Functie strict crescatoare

[Claudiu Mindrila]

Gasiti toate functiile strict crescatoare \( f:\mathbb{Z}\rightarrow\mathbb{Z} \) care satisfac conditia \( f(x^{2}+y^{2})=f^{2}(x)+f(y^{2}) \) pentru orice \( x,y\in\mathbb{Z} \).

Valentin Vornicu, lista scurta, 2005

[Marius Mainea]

\( f=\mathbb{1}_{\mathbb{Z}} \)

[Claudiu Mindrila]

Intai \( y=0\Longrightarrow f(x^{2})=f^{2}(x)\Longrightarrow f(x^{2}+y^{2})=f(x^{2})+f(y^{2}) \), iar apoi inductiv \( f(x_{1}^{2}+\ldots+x_{n}^{2})=f(x_{1}^{2})+\ldots+f(x_{n}^{2}),\ \forall a_{i}\in\mathbb{Z},\ i=\overline{1,\ n},\ n\in\mathbb{N}^{*} \).
Tot inductiv obtinem si faptul ca \( f(k^{2})=k^{2},\ \forall k\in\mathbb{Z} \). Iar daca \( n \) nu e patrat e perfect, atunci fie \( k \in \mathbb{Z} \) a. i. \( k^2<n<(k+1)^2 \Longrightarrow n=k^{2}+l,\ l\in\{ 1,\ 2,\ldots,\ 2n\} \), dar atunci
\( f(n)=(k^{2}+\underbrace{1+1+\ldots+1}_{l\ \text{ori}})=f(k^{2})+lf(1)=k^{2}+l=n. \)
Aici am folosit faptul ca \( f(1)=1 \). Intr-adevar, luand in relatia initiala \( x=y=0 \Longrightarrow f(0)=0 \). Dar \( f(x^{2})=f^{2}(x) \), deci \( f(1)=f^{2}(1)\Longrightarrow f(1)\in\{ 0,\ 1\} \Longrightarrow f(1)=1 \) (caci f e injectiva).
Asadar \( f(n)=n, \ \forall n\in\mathbb{N} \). Insa daca \( x \neq 0 \), cum \( f^{2}(-x)=f(x^{2})=f^{2}(x)\Longrightarrow f(x)\in\{ f(-x),\ -f(-x)\} \Longrightarrow f(x)=-f(-x)\Longrightarrow f(-x)=-f(x) \). Deci \( f(k)=k,\ \forall k\in\mathbb{Z} \).

[Marius Mainea]

Intai \( y=0\Longrightarrow f(x^{2})=f^{2}(x)\Longrightarrow f(x^{2}+y^{2})=f(x^{2})+f(y^{2}) \), iar apoi inductiv \( f(x_{1}^{2}+\ldots+x_{n}^{2})=f(x_{1}^{2})+\ldots+f(x_{n}^{2}),\ \forall a_{i}\in\mathbb{Z},\ i=\overline{1,\ n},\ n\in\mathbb{N}^{*} \).

Cum arati inductiv?