Let $q$ be a number. Let us consider the $q^2-1$-th line of the Pascal triangle (i.e. numbers ${{q^2-1} \choose i}$, $i=0,1,...q^2-1$. We have $q^2$ numbers.

Let us form naively a $q \times q$ matrix from them. Example for $q=2$: numbers 1,3,3,1; matrix

$\left(\matrix 1 & 3 \\ 3 & 1 \endmatrix\right)$

And let us take its determinant: -8 if $q=2$.

How to prove that if $q$ is prime then $q$ enters in this determinant with $(q+4)(q-1)/2 $-th exponent?

Example: $q=2$, $(q+4)(q-1)/2 =3$. Really, 8=2^3.

Case $q=3$: binomial coefficients are 1,8,28,56,70,56,28,8,1

The matrix:

$\left(\matrix 1 & 8 & 28 \\ 56 & 70 & 56 \\ 28 & 8 & 1 \endmatrix\right)$

det = - 2 * 3^7 * 7. Really, $(q+4)(q-1)/2 =7$.

Let $q$ be a number. Let us consider the $q^2-1$-th line of the Pascal triangle (i.e. numbers ${{q^2-1} \choose i}$, $i=0,1,...q^2-1$). We have $q^2$ numbers.

Let us form naively a $q \times q$ matrix from them. Example for $q=2$:

\[\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\]

And let us take its determinant: $-8$ if $q=2$.

How to prove that if $q$ is prime then $q$ enters in this determinant with $(q+4)(q-1)/2 $-th exponent?

Example: $q=2$, $(q+4)(q-1)/2 =3$. Really, $8=2^3$.

Case $q=3$: the binomial coefficients are $1,8,28,56,70,56,28,8,1$.

The matrix:

\[\begin{pmatrix} 1 & 8 & 28 \\ 56 & 70 & 56 \\ 28 & 8 & 1 \end{pmatrix}\]

The determinant is $2 \cdot 3^7 \cdot 7$. Really, $(q+4)(q-1)/2 =7$.