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Find all functions $ f : \mathbb R \to \mathbb R $ such that for all $ a , b , c \in \mathbb R $, if $$ a + f ( b ) + f ^ 2 ( c ) = 0 \tag {Eq1} \label {eqn1} $$ then $$ f ( a ) ^ 3 + b f ( b ) ^ 2 + c ^ 2 f ( c ) = 3 a b c \text . \tag {Eq2} \label {eqn2} $$

I've been trying to solve this "implication-based" functional equation. My first thought was to use substitution to \eqref{eqn2} from \eqref{eqn1}.

$$ f ( a ) = - f ^ 2 ( c ) - a $$ $$ f ( a ) ^ 2 = f ^ 2 ( c ) ^ 2 + 2 a f ^ 2 ( c ) + a ^ 2 $$ therefore after some manipulation: $$ f ( a ) ^ 3 + b f ^ 2 ( c ) ^ 2 + 2 a b f ^ 2 ( c ) + c ^ 2 f ( c ) = a b ( 3 c - a ) $$

Then, I tried plugging in some values, for instance the simplest case $a=b=c=1$ $$f(1)^3 + f^2(1)^2 + 2f^2(1) + f(1) = 2$$

I have not been able to deduce any useful information from this, nor from similar substitutions with 1.

I'd be grateful for your help with this problem.

Mohsen Shahriari
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texiwi
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    Do you use $f^2(c)=f(f(c))$ notation? Or it is $(f(c))^2$? – Ivan Kaznacheyeu Sep 24 '21 at 13:49
  • The former, you can tell from his working. – merelymyself Sep 24 '21 at 13:56
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    The function $f(x) = x$ is one solution. – Arthur Vause Sep 24 '21 at 15:34
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    The constant function $ f ( x ) = 0 $ is another solution. – Mohsen Shahriari Sep 24 '21 at 19:13
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    Also $f(x) = -x$ – Arthur Vause Sep 24 '21 at 20:04
  • It seems that there is a typo in the beginning of the part where you show your attempt: it should be $ \require {color} f ( { \color {red} b } ) = - f ^ 2 ( c ) - a $. Of course, the equations implied from that are affected with this mistake, too, and they all should be corrected together. I'm not sure whether you've correctly done the "manipulation" you've mentioned, either. – Mohsen Shahriari Sep 25 '21 at 00:34
  • Also, your reasoning about $ f ( 1 ) $ is invalid: you can't plug $ a = b = c = 1 $ in \eqref{eqn2} (or its consequences) unless you know beforehand that \eqref{eqn1} is true for those specific values of the variables, i.e. $ a = - f ( b ) - f ^ 2 ( c ) $. You didn't show $ 1 = - f ( 1 ) - f ^ 2 ( 1 ) $ (and in fact, you cannot show it, as it is not true for any of the known solutions of the problem), so that's not valid. – Mohsen Shahriari Sep 25 '21 at 00:34

1 Answers1

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First of all, note that the problem can be stated in the following equivalent form.

Find all functions $ f : \mathbb R \to \mathbb R $ such that $$ f \left( - f ( x ) - f ^ 2 ( y ) \right) ^ 3 + 3 x y \left( f ( x ) + f ^ 2 ( y ) \right) + x f ( x ) ^ 2 + y ^ 2 f ( y ) = 0 \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $. Here, $ f ^ 2 ( y ) $ means $ f \bigl( f ( y ) \bigr) $, and $ f ( x ) ^ 2 $ means $ f ( x ) \cdot f ( x ) $.

It's straightforward to verify that the constant zero function, the identity function and the additive inversion function all satisfy the required condition. We prove that these are the only solutions.

