### 63rd Putnam 2002

**Problem B6**

Let p be prime and q = p^{2}. Let D be the determinant:

x y z
x^{p} y^{p} z^{p}
x^{q} y^{q} z^{q}

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.)

**Solution**

Since p is prime, we have (X^{p} + Y^{p} + Z^{p}) = (X + Y + Z)^{p} mod p. Also (X + Y + Z)^{q} = (X^{p} + Y^{p} + Z^{p})^{p} = X^{q} + Y^{q} + Z^{q} mod p. Put X = ax, Y = bY, Z = cZ, for a, b, c integers. Then since a^{p} = a mod p, b^{p} = b mod p, c^{p} = c mod p, we have a x^{p} + b y^{p} + c z^{p} = (ax + by + cz)^{p} and a x^{q} + b y^{q} + c z^{q} = (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, k^{p}, k^{q}, 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 p^{2}+p+1 factors mod p:

x
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 y^{p}z^{q} in the expansion of Q comes from taking the first term of each factor, so the coefficient of x y^{p}z^{q} in Q is 1. That is also the coefficient in the expansion of D, so D = Q.

63rd Putnam 2002

© John Scholes

jscholes@kalva.demon.co.uk

11 Dec 2002