1)Let we have $f(x) < x^2+2ax$ for $x \geq 1$ ( we can start from $a=1011$) Then $x^2 \leq f(x)^2+2af(x) \to x<f(x)+a$ And $f(x)^2 \leq f(f(x))^2+2af(f(x))=x^4+2ax^2<(x^2+ax)^2 \to f(x)<x^2+ax$ So from $f(x)<x^2+2ax \to f(x)<x^2+ax \to ... \to f(x) < x^2+\frac{a}{2^n}x \to f(x) \leq x^2$ Also we have $f(x)<x^2+2ax \to f(x)<x^2+\frac{a}{2^n}{x} \to f(x)>x-\frac{a}{2^{n+1}} \to f(x) \geq x$ If $f(x)=x \to f(f(x))=x \to $ contradiction for $x>1$ If $f(x)=x^2 \to f(f(x))=x^2=f(x) \to f(x)=x \to $ contradiction for $x>1$ So $x<f(x)<x^2$ for $x>1$ ( We can have $f(1)=1$) 2. Let for every prime $p>2$ and integer $n$ we have: $f(p^{2^n})=(p^2-p)^{2^n}$ $f((p^2-p)^{2^n})=p^{2^{n+1}}$ And if number $x$ is is not in form $p^{2^n}$ or $(p^2-p)^{2^n}$ then $f(x)=x^{\sqrt{2}}$ We can check that $f(f(x))=x^2$ and $f(x) <x^2$ Also we can check that $(p^2-p)^{2^n}=q^{2^m}$ and $(p^2-p)^{2^n}=(q^2-q)^{2^m}$ have not solutions for different primes $p,q$ Also we have $\frac{f(p)}{p^2}=1-\frac{1}{p} > c$ for $p>\frac{1}{1-c}$