Letting $ k = f ( 0 ) $ and putting $ x = y = 0 $ in \eqref{0}, we get $ f \bigl( - k - f ( k ) \bigr) = 0 $. Then, plugging $ x = - k - f ( k ) $ and $ y = 0 $ in \eqref{0} we can see that $ f \bigl( - f ( k ) \bigr) = 0 $. Now, set $ x = - f ( k ) $ in \eqref{0} and consider the equation once with $ y = - k - f ( k ) $ and another time with $ y = - f ( k ) $. Comparing the two results, you can conclude $ k ^ 2 f ( k ) = 0 $, which together with $ f \bigl( - f ( k ) \bigr) = 0 $ implies $ k = 0 $.

Putting $ y = 0 $ in \eqref{0} gives $$ f \bigl( - f ( x ) \bigr) ^ 3 = - x f ( x ) ^ 2 \tag 1 \label 1 $$ for all $ x \in \mathbb R $, while plugging $ x = 0 $ in \eqref{0} yields $$ f \left( - f ^ 2 ( y ) \right) ^ 3 = - y ^ 2 f ( y ) \tag 2 \label 2 $$ for all $ y \in \mathbb R $. Letting $ x = f ( y ) $ in \eqref{1}, we get $ f \left( - f ^ 2 ( y ) \right) ^ 3 = - f ( y ) f ^ 2 ( y ) ^ 2 $, which together with \eqref{2} shows that for any $ y \in \mathbb R \setminus \{ 0 \} $ with $ f ( y ) \ne 0 $ we have $ f ^ 2 ( y ) ^ 2 = y ^ 2 $. But also note that if $ y , f ( y ) \ne 0 $ and $ f ^ 2 ( y ) = - y $, then \eqref{2} gives $ f ( y ) \in \{ - y , y \} $; $ f ( y ) = y $ cannot happen since it implies $ f ^ 2 ( y ) = y $ (which contradicts $ f ^ 2 ( y ) = - y $ as $ y \ne 0 $), and $ f ( y ) = - y $ is not possible because it implies $ f ( - y ) = f ^ 2 ( y ) = - y $, which then substituting $ - y $ for $ y $ in \eqref{2} implies $ f ( y ) = y $ (contradicting $ f ( y ) = - y $ as $ y \ne 0 $). Therefore, the case $ f ^ 2 ( y ) = - y $ is ruled out, and we must have $$ f ( y ) \ne 0 \implies f ^ 2 ( y ) = y \tag 3 \label 3 $$ for all $ y \in \mathbb R \setminus \{ 0 \} $. Now, assume that there exists $ y _ 0 \in \mathbb R \setminus \{ 0 \} $ with $ f ( y _ 0 ) \ne 0 $. Putting $ y = y _ 0 $ in \eqref{2} and using \eqref{3} we have $ f ( - y _ 0 ) ^ 3 = - y _ 0 ^ 2 f ( y _ 0 ) $. Using this and \eqref{3} and setting $ y = y _ 0 $ in \eqref{0}, we can see that if $ f ( x ) = 0 $ for some $ x \in \mathbb R $, then we must have $ x = 0 $. Hence, using \eqref{3} we can conclude $ f ^ 2 ( y ) = y $ for all $ y \in \mathbb R $. Consequently, substituting $ f ( x ) $ for $ x $ in \eqref{0}, we get $$ f ( - x - y ) ^ 3 + 3 y ( x + y ) f ( x ) + x ^ 2 f ( x ) + y ^ 2 f ( y ) = 0 \tag 4 \label 4 $$ for all $ x , y \in \mathbb R $. Interchanging $ x $ and $ y $ in \eqref{4} and comparing with \eqref{4} itself, it's straightforward to see that $$ y f ( x ) = x f ( y ) $$ for all $ x , y \in \mathbb R $. In particular, this gives $ f ( x ) = f ( 1 ) x $ for all $ x \in \mathbb R $, and as $ f ^ 2 ( 1 ) = 1 $, we have $ f ( 1 ) \in \{ - 1 , 1 \} $. Therefore, we've proven that if $ f $ differs from the constant zero function, we must either have $ f ( x ) = x $ for all $ x \in \mathbb R $ or $ f ( x ) = - x $ for all $ x \in \mathbb R $, which proves what was claimed.

Mohsen Shahriari
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