INMO 2016 proposal from IISc

This is a nice little problem. Note that from the second functional equation, we get, $(f(x))^2\le 2x(2+x)+1=2(1+x)^2-1<2(1+x)^2\implies f(x)<\sqrt{2}(1+x)$ . This in turn implies $(f(x))^2<\sqrt{2}x(x+2)+1=\sqrt{2}(x+1)^2+1-\sqrt{2}<\sqrt{2}(1+x)^2\implies f(x)<2^{1/4}(1+x)$ . Continuing in this manner we see that, for any $x\in [1,\infty]$ , $f(x)<a_n(1+x)$ , for all $n\ge 0$ , where $a_n=2^{1/2^n}$ . Hence, for any $x\in [1,\infty]$ $f(x)/(1+x)\le \inf_{n}a_n=1$ .

Now, let $\exists a\in [1,\infty]$ such that $f(a)=1+a-\epsilon_0$ for some $a\ge \epsilon_0>0$ . Then, $af(a+1)=f^2(a)-1\implies af(a+1)=a(a+2)+\epsilon_0^2-2(a+1)\epsilon_0\implies f(a+1)=a+2-\epsilon_1$ where $\epsilon_1=2(a+1)\epsilon_0/a-\epsilon_0^2/a=2\epsilon_0+\frac{2\epsilon_0-\epsilon_0^2}{a}\ge \epsilon_0(1+2/a)=\rho \epsilon_0$ where $\rho=1+2/a>1$ . Similarly, $f(a+2)=a+3-\epsilon_2$ , where $\epsilon_2\ge\rho \epsilon_1$ . Thus, we can find a sequence $\{\epsilon_n\}$ such that $f(a+n)=a+1-\epsilon_{n}, \ n\ge 0$ , where $\epsilon_n>\rho^n \epsilon_0$ . Thus, $f(a+n)<a+1-\rho^n \epsilon_0$ . But, $f(x)\ge 1$ , $\forall x\in [1,\infty]$ . This implies that $\forall n\ge 0, \rho^n \epsilon_0<a$ , which is clearly not possible, since $\rho >a$ . This gives a contradiction to the assumption that $f(a)<1+a$ .

So, there cannot be point $x$ where $f(x)<1+x$ , and hence, $f(x)=1+x\ \forall x\in [1,\infty)$ . Thus the desired answer is $\boxed{2017}$ .

If there is a polynomial function satisfying these conditions, then we have a solution and don't need to look any further. Note : This is a pragmatic approach, assuming that the solution is unique without proving it!

So let $f(x)$ be a polynomial of degree $n$ . In the second equation, the degree of the lhs is $d + 1$ and on the rhs it is $2d$ . Equation this shows $d = 1$ , so $f = ax + b$ is linear.

Substituting this in the bottom equation we get $ax^2 + (a+b)x = a^2x^2 + 2abx + b^2-1.$ Equating the coefficients gives $\begin{cases} a = a^2, \\ a+b = 2ab, \\ 0 = b^2-1. \end{cases}$ It is easy to see that $a = 1$ and $b = 1$ . Thus a polynomial function $f$ satisfying the second condition must be $f(x) = x+1.$ The condition $f(x) = x+1 \leq 2(x+1)$ is easily checked. Therefore we submit the solution $f(2016) = 2016 + 1 = \boxed{2017}.$

Edit : Proof that this is a unique solution

By popular request I will pay my debt and prove that this function is unique.

Let $f(x) = x + 1 + g(x)$ . The conditions translate into $\begin{cases} g(x) < x+1 \\ g(x+1) = \left(\frac{g(x)}x + 2 + \frac2x\right)g(x).\end{cases}$ Now suppose that for some $x$ , $g(x) > 0$ . Then the second equation implies $g(x+1) > 2g(x),$ and repeating it we get $g(x+n) > 2^ng(x).$ This exponential function grows faster than any linear function, so that from a certain value of $n$ onward, $g(x+n) > x+n+1$ , in violation of the first condition. Therefore $g(x) \leq 0$ for all $x$ .

Likewise, let $g(x) < 0$ for some $x$ . Since $f(x) \geq 1$ we know $g(x) \geq -x$ . The expression $g(x)/x + 2 + 2/x$ is positive, showing that $g(x+1) < 2g(x)$ , and likewise $g(x+n) < 2^ng(x)$ . Again, we have an exponential function; for a certain value of $n$ onward, this implies $|g(x+n)| > x+n$ , making $f(x+n) < 1$ in violation of the given domain. Therefore $g(x) \geq 0$ for all $x$ .

We have now proven that $g(x) = 0$ on the entire domain, leaving $f(x) = x+1$ as the only solution.