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Diophantine Equations with the same number of solutions

Source: 1st Brazilian MO, 1979, Problem 4 (from Kalva backup)

December 23, 2017
Brazilian Math Olympiad 1979number theoryDiophantine equationBrazilian Math Olympiad

Problem Statement

Show that the number of positive integer solutions to x1+23x2+33x3+…+103x10=3025x_1 + 2^3x_2 + 3^3x_3 + \ldots + 10^3x_{10} = 3025 (*) equals the number of non-negative integer solutions to the equation y1+23y2+33y3+…+103y10=0y_1 + 2^3y_2 + 3^3y_3 + \ldots + 10^3y_{10} = 0. Hence show that (*) has a unique solution in positive integers and find it.
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