Putnam 2002, Problem B-6

Let $p$ be a prime number. Prove that the determinant of the matrix
$$\begin{pmatrix}
x & y & z \\
x^p & y^p & z^p \\
x^{p^2} & y^{p^2} & z^{p^2}
\end{pmatrix}$$
is congruent modulo $p$ to a product of polynomials of the form $ax + by + cz$, where $a,b,c$ are integers. (We say two integer polynomials are congruent modulo $p$ if corresponding coefficients are congruent modulo $p$.)


Solution:
We prove that the determinant is congruent modulo \( p \) to  
\[ x \prod_{i=0}^{p-1} (y + ix) \prod_{i,j=0}^{p-1} (z + ix + jy). \tag{3} \]  

We first check that  
\[ \prod_{i=0}^{p-1} (y + ix) \equiv y^p - x^{p-1}y \pmod{p}. \tag{4} \]  

Since both sides are homogeneous as polynomials in \( x \) and \( y \), it suffices to check (4) for \( x = 1 \), as a congruence between polynomials. Now note that the right side has \( 0, 1, \ldots, p - 1 \) as roots modulo \( p \), as does the left side. Moreover, both sides have the same leading coefficient. Since they both have degree only \( p \), they must then coincide.  

We thus have  
\[ x \prod_{i=0}^{p-1} (y + ix) \prod_{i,j=0}^{p-1} (z + ix + jy) \]  
\[ \equiv x(y^p - x^{p-1}y) \prod_{j=0}^{p-1} ((z + jy)^p - x^{p-1}(z + jy)) \]  
\[ \equiv (xy^p - x^p y) \prod_{j=0}^{p-1} (z^p - x^{p-1}z + jy^p - jx^{p-1}y) \]  
\[ \equiv (xy^p - x^p y)((z^p - x^{p-1}z)^p - (y^p - x^{p-1}y)^{p-1}(z^p - x^{p-1}z)) \]  
\[ \equiv (xy^p - x^p y)(z^{p^2} - x^{p^2 - p}z^p) \]  
\[ - x(y^p - x^{p-1}y)(z^p - x^{p-1}z)^p \]  
\[ \equiv xy^p z^{p^2} - x^p y z^{p^2} - x^{p^2 - p + 1} y^p z^p + x^{p^2} y z^p \]  
\[ - xy^p z^p + x^p y z^p + x^{p^2 - p + 1} y^p z - x^{p^2} y z \]  
\[ \equiv xy^p z^{p^2} + y z^p x^{p^2} + z x^p y^{p^2} \]  
\[ - x z^p y^{p^2} - y x^p z^{p^2} - z y^p x^{p^2}, \]  
which is precisely the desired determinant.  

Note: a simpler conceptual proof is as follows. (Everything in this paragraph will be modulo \( p \).) Note that for any integers \( a, b, c \), the column vector \( [ax + by + cz, (ax + by + cz)^p, (ax + by + cz)^{p^2}] \) is a linear combination of the columns of the given matrix. Thus \( ax + by + cz \) divides the determinant. In particular, all of the factors of (3) divide the determinant; since both (3) and the determinant have degree \( p^2 + p + 1 \), they agree up to a scalar multiple. Moreover, they have the same coefficient of \( z^{p^2} y^p x \) (since this term only appears in the expansion of (3) when you choose the first term in each factor). Thus the determinant is congruent to (3), as desired.  

Either argument can be used to generalize to a corresponding \( n \times n \) determinant, called a Moore determinant; we leave the precise formulation to the reader. Note the similarity with the classical Vandermonde determinant: if \( A \) is the \( n \times n \) matrix with \( A_{ij} = x_i^j \) for \( i, j = 0, \ldots, n - 1 \), then  
\[ \det(A) = \prod_{1 \leq i < j \leq n} (x_j - x_i). \]