Problem:
Let
p(x,y)=a0+a1x+a2y+a3x2+a4xy+a5y2+a6x3+a7x2y+a8xy2+a9y3.
Suppose that
p(0,0)=p(1,0)=p(−1,0)=p(0,1)=p(0,−1)=p(1,1)=p(1,−1)=p(2,2)=0
There is a point (ca,cb) for which p(ca,cb)=0 for all such polynomials, where a,b, and c are positive integers, a and c are relatively prime, and c>1. Find a+b+c.
Solution:
Applying the first five conditions in order yields a0=0, and then a1+a3+ a6=0,a1−a3+a6=0,a2+a5+a9=0,a2−a5+a9=0, which imply that a3=a5=0 and a6=−a1,a9=−a2. Thus p(x,y)=a1(x−x3)+a2(y− y3)+a4xy+a7x2y+a8xy2. Similarly, the next two conditions imply that a8=0 and a7=−a4, so that p(x,y)=a1(x−x3)+a2(y−y3)+a4(xy−x2y). The last condition implies that −6a1−6a2−4a4=0, so that a4=−23(a1+ a2). Thus p(x,y)=a1(x−x3−23(xy−x2y))+a2(y−y3−23(xy−x2y)). If p(r,s)=0 for every such polynomial, then
00=r−r3−23(rs−r2s)=21r(r−1)(3s−2r−2), and =s−s3−23(rs−r2s)=21s(2−2s2−3r+3r2).
The solutions to the first equation are r=0 or 1, or s=32(r+1). Substituting r=0 or 1 into the second equation implies that s=0,1, or −1. If s=0 and 3s−2r−2=0, then r=−1. These solutions represent the first seven points. Finally, if s=0 and s=32(r+1), the second equation reduces to
0=2−2s2−3r+3r2=2−2⋅94(r2+2r+1)−3r+3r2=910−43r+19r2.
Because 10−43r+19r2=(2−r)(5−19r), it follows that r=2 or r=195. In the first case s=2, which produces the point (2,2). The second case yields s=1916. Thus a+b+c=5+16+19=40.
The problems on this page are the property of the MAA's American Mathematics Competitions