, 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
$$f(x) + \frac{mx^5 f(y_0)}{f(x)^2} = 0,$$
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
$$f(x + f(y)) = f(x) + \frac{2x^5 f(y)}{f(x)^2} + f(y^2), \quad \forall x, y > 0.$$
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
$$f(x + n^2) = f(x) + \frac{2x^5 n^2}{f(x)^2} + n^4 \geq (x + n^2)^2.$$
By factoring this inequality, one can get
$$(f(x) - x^2) \left(1 - \frac{2x \cdot n^2 \cdot (f(x) + x^2)}{f(x)^2}\right) \geq 0, \quad \forall x > 0.$$
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
$$1 - \frac{2x_0 \cdot n^2 \cdot (f(x_0) + x_0^2)}{f(x_0)^2} \geq 0, \quad \forall n \in \mathbb{Z}^+,$$
which implies that
$$n^2 \leq \frac{f(x_0)^2}{2x_0(f(x_0) + x_0^2)}.$$
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. $\square$