Problem:
Find the number of second-degree polynomials f(x) with integer coefficients and integer zeros for which f(0)=2010.
Solution:
If f(x)=c(x−r1​)(x−r2​), the product cr1​r2​ must be 2010=2⋅3⋅5⋅67. For 1≤k≤4, if c has 4−k prime factors, there are (k4​) choices for the k prime factors of 2010 that divide r1​r2​. Of these, there are 2k choices for the factors dividing r1​; the others must divide r2​. The roots of each polynomial obtained in this way are distinct and each possible pair of roots is counted exactly twice. Therefore there are (k4​)⋅2k−1 choices for the two roots, up to sign. Furthermore, an even number of {c,r1​,r2​} must be negative. This gives (03​)+(23​)=4 possible assignments of signs for each of the ∑k=14​(k4​)⋅2k−1=4⋅1+6⋅2+4⋅4+1⋅8=40 choices of {∣c∣,∣r1​∣,∣r2​∣}. Finally, if ∣c∣=2010 then ∣r1​∣=∣r2​∣=1. There are only three sign assignments that give rise to distinct polynomials in this case, because both cases in which r1​=−r2​ give rise to the same polynomial. Combining this with the prior discussion, there are 4⋅40+3=163​ such polynomials in all.
The problems on this page are the property of the MAA's American Mathematics Competitions