Let be a prime number. Prove that the determinant of the matrix is congruent modulo to a product of polynomials of the form , where , , and are integers. (We say two integer polynomials are congruent modulo if corresponding coefficients are congruent modulo .)
Problem 1516
Official solution
To prove that the determinant of the matrix
is congruent modulo to a product of polynomials of the form , where , , and are integers, we will use properties derived from Fermat's Little Theorem.
1. Fermat's Little Theorem:
Fermat's Little Theorem states that for any integer and a prime ,
This implies that for any integer not divisible by ,
2. Matrix Setup:
Consider the matrix
We need to show that the determinant of can be factored into linear factors modulo .
3. Row Operations:
Perform the following row operations on :
- Subtract times the first row from the second row.
- Subtract times the second row from the third row.
After these operations, the matrix becomes:
4. Simplification Using Fermat's Little Theorem:
Let and . Using Fermat's Little Theorem, we know:
5. Determinant Calculation:
The matrix now looks like:
The determinant of this matrix is:
6. Factorization:
We need to factor . Notice that:
By Fermat's Little Theorem, and . Thus:
7. Final Factorization:
Therefore, the determinant can be written as:
Each of these terms , , , and can be expressed as linear polynomials in , , and .
8. Conclusion:
The determinant of the matrix is congruent modulo to a product of polynomials of the form , where , , and are integers.