Puzzling Polynomial

First we need the order of the polynomial. If the leading term is $ax^{n}$ then the leading term of the differential is $anx^{n-1}$ meaning that $a^{2}n^{2}x^{2n-2}=ax^{n}$ . Therefore 2n-2=n (n=2).

From this we can also tell that $4a^{2}=a$ Which implies a=0, 0.25 a cannot be 0 as otherwise n would be 1 so a=0.25.

Now we have $f(x)=\frac{x^{2}}{4} +bx+c$ = $f'(x)^{2}$ = $\frac{x^{2}}{4}+bx+b^{2}$ , $b^{2}=c=1$ (as f(0)=1). So $f(x)=\frac{x^{2}}{4} +x+1$ , f(9)=121/4.

121+4=125

Letting $y=f(x)$ and $\dfrac{dy}{dx}=f'(x)$ , we have $f(x)=(f'(x))^2 \Longrightarrow y=\left(\frac{dy}{dx}\right)^2 \Longrightarrow \frac{dy}{dx}=\pm \sqrt{y} = \sqrt{y}$ Here, we only take the positive value of $\sqrt{y}$ , for since the coefficients of $f(x)$ are all positive, then $f(x)>0$ and $f'(x) > 0$ for all real $x\geqslant 0$ . By separation of variables, we have $\begin{aligned} \frac{1}{\sqrt{y}} dy = y^{-1/2} dy &= dx \\ \int_{y=1}^{y=f(9)} y^{-1/2} dy &= \int_{x = 0}^{x = 9} dx\\ \left[2y^{1/2}\right]_1^{f(9)} &= [x]_0^9 \\ 2\sqrt{f(9)} - 2 &= 9-0 \\ \sqrt{f(9)} &= \frac{11}{2} \\ f(9) &= \frac{121}{4} \end{aligned}$ Therefore $a+b=121+4=\boxed{125}$ .

Let $d$ denote the degree of $f$ . Then note that the degree of the RHS is $2(d-1)$ as differentiation lowers the degree of a (non-zero) polynomial by $1$ , and squaring doubles the degree. So $d=2(d-1)$ (by the given equation) so $d=2$ .

Since $d=2$ , then we can write $f(x)$ as $ax^2+bx+c$ with $a \neq 0$ , and since $f(0)=1$ , in fact $f(x)=ax^2+bx+1$ .

So $f'(x)=2ax+b$ , and $(f'(x))^2=4a^2x^2+4abx+b^2$

By the equation, $ax^2+bx+1=4a^2x^2+4abx+b^2$ , so equating coefficients:

$(^1) a=4a^2$ , $(^2) b=4ab$ and $(^3) 1=b^2$

By $(^1)$ , $a=0$ or $a=\frac{1}{4}$ , but since $a \neq 0$ , $a=\frac{1}{4}$ .

By $(^3)$ , $b=1$ or $b=-1$ , but since all coefficients of $f$ are positive, $b=1$ .

So $f(x)=\frac{1}{4}x^2+x+1$ , and $f(9)=\frac{81}{4}+10=\frac{121}{4}$ , so $a+b=121+4=\boxed{125}$