Problem:
Let a>0, and let P(x) be a polynomial with integer coefficients such that
P(1)=P(3)=P(5)=P(7)=a, and P(2)=P(4)=P(6)=P(8)=−a.
What is the smallest possible value of a ?
Answer Choices:
A. 105
B. 315
C. 945
D. 7!
E. 8!
Solution:
Because 1,3,5, and 7 are roots of the polynomial P(x)−a, it follows that
P(x)−a=(x−1)(x−3)(x−5)(x−7)Q(x)
where Q(x) is a polynomial with integer coefficients. The previous identity must hold for x=2,4,6, and 8 , thus
−2a=−15Q(2)=9Q(4)=−15Q(6)=105Q(8)
Therefore 315=lcm(15,9,105) divides a, that is a is an integer multiple of 315. Let a=315A. Because Q(2)=Q(6)=42A, it follows that Q(x)− 42A=(x−2)(x−6)R(x) where R(x) is a polynomial with integer coefficients. Because Q(4)=−70A and Q(8)=−6A it follows that −112A=−4R(4) and −48A=12R(8), that is R(4)=28A and R(8)=−4A. Thus R(x)=28A+ (x−4)(−6A+(x−8)T(x)) where T(x) is a polynomial with integer coefficients. Moreover, for any polynomial T(x) and any integer A, the polynomial P(x) constructed this way satisfies the required conditions. The required minimum is obtained when A=1 and so a=315.
The problems on this page are the property of the MAA's American Mathematics Competitions