Problem: Factor $\begin{vmatrix} x & y & z \\ x^p & y^p & z^p \\ x^{p^2} & y^{p^2} & z^{p^2} \end{vmatrix}$ over $\mathbb{Z}_p$ as a product of polynomials of the form $ax+by+cz$ with $a,b,c$ constants.
My solution: The determinant can be treated as a polynomial of degree $p^2$ in $x$ with coefficients in $\mathbb{Z}_p[y,z].$ If $x=ay+bz,$ then $x^p = (ay+bz)^p \equiv a^p z^p + b^p z^p \equiv az^p + bz^p \mod p$ and $x^{p^2} = (x^p)^p \equiv (az^p+bz^p)^p \equiv az^{p^2} + by^{p^2} \mod p$ similarly, so $$\begin{vmatrix} x & y & z \\ x^p & y^p & z^p \\ x^{p^2} & y^{p^2} & z^{p^2} \end{vmatrix} \equiv \begin{vmatrix} x & y & z \\ ay^p+bz^p & y^p & z^p \\ ax^{p^2}+by^{p^2} & y^{p^2} & z^{p^2} \end{vmatrix} = 0 \mod p.$$ Thus, the polynomial is divisible by $x-ay-bz$ for $0 \le a, b < p.$ This leads to $p^2$ factors, so we have accounted for all the factors of the polynomial and it remains to factor the leading coefficient, which is $\begin{vmatrix} y & z \\ y^p & z^p\end{vmatrix}.$ Similarly, we can treat this as a degree $p$ polynomial in $y$ with coefficients in $\mathbb{Z}_p[z]$ and find that the determinant is $0 \mod p$ when $y=az,$ so $y-az$ is a factor for $0 \le a < p.$ This accounts for all $p$ factors, and the leading coefficient is $-z,$ finally leading to the factorization $$-z\prod\limits_{k=0}^{p-1} (y-kz) \prod\limits_{0 \le i,j < p} (x-iy-jz).$$
This problem appeared on the Putnam, which is known for strict grading, so I wish to be careful that no major details are left out. During the competition, $17$ people solved (8,9 or 10 points) this problem and $34$ people attempted (0,1, or 2 points) it unsuccessfully. This leads me to believe that lots of people who thought they solved the problem were mistaken. If this solution is valid, how did 2/3 competitors who wrote down something mess up their solution?
I get the determinant to be the negative of yours. If you look at the coefficient of $x^{p^2}y^pz$ in your solution, it is $1$ but in the determinant it is $-1$.
Anyway, the determinant is a homogeneous polynomial of degree $p^2+p+1$. I claim that every linear form $ax+by+cz$ with $(a,b,c)$ a nonzero vector over $\Bbb Z_p$ is a factor of the determinant. You can prove it your way, or observe that if $ax+by+cz=0$ in a field of characteristic $p$, then $ax^p+by^p+cz^p=0$ and $ax^{p^2}+by^{p^2}+cz^{p^2}=0$ so that $(a.b.c)^t$ is in the nullspace of the matrix.