\forall x, y, z \in \mathbb R$ $$f(x;f(y;z))=2021f(f(x;y);z)$$ Let $(a, b) \in \mathbb{R}^2$, then thanks to (1) there exists $(x, y) \in \mathbb{R}^2$ such that $b=f(x, y) $. Thus
$$f(a, b)=f(a, f(x, y))...(4)$$ Now, we have $$\begin{array}{ccc} f(a, f(x, y))&=&-f(f(x, y), a) \hspace{0.5cm}\text{(from(2))} \\ &=&-\dfrac{1}{2021} f(x, f(y, a)) \hspace{0.5cm}\text{(from(3))} \\ &=&\dfrac{1}{2021} f(f(y, a),x) \hspace{0.5cm}\text{(from(2))} \\ &=&\dfrac{1}{2021^2} f(y, f(a,x)) \hspace{0.5cm}\text{(from(3))} \\ &=&-\dfrac{1}{2021^2} f(f(a,x),y) \hspace{0.5cm}\text{(from(2))} \\ &=&-\dfrac{1}{2021^3} f(a,f(x, y)) \hspace{0.5cm}\text{(from(3))} \\ \end{array} $$ Then $$ \left(1+\dfrac{1}{2021^3} \right) f(a,f(x, y))=0 $$ which gives $$ f(a,f(x, y))=0 $$ Now, $(4)$ allows to say that $$ \forall (a, b) \in \mathbb{R}^2; f(a, b)=0$$ but this function doesn't satisfy (1), hence we have no solutions.

Ahmed Bachir
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