Problema rezolvată #23 – ONGM 2020-2021

 f(x^3 + f(y))=x\cdot f^2(x)+y, (\forall)x, \ y \in  \mathbf{N}

\textbf{ Dacă  x = 0  } \text{observăm că: }\qquad \qquad \qquad \qquad \qquad\qquad\qquad \qquad\\ f( 0 + f(y))= 0\cdot f^2(0)+y\implies \\ \implies f(f(y)) = y, (\forall) \ y \in  \mathbf{N}.   \qquad

 \text{ Fie } z \in  \mathbf{N} \text{ a.î. }  f(z) = 0  \overset{\circ f}  {\implies}  \\  \overset{\circ f}{\implies}  f(f(z))= f(0) \implies\\  \implies f(0)= \ z \\ \; \\
    f(z^3)=z\cdot f^2(z) + z \implies \\  \implies f(z^3) = z \cdot 0 +z \implies \\  \implies f(z^3) = z \overset{\circ f}{\implies}  f(f(z^3)) = f(z)  \implies \\  \implies  z^3 = f(z) \implies z^3 = 0  \implies  \\  \implies z = 0 \implies f(0) = 0 \implies  \newline  \newline \implies   f(x^3)=x\cdot f^2(x), \ \  (\forall)x, \ \in  \mathbf{N}\\ \;\\
  f(1^3) = 1 \cdot f^2(1)  \implies \\ \implies f(1) = 1 \cdot f^2(1) \implies   \\  \implies f^2(1)  - f(1) = 0 \Leftrightarrow \\ \Leftrightarrow f(1)\cdot (f(1) -1) = 0 \newline \newline

 \textbf{ Vom analiza două cazuri}  f(1) = 0 \textbf{ și } f(1) -1 = 0 \qquad \qquad \qquad \qquad \qquad 

f(1) = 0 \implies f(f(1)) = f(0) \implies  \\ \implies f(0) = 1 \text{ imposibil} f(0) = 0

f(1) - 1 = 0 \implies f(1) = 1\\\;\\
  f(2) = f(1 + f(1)) = 1\cdot f^2(1) + 1 \implies  \\ \implies f(2) = 1 \cdot 1 + 1 \implies  f(2) = 2 

 \left .
  \begin{array}{ll}
  f(n+1)= f(1+n) =f(1^3+n) \\
  f(x^3 + f(y))=x\cdot f^2(x)+y\\
  \end{array}
  \right \} \implies \\\;\\ \implies f(1^3+n) = 1 \cdot f^2(1) + n \overset{f(1) =1}{\implies} \\\;\\ \overset{f(1) =1}{\implies}f(1+n) = 1 \cdot 1 + n  \implies \\\;\\ \implies  f(n+1) = n+1  \implies \\\;\\ \implies f(n) = n, \  (\forall)n, \ \in  \mathbf{N} \implies \\\;\\ 
  \implies f(2021) = 2021.