Competition problem — Number theory

Number theory Difficulty 5.2 AIME, harder Prove it Brazil

Show that the number of positive integer solutions to
$x_1 + 2^3 x_2 + 3^3 x_3 + \cdots + 10^3 x_{10} = 3025 \quad (*)$
equals the number of non-negative integer solutions to the equation
$y_1 + 2^3 y_2 + 3^3 y_3 + \cdots + 10^3 y_{10} = 0$
Hence show that (*) has a unique solution in positive integers and find it.

Solution

We have $1^3 + 2^3 + \cdots + 10^3 = 3025$ . Now $x_i$ is a positive integer solution to $(*)$ iff $y_i = x_i - 1$ are all non-negative and satisfy $(y_1+1)+2^3(y_2+1)+3^3(y_3+1)+\cdots+10^3(y_{10}+1) = 3025$ and hence $y_1+2^3y_2+3^3y_3+\cdots+10^3y_{10} = 0$ . But that clearly has the unique solution $y_i = 0$ , so the unique solution to $(*)$ is $x_i = 1$ .

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.