Problem:
Let G be the set of polynomials of the form
P(z)=zn+cn−1​zn−1+⋯+c2​z2+c1​z+50
where c1​,c2​,…,cn−1​ are integers and P(z) has n distinct roots of the form a+ib with a and b integers. How many polynomials are in G ?
Answer Choices:
A. 288
B. 528
C. 576
D. 992
E. 1056
Solution:
Let P(z) be a polynomial in G. Because the coefficients of P(z) are real, it follows that the nonreal roots of P(z) must be paired by conjugates; that is, if a+ib is a root, then a−ib is a root as well. In particular, P(z) can be factored into the product of pairwise different linear polynomials of the form (z−c) with c∈Z and quadratic polynomials of the form (z−(a+ib))(z−(a− ib))=z2−2az+(a2+b2) with a,b∈Z and bî€ =0. Moreover, the product of the independent terms of these polynomials must be equal to 50 , so each of a2+b2 or c must be a factor of 50. Call these linear or quadratic polynomials basic and for every d∈{1,2,5,10,25,50}, let Bd​ be the set of basic polynomials with independent term equal to ±d.
The equation a2+b2=1 has a pair of conjugate solutions in integers with bî€ =0, namely (a,b)=(0,±1). Thus there is only 1 basic quadratic polynomial with independent term of magnitude 1:(z−i)(z+i)=z2+1. Similarly, the equation a2+b2=2 has 2 pairs of conjugate solutions with bî€ =0,(a,b)=(1,±1) and (−1,±1). These give the following 2 basic polynomials with independent term ±2:(z−1−i)(z−1+i)=z2−2z+2 and (z+1+i)(z+1−i)=z2+2z−2. In the same way the equations a2+b2=5,a2+b2=10,a2+b2=25, and a2+b2=50 have 4,4,5, and 6 respective pairs of conjugate solutions (a,b). These are (2,±1),(−2,±1),(1,±2), and (−1,±2);(3,±1),(−3,±1), (1,±3), and (−1,±3);(3,±4),(−3,±4),(4,±3),(−4,±3), and (0,±5); and (7,±1),(−7,±1),(1,±7),(−1,±7),(5,±5), and (−5,±5). These generate all possible basic quadratic polynomials with nonreal roots and independent term that divides 50 . The basic linear polynomials with real roots are z−c where c∈{±1,±2,±5,±10,±25,±50}. Thus the linear basic polynomials contribute 2 to ∣Bd​∣. It follows that ∣B1​∣=3,∣B2​∣=4,∣B5​∣=6,∣B10​∣=6,∣B25​∣=7, and ∣B50​∣=8.
Because P has independent term 50, there are either 8 choices for a polynomial in B50​, or 7⋅4 choices for a product of two polynomials, one in B25​ and the other in B2​, or 6⋅6 choices for a product of two polynomials, one in B10​ and the other in B5​, or 4⋅(62​) choices for a product of three polynomials, one in B2​ and the other two in B5​. Finally, each of the polynomials z+1 and z2+1 in B1​ can appear or not in the product, but the presence of the polynomial z−1 is determined by the rest: if the product of the remaining independent terms is -50 , then it has to be present, and if the product is 50 , then it must not be in the product. Thus, the grand total is
22(8+7⋅4+6⋅6+4⋅(62​))=22(8+28+36+60)=4⋅132=528​
The problems on this page are the property of the MAA's American Mathematics Competitions