Problem:
Let G be the set of polynomials of the form
P(z)=zn+cnβ1βznβ1+β―+c2βz2+c1βz+50
where c1β,c2β,β¦,cnβ1β 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 aβib is a root as well. In particular, P(z) can be factored into the product of pairwise different linear polynomials of the form (zβc) with cβZ and quadratic polynomials of the form (zβ(a+ib))(zβ(aβ ib))=z2β2az+(a2+b2) with a,bβZ and bξ =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 dβ{1,2,5,10,25,50}, let Bdβ be the set of basic polynomials with independent term equal to Β±d.
The equation a2+b2=1 has a pair of conjugate solutions in integers with bξ =0, namely (a,b)=(0,Β±1). Thus there is only 1 basic quadratic polynomial with independent term of magnitude 1:(zβi)(z+i)=z2+1. Similarly, the equation a2+b2=2 has 2 pairs of conjugate solutions with bξ =0,(a,b)=(1,Β±1) and (β1,Β±1). These give the following 2 basic polynomials with independent term Β±2:(zβ1βi)(zβ1+i)=z2β2z+2 and (z+1+i)(z+1βi)=z2+2zβ2. 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,Β±1),(β2,Β±1),(1,Β±2), and (β1,Β±2);(3,Β±1),(β3,Β±1), (1,Β±3), and (β1,Β±3);(3,Β±4),(β3,Β±4),(4,Β±3),(β4,Β±3), and (0,Β±5); and (7,Β±1),(β7,Β±1),(1,Β±7),(β1,Β±7),(5,Β±5), and (β5,Β±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 zβc where cβ{Β±1,Β±2,Β±5,Β±10,Β±25,Β±50}. Thus the linear basic polynomials contribute 2 to β£Bdββ£. It follows that β£B1ββ£=3,β£B2ββ£=4,β£B5ββ£=6,β£B10ββ£=6,β£B25ββ£=7, and β£B50ββ£=8.
Because P has independent term 50, there are either 8 choices for a polynomial in B50β, or 7β
4 choices for a product of two polynomials, one in B25β and the other in B2β, or 6β
6 choices for a product of two polynomials, one in B10β and the other in B5β, or 4β
(62β) choices for a product of three polynomials, one in B2β and the other two in B5β. Finally, each of the polynomials z+1 and z2+1 in B1β can appear or not in the product, but the presence of the polynomial zβ1 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+7β
4+6β
6+4β
(62β))=22(8+28+36+60)=4β
132=528β
The problems on this page are the property of the MAA's American Mathematics Competitions