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Brazilian Mathematical Olympiad

Number theory Difficulty 5.2 AIME, harder Prove it Brazil

Show that the number of positive integer solutions to
x1+23x2+33x3+⋯+103x10=3025(∗) 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
y1+23y2+33y3+⋯+103y10=0 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 13+23+⋯+103=30251^3 + 2^3 + \cdots + 10^3 = 3025. Now xix_i is a positive integer solution to (∗)(*) iff yi=xi−1y_i = x_i - 1 are all non-negative and satisfy (y1+1)+23(y2+1)+33(y3+1)+⋯+103(y10+1)=3025(y_1+1)+2^3(y_2+1)+3^3(y_3+1)+\cdots+10^3(y_{10}+1) = 3025 and hence y1+23y2+33y3+⋯+103y10=0y_1+2^3y_2+3^3y_3+\cdots+10^3y_{10} = 0. But that clearly has the unique solution yi=0y_i = 0, so the unique solution to (∗)(*) is xi=1x_i = 1.

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