Maths Olympiad Prep

Track / Stage 7 / 116 of 300 #1516 of 1964

Problem 1516

National olympiad second round; IMO P1/P4
Algebra Difficulty 7.3 Prove it

Let pp be a prime number. Prove that the determinant of the matrix [xyzxpypzpxp2yp2zp2] \begin{bmatrix}x & y & z\\ x^p & y^p & z^p \\ x^{p^2} & y^{p^2} & z^{p^2} \end{bmatrix} is congruent modulo pp to a product of polynomials of the form ax+by+czax+by+cz, where aa, bb, and cc are integers. (We say two integer polynomials are congruent modulo pp if corresponding coefficients are congruent modulo pp.)

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

To prove that the determinant of the matrix
[xyzxpypzpxp2yp2zp2] \begin{bmatrix} x & y & z \\ x^p & y^p & z^p \\ x^{p^2} & y^{p^2} & z^{p^2} \end{bmatrix}
is congruent modulo p p to a product of polynomials of the form ax+by+cz ax + by + cz , where a a , b b , and c c 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 a a and a prime p p ,
apa(modp). a^p \equiv a \pmod{p}.
This implies that for any integer a a not divisible by p p ,
ap11(modp). a^{p-1} \equiv 1 \pmod{p}.

2. Matrix Setup:
Consider the matrix
A=[xyzxpypzpxp2yp2zp2]. A = \begin{bmatrix} x & y & z \\ x^p & y^p & z^p \\ x^{p^2} & y^{p^2} & z^{p^2} \end{bmatrix}.
We need to show that the determinant of A A can be factored into linear factors modulo p p .

3. Row Operations:
Perform the following row operations on A A :
- Subtract xp1 x^{p-1} times the first row from the second row.
- Subtract xp2p x^{p^2-p} times the second row from the third row.

After these operations, the matrix becomes:
[xyz0y(yp1xp1)z(zp1xp1)0yp(yp2pxp2p)zp(zp2pxp2p)]. \begin{bmatrix} x & y & z \\ 0 & y(y^{p-1} - x^{p-1}) & z(z^{p-1} - x^{p-1}) \\ 0 & y^p(y^{p^2-p} - x^{p^2-p}) & z^p(z^{p^2-p} - x^{p^2-p}) \end{bmatrix}.

4. Simplification Using Fermat's Little Theorem:
Let u=y(yp1xp1) u = y(y^{p-1} - x^{p-1}) and v=z(zp1xp1) v = z(z^{p-1} - x^{p-1}) . Using Fermat's Little Theorem, we know:
upyp(yp1xp1)(modp), u^p \equiv y^p(y^{p-1} - x^{p-1}) \pmod{p},
vpzp(zp1xp1)(modp). v^p \equiv z^p(z^{p-1} - x^{p-1}) \pmod{p}.

5. Determinant Calculation:
The matrix now looks like:
[xyz0uv0upvp]. \begin{bmatrix} x & y & z \\ 0 & u & v \\ 0 & u^p & v^p \end{bmatrix}.
The determinant of this matrix is:
det(A)=x(uvpvup). \det(A) = x \cdot (uv^p - vu^p).

6. Factorization:
We need to factor uvpvup uv^p - vu^p . Notice that:
uvpvup=uv(vp1up1). uv^p - vu^p = uv(v^{p-1} - u^{p-1}).
By Fermat's Little Theorem, vp1v(modp) v^{p-1} \equiv v \pmod{p} and up1u(modp) u^{p-1} \equiv u \pmod{p} . Thus:
vp1up1vu(modp). v^{p-1} - u^{p-1} \equiv v - u \pmod{p}.

7. Final Factorization:
Therefore, the determinant can be written as:
det(A)=xuv(vu). \det(A) = x \cdot uv \cdot (v - u).
Each of these terms x x , u u , v v , and vu v - u can be expressed as linear polynomials in x x , y y , and z z .

8. Conclusion:
The determinant of the matrix is congruent modulo p p to a product of polynomials of the form ax+by+cz ax + by + cz , where a a , b b , and c c are integers.

\blacksquare

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.