Fractions Fun with Functions

First, since $\dfrac{f(f(x))}{f(x)}$ is always positive, we can square both sides to obtain $\dfrac{f(x)^2}{x^2}=\dfrac{f(f(x))}{f(x)}$ . Clearing denominators gives $f(x)^3=x^2f(f(x)) \qquad (*)$

Let $\text{deg}(f(x))=d$ . We can see that $3d=2+d^2$ , or $d^2-3d+2=0$ This factors as $(d-1)(d-2)=0$ , so $d=1,2$

We can now let $f(x)=ax^2+bx+c$ . With this format, we can cover all polynomials with degree $\le 2$ .

First, notice that the constant term of the $LHS$ of $(*)$ is $c^3$ , while the constant term of the $RHS$ is $0$ . Thus, we are forced to make $c=0$

Thus $f(x)=ax^2+bx$ . We cannot compare the corresponding coefficients easily anymore, so we are forced to just expand it.

Plugging in $f(x)=ax^2+bx$ , we see that $(ax^2+bx)^3=x^2(a(ax^2+bx)^2+b(ax^2+bx))$ Expanding gives $a^3x^6+3a^2bx^5+3ab^2x^4+b^3x^3=a^3x^6+2a^2bx^5+ab(b+1)x^4+b^2x^3$

Looking at the $x^3$ term, we see that $b^3=b^2$ , so $b=0,1$

If $b=1$ , then looking at the $x^5$ term tells us that $3a^2=2a^2$ , so $a=0$ Thus one solution is $f(x)=x$

If $b=0$ , then everything expect for the $x^6$ term disappears, and we are left with $a^3x^6=a^3x^6$ which is always true. Thus, $a$ can be any integer. Our second family of solutions is thus $f(x)=ax^2$

Since the polynomial satisfies $f:\mathbb{R}^+\to \mathbb{R}^+$ , we must have $a\in \mathbb{Z}^+$ .

Now all we need to do is find the sum of all possible values of $f(3)$ under $100$ . Clearly, if $f(x)=x$ , then $f(3)=3$ . If $f(x)=ax^2$ , then $f(3)=9a$ . Since $9a < 100$ , we must have $a\le 11$ . Thus, any polynomial $f(x)=ax^2$ with $a=1\to 11$ works.

We sum up all of the $f(3)$ from this family of solutions: $9(1+2+\cdots +10+11)=9\cdot \dfrac{11\cdot 12}{2}=594$ . Thus, our answer is $594+3=\boxed{597}$ .