Saudi Arabia booklet 2024

If there exist $y_0$ such that $f(y_0)<y_0^2$ then taking $x=y_0^2-f(y_0)$ to get $$f(x+f(y_0))=f(y_0^2).$$ From this, one can get $$f(x)+\frac{m{x}^{5}f(y_0)}{f(x{)}^2}=0,$$ which is contradiction. So $f(y)\ge y^2$ for $y>0$. Put $x=y=1$ then $$f(1+f(1))=2f(1)+\frac{m}{f(1)}\ge (1+f(1){)}^2.$$ Since $f(1)\ge 1$ then we have $m\ge f(1)+f(1{)}^3\ge 2$. So the least value of $m$ is 2. For $m=2$, it is easy to verify that $f(1)=1$ and $$f(x+f(y))=f(x)+\frac{2x^{5}f(y)}{f(x{)}^2}+f(y^2),\forall x,y>0.$$ Substitute $y=1$ then $f(x+1)=f(x)+\frac{2x^5}{f(x{)}^2}+1$ for all $x>0$. Put $x=1$ then $f(2)=f(1)+\frac{2}{f(1)}+1=4$. Continue to put $x=2,3,\dots$ then by induction, one can get $f(n)=n^2$ for all positive integers $n$. Put $x=n$ into the condition then $$f(x+n^2)=f(x)+\frac{2x^5{n}^2}{f(x{)}^2}+n^4\ge (x+n^2{)}^2.$$ By factoring this inequality, one can get $$(f(x)-x^2)\left(1-\frac{2x\cdot n^2\cdot (f(x)+x^2)}{f(x{)}^2}\right)\ge 0,\forall x>0.$$ If $f(x)=x^2$ for all $x>0$ then we can check this is a solution. Otherwise, there exists $x_0$ such that $f(x_0)>x_0^2$ then we must have $$1-\frac{2x_0\cdot n^2\cdot (f(x_0)+x_0^2)}{f(x_0{)}^2}\ge 0,\forall n\in {\mathbb{Z}}^{+},$$ which implies that $$n^2\le \frac{f(x_0{)}^2}{2x_0(f(x_0)+x_0^2)}.$$ This is a contradiction when we take $n\to +\infty$. Thus for $m=2$, function $f(x)=x^2$ is the unique function satisfying the condition. Therefore, the least value of $m$ that need to find is 2. $\square$