There is only one such function, that is $f(x) = x^2 + x + 1$.

Writing $x = 0$ in (i) gives $f(0) = f(0)^2$. As $f(x) > 0$, $f(0) = 1$.

Plugging in $-x$ in (i) gives $f(x)^2 - 2x f(x) = f(-x)^2 + 2x f(-x)$, that is $(f(x) - f(-x))(f(x) + f(-x)) = 2x(f(x) + f(-x))$. Again as $f(x) > 0$, we obtain $f(x) = f(-x) + 2x$.

Then by (ii) we have $f(x) = 2x + f(x - 1)$ (*).

By induction on $n$ we can show that $f(n) = n^2 + n + 1$ for all non-negative integer $n$.

By (*) applying induction on $n$ gives $f(x+n) = f(x) + 2x n + n^2 + n$ (**) for all real number $x$ and non-negative integer $n$.

Let $x = \frac{m}{n}$ where $m$ and $n$ are positive integers. Then by (**) we have $f((x+n)^2) = f(x^2 + 2m + n^2) = f(x^2) + 2x^2(2m + n^2) + (2m + n^2)^2 + 2m + n^2$.

On the other hand by (i) and (**) we get $f((x+n)^2) = (f(x) + 2m + n^2 + n)^2 - 2(x+n)(f(x) + 2x n + n^2 + n)$.

The last two equations conclude that $f(x) = x^2 + x + 1$ for all positive rational number $x$.

Since both $f(x)$ and $x^2 + x + 1$ are strictly increasing on $(1, \infty)$ and there exists a rational number in any interval, we can easily show that $f(x) = x^2 + x + 1$ for every real number $x > 1$.

Using (*) and (iii) we can show that $f(x) = x^2 + x + 1$ for all real number $x$ and it satisfies all three conditions.