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2015! divides determinant of integer powers

Source: VTRMC 2015 P3

May 5, 2021
Matriceslinear algebra

Problem Statement

Let (ai)1≤i≤2015(a_i)_{1\le i\le2015} be a sequence consisting of 20152015 integers, and let (ki)1≤i≤2015(k_i)_{1\le i\le2015} be a sequence of 20152015 positive integers (positive integer excludes 00). Let A=(a1k1a1k2⋯a1k2015a2k1a2k2⋯a2k2015⋮⋮⋱⋮a2015k1a2015k2⋯a2015k2015).A=\begin{pmatrix}a_1^{k_1}&a_1^{k_2}&\cdots&a_1^{k_{2015}}\\a_2^{k_1}&a_2^{k_2}&\cdots&a_2^{k_{2015}}\\\vdots&\vdots&\ddots&\vdots\\a_{2015}^{k_1}&a_{2015}^{k_2}&\cdots&a_{2015}^{k_{2015}}\end{pmatrix}.Prove that 2015!2015! divides det⁡A\det A.
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