Romanian Mathematical Olympiad

a) For $a=b$ we have $f(0)=0$. Then $f(a^2)=f(a{)}^2$ for all $a\in \mathbb{N}$, implying $f(1)=f(1{)}^2$ and then $f(1)\in \{0,1\}$. Both following cases hold, $f_0(1)=0$ for $f_0\equiv 0\in \mathcal{F}$, and $f_1(1)=1$ for $f_1=1_{\mathbb{N}}\in \mathcal{F}$. The requested set is $\{0,1\}$.

b) Let $a,b\in \mathbb{N}$, $a\ge b$ and $f\in \mathcal{F}$. Then $f(a),f(b)\in \mathbb{N}$ and $f(a{)}^2-f(b{)}^2=f(a^2-b^2)\ge 0$, which yields $f(a)\ge f(b)$, establishing that $f$ is non-decreasing. Suppose $f(1)=0$. From $f(4)=f(2{)}^2=f(2{)}^2-f(1{)}^2=f(3)$ we obtain $f(7)=f(4{)}^2-f(3{)}^2=0$. By induction, $f(7^{2^n})=0$. Since $f$ is monotonic, it follows that $f=f_0$. Suppose $f(1)=1$ and set $f(2)=x$. Then $f(4)=x^2$, $f(3)=x^2-1$, $f(5)=f(3{)}^2-f(2{)}^2=x^4-3x^2+1$. As $f(16)=f(5{)}^2-f(3{)}^2$, we obtain $x^4=(x^4-3x^2+1{)}^2-(x^2-1{)}^2$, which rewrites as $x^2(x^2-1{)}^2(x^2-4)=0$. The function $f$ is increasing ($a,b\in \mathbb{N}$, $a>b$ implies $f(a{)}^2-f(b{)}^2\ge f(1)>0$), so $x=2$. By induction we establish that $f(2^{2^n})=2^{2^n}$ and then, using the monotony, $f=f_1$.