Find all functions $ f : \mathbb R \to \mathbb R $ satisfying $$ f \left( x + y ^ 2 \right) = f ( x ) +f ( y ) ^ 2 $$ for all $ x , y \in \mathbb R $.
Attemp:
$ f $ is non-decreasing:
In fact, $ f ( y ) ^ 2 \ge 0 $, so $ f ( x + \epsilon ) \ge f ( x ) $ for all $ \epsilon > 0 $.$ f ( 0 ) = 0 $:
Just set $ x = y = 0 $.$ f \left( y ^ 2 \right) = f ( y ) ^ 2 $, for all real $ y $:
Just set $ x = 0 $.$ f $ is odd:
Setting $ x = - y ^ 2 $, we have $ f \left( - y ^ 2 \right) + f ( y ) ^ 2 = f \left( y ^ 2 \right) + f \left( - y ^ 2 \right) = 0 $. So $ f ( r ) = - f ( - r ) $ for every real $ r $.$ f $ satisfies the Cauchy functional equation:
From the previous items, it's easy to see that $ f ( x + a ) = f ( x ) + f ( a ) $ and $ f ( x - a ) = f ( x ) + f ( - a ) $ for all non-negative reals $ a $ and all reals $ x $.$ f $ is increasing or $ f \equiv 0 $:
Suppose that $ 0 $ is not the only solution of $ f ( x ) = 0 $ and thus $ f ( \epsilon ) = 0 $ for some $ \epsilon > 0 $. from $ f $ satisfying the Cauchy equation we have that $ f ( q ) = f ( 1 ) q $ for any rational $ q $. In addition, $ f \left( 1 ^ 2 \right) = f ( 1 ) ^ 2 $, so that $ f ( 1 ) = 1 $ or $ f ( 1 ) = 0 $. If $ f ( 1 ) = 0 $ we have $ f ( q ) = 0 $ for all rational $ q $, and this with the condition of being non-decreasing would guarantee that $ f ( x ) = 0 $ for all real $ x $. If $ f ( 1 ) = 1 $ take rationals $ \alpha $ and $ \beta $ with $ \alpha < \epsilon < \beta $. We have $ 0 < \alpha \le f ( \epsilon ) \le \beta $, which is absurd, so $ f (x) = 0 $ if and only if $ x = 0 $. In conclusion, if $ f \ne 0 $, $ f ( y ) ^ 2 = f \left( y ^ 2 \right) > 0 $ holds for all non-zero $ y $ and $ f ( x + \epsilon ) > f ( x ) $ holds for all real $ x $ and $ \epsilon > 0 $.
These conditions together with the Cauchy equation guarantee that in the non-zero case the solution is $ f ( x ) = x $ for all real $ x $.
Answer: $ f = 0 $ or $ f ( x ) = x $.
Am I right?