Do the root shuffle

Since $f(x)$ is a monic quadratic polynomial whose roots are $a_1$ and $a_2$ , $f(x) = (x - a_1)(x - a_2).$ Similarly, $g(x) = (x - b_1)(x - b_2)$ .

Since $\{f(b_1), f(b_2)\} = \{b_1,b_2\}$ , either $f(b_1) = b_1$ and $f(b_2) = b_2$ , or $f(b_1) = b_2$ and $f(b_2) = b_1$ .

Case 1 : $f(b_1) = b_1$ and $f(b_2) = b_2$ .

In this case, we see that $f(x) - x$ is a monic quadratic whose roots are $b_1$ and $b_2$ , so it coincides with $g(x)$ . Then $g(a_1) = f(a_1) - a_1 = -a_1$ , and $g(a_2) = f(a_2) - a_2 = -a_2$ . But $\{g(a_1), g(a_2)\} = \{a_1,a_2\}$ , so we must have $a_2 = -a_1$ . Then $f(x) = (x - a_1)(x + a_1) = x^2 - a_1^2,$ and $g(x) = f(x) - x = x^2 - x - a_1^2.$

From the equation $f(1) g(1) = 132$ , we get $(1 - a_1^2)(-a_1^2) = 132$ , which gives us $(a_1^2 - 12)(a_1^2 + 11) = 0.$ Since $a_1$ is real, we must have $a_1^2 = 12$ . Then $f(2) g(2) = (4 - a_1^2)(2 - a_1^2) = (4 - 12)(2 - 12) = 80.$

Case 2 : $f(b_1) = b_2$ and $f(b_2) = b_1$ .

In this case, we see that $f(x) + x - b_1 - b_2$ is a monic quadratic whose roots are $b_1$ and $b_2$ , so it coincides with $g(x)$ . Then $g(a_1) = f(a_1) + a_1 - b_1 - b_2 = a_1 - b_1 - b_2$ , and $g(a_2) = f(a_2) + a_2 - b_1 - b_2 = a_2 - b_1 - b_2$ . But $\{g(a_1), g(a_2)\} = \{a_1,a_2\}$ , so $g(a_1) + g(a_2) = a_1 + a_2$ , which gives us $a_1 + a_2 - 2b_1 - 2b_2 = a_1 + a_2,$ or $b_1 + b_2 = 0$ . Hence, $f(x) + x = g(x)$ for all $x$ , and we can then proceed as in Case 1, with the roles of $f$ and $g$ reversed.

Therefore, in either case, $f(2) g(2) = 80$ .

Since $f$ and $g$ are monic quadratic, we have $f(x) = (x-a_1)(x-a_2)$ and $g(x)=(x-b_1)(x-b_2)$ from [1] and [2]. From [3] we know that $f(x)-g(x)$ is a linear polynomial which maps $\{b_1,b_2\}$ to $\{b_1,b_2\}$ : either $f(x)-g(x) = x$ or $f(x)-g(x) = b_1+b_2-x$ . Likewise [4] tells us $g(x)-f(x) = x$ or $g(x)-f(x) = a_1 + a_2 - x$ . Comparing the last two facts and switching $f,g$ and $a_i,b_i$ if necessary, we have WLOG that $f(x)-g(x) = x$ and $a_1+a_2 = 0$ . Thus $f(x) = x^2-a_1^2$ and $g(x) = x^2-x-a_1^2$ . By [5], $(1-a_1^2)(-a_1^2) = a_1^4 - a_1^2 = 132$ , so $a_1^2-12)(a_1^2+11) = 0$ . Since $a_1$ is real, $a_1^2=12$ , giving $f(2)g(2) = (-8)(-10) = 80$ .