function with restrictions in finding solutions

Question

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How to prove the following functional equation, which has no solution.

$A + B + C + (A + B)C - 2 = 0$ has no solution, where $A = f(x)$, $B = g(x)$ and $C = h(x)$. Here $A, B$ and $C$ are real functions and knowing that all functions are non-linear with the following conditions:

1) $2A = A + (B - C) A$

2) $4B = B + (C- A) B$

3) $6C = C + (A - B) C$

4) $A, B$ and $C$ are not equal to zero.

Prove that $A + B + C + (A + B)C - 2 = 0$ has no solution with respect to the above conditions.

Answer 1

Part I: At least one has to be zero

Let $K_A := \{x \in \mathbb{R} | f(x) \neq 0\}$, $K_B := \{x \in \mathbb{R} | g(x) \neq 0\}$ and $K_C := \{x \in \mathbb{R} | h(x) \neq 0\}$

Condition 1

$\begin{align*} 2A &= A + (B-C) \cdot A \;\;\;\forall x \in \mathbb{R}\\ \Leftrightarrow A &= (B-C) \cdot A \;\;\;\forall x \in \mathbb{R}\\ \Rightarrow 1 &= B-C \;\;\;\forall x \in K_A \end{align*}$

Condition 2

$\begin{align*} 4B &= B + (C-A) \cdot B \;\;\;\forall x \in \mathbb{R}\\ \Leftrightarrow 3B &=(C-A) \cdot B \;\;\;\forall x \in \mathbb{R}\\ \Rightarrow 3 &= C-A\;\;\;\forall x \in K_B\\ \end{align*}$

Condition 3

$\begin{align*} 6C &= C + (A-B) \cdot C \;\;\;\forall x \in \mathbb{R}\\ \Leftrightarrow 5C &=(A-B) \cdot C \;\;\;\forall x \in \mathbb{R}\\ \Rightarrow 5 &= A-B\;\;\;\forall x \in K_C\\ \end{align*}$

Conclusion

$\forall x \in (K_A \cap K_B \cap K_C):$

$\begin{align} C &= B - 1\\ \Rightarrow 3 &= (B-1) - A\\ \Rightarrow 5 &= A - B\\ \\ B &= 4 + A\\ \Rightarrow 5 &= A - (4+A) = -4\\ \end{align} $

This is a contradiction $\Rightarrow K_A \cap K_B \cap K_C = \emptyset$

This means, there is no $x \in \mathbb{R}$ such that $f(x) \neq 0$ and $g(x) \neq 0$ and $h(x) \neq 0$.

Part II: Check cases

Case 1: $\{x \in \mathbb{R} | C(x) = 0\}$

Condition is $A + B = 2$

$\begin{align} 2A &= A + AB \\ \Leftrightarrow A &= AB\\ \\ 4B &= B - AB\\ \Leftrightarrow -3B &= AB\\ \\ A &= -3B \end{align}$

Now you can simplify the condition:

$\begin{align} -3B + B = -2B &= 2\\ \Leftrightarrow B &= -1\\ \Rightarrow A &= 3 \end{align}$

Case 2: $\{x \in \mathbb{R} | B(x) = 0\}$

Condition is $A + C + AC -2 = 0$

$\begin{align} 2A &= A - AC \\ \Leftrightarrow -A &= AC\\ \\ 6C &= C + AC\\ \Leftrightarrow 5C &= AC\\ \\ \Rightarrow -A &= 5C\\ \Leftrightarrow A &= -5C \end{align}$

Now you can simplify the condition:

$\begin{align} A + C + AC -2 &= 0\\ \Leftrightarrow -5C + C + (-5C)C -2 &= 0\\ \Leftrightarrow -5 C^2 -4C -2 &= 0\\ \Leftrightarrow C_{1,2} = \frac{1}{2 \cdot (-5)} \cdot \left (-4 \pm \sqrt{16 - 4 \cdot (-5) \cdot (-2)} \right ) \end{align}$

This means $C$ has no solution in $\mathbb{R}$. Hence $\{x \in \mathbb{R} | B(x) = 0\} = \emptyset$

Case 3: $\{x \in \mathbb{R} | A(x) = 0\}$

Condition is $B+C+BC-2 = 0$

$\begin{align} 4B &= B + BC\\ \Leftrightarrow 3B &= BC\\ \\ 6C &= C - BC\\ \Leftrightarrow 5C &= -BC\\ \Leftrightarrow -5C &= BC\\ \\ 3B &= -5C\\ \Leftrightarrow B &= -\frac{5}{3}C \end{align}$

Now you can simplify the condition:

$\begin{align} -\frac{5}{3}C+C+(-\frac{5}{3}C)C-2 &= 0\\ -\frac{5}{3}C^2-\frac{2}{3}C-2 &= 0\\ \frac{5}{3}C^2+\frac{2}{3}C+2 &= 0\\ \Rightarrow C_{1,2} &= \frac{1}{2 \frac{5}{3}} \cdot \left (- \frac{2}{3} \pm \sqrt{\frac{4}{9} - 4 \cdot \frac{5}{3} 2} \right )\\ &= \frac{3}{10} \cdot \left (- \frac{2}{3} \pm \sqrt{\frac{4}{9} - \frac{40}{3}} \right) \end{align}$

This means $C$ has no solution in $\mathbb{R}$. Hence $\{x \in \mathbb{R} | A(x) = 0\} = \emptyset$

Bring it together

• $\forall x \in \mathbb{R}: f(x) = 0 \text{ or } g(x) = 0 \text{ or } h(x) = 0$
• $\forall x \in \mathbb{R}: f(x) \neq 0 \text{ and } g(x) \neq 0$
• $\Rightarrow \forall x \in \mathbb{R} : h(x) = 0$
• $\Rightarrow h = 0 \Leftrightarrow C = 0$
• $\Rightarrow$ contradiction
• $\Rightarrow$ there are no such real functions $f,g, h \blacksquare$

Answer 2

The first three conditions can be written as $$A\bigl(1-(B-C)\bigr)=0,\quad B\bigl(3-(C-A)\bigr)=0,\quad C\bigl(5-(A-B)\bigr)=0\ .\tag{1}$$ According to the fourth condition $A$, $B$, $C$ are all nonzero. Therefore from $(1)$ we can conclude that $$1-(B-C)=0,\quad 3-(C-A)=0,\quad 5-(A-B)=0\ .$$ Adding these three equations together gives $9=0$. This signifies that your four conditions cannot be fulfilled simultaneously.