Do the root shuffle

From the 3rd condition we have either $f(b_{1})=b_{1}$ , $f(b_{2})=b_{2}$ or $f(b_{1})=b_{2}$ , $f(b_{2})=b_{1}$ . We'll study each case individually :

1) $f(b_{1})=b_{1}$ and $f(b_{2})=b_{2}$

$f(b_{1})=b_{1} \Leftrightarrow f(b_{1})-b_{1}=0$ , $f(b_{2})=b_{2} \Leftrightarrow f(b_{2})-b_{2}=0$ .

Now let $h(x)=f(x)-x$ then $h(b_{1})=0$ and $h(b_{2})=0$ , thus $h(x)$ has the same roots as $g(x)$ . It must follow by Vieta's Formulas that $h(x)=g(x)$ for every real number $x$ , or $f(x)-x=g(x) \forall x$ (Eq1) .

Taking the 4th condition into account we can only have $g(a_{1})=a_{2}$ and $g(a_{2})=a_{1}$ otherwise we'd arrive at $g(x)-x=f(x) \forall x$ (by the same argument as before) , which obviously contradicts $f(x)-x=g(x) \forall x$ .

Thus $g(a_{1})=a_{2} \Leftrightarrow g(a_{1})-a_{2}=0$ and $g(a_{2})=a_{1} \Leftrightarrow g(a_{2})-a_{1}=0$ .

Let $h'(x)=g(x)-(a_{1}+a_{2}-x)$ , $h'(a_{1})=0$ and $h'(a_{2})=0 \Rightarrow h'(x)=f(x)\Rightarrow$ $g(x)-(a_{1}+a_{2}-x)=f(x)\forall x$ (Eq2) .

From Eq1 and Eq2 we get (by substitution) that $a_{1}+a_{2}=0$ . $-(a_{1}+a_{2})$ is the coefficient of $x$ in the quadratic polynomial $f(x)$ ( Vieta's Formulas ) , and since $a_{1}+a_{2}=0 \Rightarrow f(x)$ has the form $x^2-m$ , where m is nonnegative (the roots of $f(x)$ must be real) . $g(x)=x^2-x-m$ (from Eq1 ) $\Rightarrow f(1)=1-m$ and $g(1)=-m \Rightarrow f(1)g(1)=m^2-m$ .

Also $f(1)g(1)=132 \Rightarrow m^2-m-132=0$ . The only nonnegative root is $m=12$ , we see that this solution meets the conditions $\Rightarrow f(x)=x^2-12 , g(x)=x^2-x-12 \Rightarrow f(2)g(2)=80$

2) $f(b_{1})=b_{2}$ and $f(b_{2})=b_{1}$

Using the previous strategy : $f(x)-(b_{1}+b_{2}-x)=g(x)$ and $g(x)-x=f(x) \Rightarrow$ same thing as before only $f(x)$ and $g(x)$ are switched , not surprisingly this doesn't affect their product at all.

$\Rightarrow f(2)g(2)=80$

That literally take me whole day to do this!

No calculus involved!

I'll call the word same $f$ if $f(x_{1}) = x_{1}$ and $f(x_{2}) = x_{2}$ , and the word different $f$ if $f(x_{1}) = x_{2}$ and $f(x_{2}) = x_{1}$ .

We'll prove that either same $f$ different $g$ , or different $f$ same $g$ satisfy the conditions.

We can construct

$f(x) = x^{2}-(a_{1}+a_{2})+a_{1}a_{2}$ and

$g(x) = x^{2}-(b_{1}+b_{2})+b_{1}b_{2}$ .

Substitute all the cases possible, and we'll see that

If same $f$ , we get $(b_{1}+b_{2}) - (a_{1}+a_{2}) = 1$ . (1.1)

If different $f$ , we get $(b_{1}+b_{2}) - (a_{1}+a_{2}) = -1$ . (1.2)

If same $g$ , we get $(a_{1}+a_{2}) - (b_{1}+b_{2}) = 1$ . (2.1)

If different $g$ , we get $(a_{1}+a_{2}) - (b_{1}+b_{2}) = -1$ . (2.2)

We see that only pair ((1.1),(2.2)) and ((1.2),(2.1)) works.

WLOG that $f(b_{1}) = b_{1}, f(b_{2}) = b_{2}, g(a_{1}) = a_{2}, g(a_{2}) = a_{1}$ .

We know that $f(x) = x$ has roots $b_{1},b_{2}$ and $f(x)$ is monic, we get

$f(x) = (x-b_{1})(x-b_{2})+x$ (3)

Also, $g(x) = 0$ has roots $b_{1},b_{2}$ , and $g(x)$ is monic, we get

$g(x) = (x-b_{1})(x-b_{2})$ . Substitute in (3) we get

$f(x) = x + g(x)$ (#)

Plug in $x = 1$ we get

$f(1) = g(1)+1$ . (4)

From condition $5$ and (4), we get $(f(1),g(1)) = (12,11)$ (6.1) or $(-11,-12)$ (6.2)

From (#), plug in $x = a_{1}$ we get

$f(a_{1}) = a_{1}+g(a_{1})$

$a_{1}+a_{2} = 0$ .

Which makes $f(x) = x^{2}+a_{1}a_{2}$ .

Substitute from (6) we get $a_{1}a_{2} = 11$ (6.1) or $a_{1}a_{2} = -12$ (6.2)

Obviously (6.1) is impossible because $a_{1}+a_{2} = 0$ .

Therefore, $f(x) = x^{2}-12$ .

Which makes $g(x) = x^{2}-x-12$ from (#).

Substitute and we get $f(2)g(2) = (-8)(-10) = \boxed{80}$ ~~~

Oh my dog this is almost 1 AM right now, too lazy to sleep. Lazy bum.