Solving the functional equation $f\left(x^2+f(y)\right)=f(x)^2+y^4+2f(xy)$

Question

Problem: find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $$f\left(x^2+f(y)\right)=f(x)^2+y^4+2f(xy),\ \ \ \forall x,y\in\mathbb{R}\text.$$

Answer 1

It can be shown that the only function $ f : \mathbb R \to \mathbb R $ satisfying $$ f \left( x ^ 2 + f ( y ) \right) = f ( x ) ^ 2 + y ^ 4 + 2 f ( x y ) \tag 0 \label 0 $$ for all $ x , y \in \mathbb R $ is $ f ( x ) = x ^ 2 $. It's straightforward to check that this is indeed a solution. We prove that it's the only one.

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General Properties of $ f $:

Let $ a = f ( 0 ) $ and $ b = f ( 1 ) $. By letting $ y = 0 $ and $ y = 1 $ in \eqref{0}, we respectively get the following: $$ f \left( x ^ 2 + a \right) = f ( x ) ^ 2 + 2 a \text ; \tag 1 \label 1 $$ $$ f \left( x ^ 2 + b \right) = f ( x ) ^ 2 + 2 f ( x ) + 1 \text . \tag 2 \label 2 $$ Substituting $ - x $ for $ x $ in \eqref{1} and comparing with \eqref{1} itself, we have $ f ( - x ) ^ 2 = f ( x ) ^ 2 $. Then, substituting $ - x $ for $ x $ in \eqref{2} and comparing with \eqref{2} itself we get $ f ( - x ) = f ( x ) $; i.e. $ f $ is an even function. Letting $ x = 0 $ in \eqref{0} we have $$ f \big( f ( y ) \big) = y ^ 4 + a ^ 2 + 2 a \text . \tag 3 \label 3 $$

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Showing $ f ( 0 ) = 0 $ and $ f ( 1 ) = 1 $:

We can use \eqref{1} and \eqref{3} together to get $$ f \left( x ^ 4 + 2 a x ^ 2 + a ^ 2 + a \right) = f \left( x ^ 2 + a \right) ^ 2 + 2 a \\ = f ( x ) ^ 4 + 4 a f ( x ) ^ 2 + 4 a ^ 2 + 2 a = f \Big( f \big( f ( x ) \big) \Big) + 4 a f ( x ) ^ 2 + 3 a ^ 2 \\ = f \left( x ^ 4 + a ^ 2 + 2 a \right) + 4 a f ( x ) ^ 2 + 3 a ^ 2 \text , $$ which by letting $ x = \sqrt { \frac 1 2 } $ yields $ 4 a f \left( \sqrt { \frac 1 2 } \right) ^ 2 + 3 a ^ 2 = 0 $, and that implies $ a \le 0 $. We can then use \eqref{1} and the fact that $ f $ is even to get $$ f \left( \sqrt { - \frac a 2 } \right) ^ 2 + 2 a = f \left( \frac a 2 \right) = f \left( - \frac a 2 \right) = f \left( \sqrt { - \frac { 3 a } 2 } \right) ^ 2 + 2 a \text . $$ On the other hand, substituting $ f ( x ) $ for $ x $ in \eqref{2} we can use \eqref{3} and see that $$ f \left( f ( x ) ^ 2 + b \right) = \Big( f \big( f ( x ) \big) + 1 \Big) ^ 2 = x ^ 8 + 2 ( a + 1 ) ^ 2 x ^ 4 + ( a + 1 ) ^ 4 \text . $$ As we know that $ f \left( \sqrt { - \frac a 2 } \right) ^ 2 = f \left( \sqrt { - \frac { 3 a } 2 } \right) ^ 2 $, we can conclude that $$ \left( \sqrt { - \frac a 2 } \right) ^ 8 + 2 ( a + 1 ) ^ 2 \left( \sqrt { - \frac a 2 } \right) ^ 4 = \left( \sqrt { - \frac { 3 a } 2 } \right) ^ 8 + 2 ( a + 1 ) ^ 2 \left( \sqrt { - \frac { 3 a } 2 } \right) ^ 4 \text , $$ which can be simplified to $ a ^ 2 \left( 5 a ^ 2 + 4 ( a + 1 ) ^ 2 \right) = 0 $, and thus $ a = 0 $. Now, putting $ x = 1 $ in \eqref{1} we have $ b ^ 2 = b $. But if $ b = 0 $, putting $ x = 0 $ in \eqref{2} leads to a contradiction. Hence $ b $ can't be equal to $ 0 $, and we must have $ b = 1 $.

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Proving $ f ( x ) = x ^ 2 $:

Using \eqref{0}, \eqref{1}, \eqref{3} and $ a = 0 $, we have $$ f \left( x ^ 4 + y ^ 4 \right) = f \Big( x ^ 4 + f \big( f ( y ) \big) \Big) = f \left( x ^ 2 \right) ^ 2 + f ( y ) ^ 4 + 2 f \big( x ^ 2 f ( y ) \big) \\ = f ( x ) ^ 4 + f ( y ) ^ 4 + 2 f \big( x ^ 2 f ( y ) \big) \text , $$ which by symmetry yields $$ f \big( x ^ 2 f ( y ) \big) = f \big( y ^ 2 f ( x ) \big) \text . \tag 4 \label 4 $$ Now, putting $ y = 1 $ in \eqref{4} and using \eqref{3}, $ a = 0 $ and $ b = 1 $ we get $ f \left( x ^ 2 \right) = f \big( f ( x ) \big) = x ^ 4 $, and thus $ f ( x ) = x ^ 2 $ for all $ x \ge 0 $. As $ f $ is even, we conclude that $ f ( x ) = x ^ 2 $ for all $ x \in \mathbb R $.