Can identity help us?

Let $\ x = A - B \$ and $\ y = AB \$ . Then $\begin{aligned} x^2 &= (A - B)^2 \\ &= A^2 - 2AB + B^2 \\ & = A^2 - 2y + B^2 \end{aligned}$ and $\begin{aligned} x^3 &= (A - B)^3 \\ &= A^3 - 3A^2B + 3AB^2 - B^3 \\ &= A^3 - 3AB(A - B) - B^3 \\ &= A^3 - 3xy - B^3 \end{aligned}$ so that $\ A^2 + B^2 = x^2 + 2y \$ and $\ A^3 - B^3 = x^3 + 3xy$ .

Then $\begin{aligned} x^2 + 2y &= 19610 \qquad &&\small \color{#3D99F6} \text{[Multiply by }3x] \\ x^3 + 3xy &= 117628 \qquad &&\small \color{#3D99F6} \text{[Multiply by -}2] \\ \\ \end{aligned}$ $\begin{aligned} 3x^3 + 6xy &= 58830x \\ \text{-}2x^3 - 6xy &= \text{-}235256 \\ \hline x^3 &= 58830x - 235256 \end{aligned}$

To solve this cubic equation, we factor $\ 235256 = 2^3 \cdot 7 \cdot 4201$ ; applying the integral zero theorem, we soon find that $\ 4\$ is a root.

$\begin{aligned} x^3 - 58830x + 235256 &= 0 \\ (x - 4)(x^2 + 4 x - 58814) &= 0 \end{aligned}$

and checking the discriminant of the above quadratic reveals no other integral roots. So $\ x = 4 \$ which gives us $\ y = 9797$ .

Now to reverse-engineer $\ A \$ and $\ B$ :

$\begin{aligned} (A + B)^2 &= A^2 + 2AB + B^2 \\ &= A^2 + B^2 + 2y \\ &= 19610 + 2(9797) = 39204 \\ \\ A + B &= 198 \end{aligned}$

Combining this with $\ A - B = x$ :

$\begin{aligned} A + B &= 198 \\ A - B &= 4 \\ \hline 2A &= 202 \end{aligned}$

so $\ A = 101 \$ and $\ B = 97 \$ . Finally our answer is $\ \sqrt{A^2 - (B + 2)^2} = \sqrt{101^2 - 99^2} = \sqrt{(101 - 99)(101 + 99)} = \sqrt{2 \cdot 200} = \sqrt{400}=\boxed{20}$

A^3 - B^3 = (A-B)(A^2 + AB + B^2), 117628 = 4. (19610 + AB) , 9797 = A.B. Find two numbers which difference is 4, and the product is 9797. Then we find A=101 and B=97. So the answer is 20