63rd Putnam 2002

Problem B6

Let p be prime and q = p2. Let D be the determinant:

   x     y     z

   xp    yp    zp

   xq    yq    zq

Show that we can find a set of linear polynomials ax + by + cz (with a, b, c integers) whose product Q equals D mod p. (In other words, after expansion the corresponding coefficients of Q and D are equal mod p.)



Since p is prime, we have (Xp + Yp + Zp) = (X + Y + Z)p mod p. Also (X + Y + Z)q = (Xp + Yp + Zp)p = Xq + Yq + Zq mod p. Put X = ax, Y = bY, Z = cZ, for a, b, c integers. Then since ap = a mod p, bp = b mod p, cp = c mod p, we have a xp + b yp + c zp = (ax + by + cz)p and a xq + b yq + c zq = (ax + by + cz)q.

So if we take as the new first column of the determinant a x first col + b x second col + c x third col, then mod p the new first column is k, kp, kq, where k = ax + by + cz. So the determinant has a factor ax + by + cz. This is true for all a, b, c. Now consider the following p2+p+1 factors mod p:


y, y + x, y + 2x, y + 3x, ... , y + (p-1)x

z, z + x, z + 2x, z + 3x, ... , z + (p-1)x

z+y, z+y+x, z+y+2x, ... , z+y+(p-1)x

z+2y, z+2y+x, z+2y+2x, ... , z+2y+(p-1)x


z+(p-1)y, z+(p-1)y+x, z+(p-1)y+2x, ... , z+(p-1)y+(p-1)x

Evidently none is of these is a multiple of another, so their product Q must divide D. But D has total degree 1+p+q, which is the same as the total degree of Q. So D must be a multiple of Q. But the only term x ypzq in the expansion of Q comes from taking the first term of each factor, so the coefficient of x ypzq in Q is 1. That is also the coefficient in the expansion of D, so D = Q.



63rd Putnam 2002

© John Scholes
11 Dec 2002