Problem:
Let
p(x,y)=a0​+a1​x+a2​y+a3​x2+a4​xy+a5​y2+a6​x3+a7​x2y+a8​xy2+a9​y3.
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)+a4​xy+a7​x2y+a8​xy2. 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)=21​r(r−1)(3s−2r−2), and =s−s3−23​(rs−r2s)=21​s(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