Problem:
Find the number of second-degree polynomials with integer coefficients and integer zeros for which .
Solution:
If , the product must be . For , if has prime factors, there are choices for the prime factors of that divide . Of these, there are choices for the factors dividing ; the others must divide . The roots of each polynomial obtained in this way are distinct and each possible pair of roots is counted exactly twice. Therefore there are choices for the two roots, up to sign. Furthermore, an even number of must be negative. This gives possible assignments of signs for each of the choices of . Finally, if then . There are only three sign assignments that give rise to distinct polynomials in this case, because both cases in which give rise to the same polynomial. Combining this with the prior discussion, there are such polynomials in all.
The problems on this page are the property of the MAA's American Mathematics Competitions