Problem:
For polynomial P(x)=1−31x+61x2, define
Q(x)=P(x)P(x3)P(x5)P(x7)P(x9)=i=0∑50aixi
Then ∑i=050∣ai∣=nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
The polynomial P(−x)=1+31x+61x2 has nonnegative coefficients equal in absolute value to the coefficients of P(x). The coefficients of Q(−x)= P(−x)P(−x3)P(−x5)P(−x7)P(−x9) are nonnegative as well because Q(−x) is a product of five polynomials with nonnegative coefficients. Thus the sum of the absolute values of the coefficients of Q(x) is equal to the sum of the coefficients of Q(−x), which is Q(−1)=P(−1)5=(23)5=32243. The requested sum is 243+32=275.
The problems on this page are the property of the MAA's American Mathematics Competitions