Common Roots

Define $f(x)=x^2+ax+b$ . Note that $x^2f(\frac{1}{x})=bx^2+ax+1$ . We want $f(x)=0$ and $x^2f(\frac{1}{x})=0$ for some number $x$ . Note that $x\ne 0$ because $0$ is not a root of $bx^2+ax+1$ , so we can safely divide both sides of the second equation by $x^2$ . We now have the following two equations: $\left\{\begin{array}{l}f(x)=0\\ f\left(\frac{1}{x}\right)=0 \end{array}\right.$

There are now two cases.

Case 1: the roots of * $f(x)$ are * $x,\frac{1}{x}$

Since the roots of $f(x)$ are $x,\frac{1}{x}$ , the product of the roots is $1$ , so $b=1$ by Vieta's. We can see that $f(x)=f(\frac{1}{x})=x^2+ax+1$ , so we can choose any $a$ from $a=-61\to 61$ .

*Case 2: * $x=\frac{1}{x}$

In this case, solving gives $x=1,-1$ .

*Case 2a: * $x=1$

Because $x=1$ , $f(x)=x^2+ax+b=(x-1)(x-b)=x^2+(-b-1)x+b$ so $a=-b-1$ .

*Case 2b: * $x=-1$

Because $x=-1$ , $f(x)=x^2+ax+b=(x+1)(x+b)=x^2+(b+1)x+b$ so $a=b+1$

Combining these two cases gives $b+1=\pm a$ .

Therefore in case 2, there are two solutions for all $b$ where $b=-61\to 60$ ( $b=61$ doesn't work because then $|a|=62$ which doesn't meet the problem requirements)

Putting it all together

In case 1, we have $123$ solutions. In case 2, we have $2\times 122=244$ solutions. However, we have counted the solutions $(a,b)=(2,1),(-2,1)$ twice: one time in case 1, and one time in case 2. In addition, we have counted the solution $(a,b)=(0,-1)$ twice: one time in case 2a, one time in case 2b. Therefore we subtract $3$ from our total number of solutions, so our final answer is $123+244-3=\boxed{364}$ solutions.

Let $\alpha$ be the common root

As it is a common root..it must satisfy both equations

$\Rightarrow \alpha^2+a\alpha+b=0............(1)$

and

$b\alpha^2+a\alpha+1=0...........(2)$

on subtracting eqn $1$ from eqn $2$

we get

$(\alpha^2-1)(b-1)=0$

$\Rightarrow$ Either $\alpha=1,-1$

or $b=1$

Now consider the following cases

CASE 1 ( $b=1)$

if $b=1$ ...for each integer value of $a$ we will have an ordered pair

$\Rightarrow$ Number of Ordered Pairs from CASE 1 =123

CASE 2 ( $\alpha=1)$

Substituting the value of $\alpha$ in either of the equations.. we get

$a+b=-1..............(3)$

Note here that $b$ can't be $61$

Also see that we have already counted the pair $(-2,1)$ in CASE 1

Hence we have 121 Ordered Pairs from CASE 2

Similarly... CASE 3 ( $\alpha=-1)$

$a-b=1...............(4)$

Here too... $b$ can't be $61$ ....and we have already counted the pair $(2,1)$

Hence we again have 121 Ordered Pairs from CASE 3

BUT WAIT!......what if equations $3$ and $4$ are satisfied simultaneously

This means we have counted $(0,-1)$ TWICE

Hence Total number of Ordered Pairs $=123+(2 \times 121)-1=\boxed{364}$