Maths Olympiad Prep

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Number theory Difficulty 5.4 AIME, harder Prove it Ukraine

Solve the system of equations for positive integer numbers x,y,zx, y, z:
{x3−6y2+27z=132,y3−9z2+3x=125,z3−3x2+12y=−68. \begin{cases} x^3 - 6y^2 + 27z = 132, \\ y^3 - 9z^2 + 3x = 125, \\ z^3 - 3x^2 + 12y = -68. \end{cases}

Solution

Let's add up all three equations:

(x3−3x2+3x)+(y3−6y2+12y)+(z3−9z2+27z)=189⇔(x3−3x2+3x−1)+(y3−6y2+3y−27)+(z3−9z2+27z−27)=153⇔(x−1)3+(y−2)3+(z−3)3=153. (x^3 - 3x^2 + 3x) + (y^3 - 6y^2 + 12y) + (z^3 - 9z^2 + 27z) = 189 \Leftrightarrow \\ (x^3 - 3x^2 + 3x - 1) + (y^3 - 6y^2 + 3y - 27) + (z^3 - 9z^2 + 27z - 27) = 153 \Leftrightarrow \\ (x-1)^3 + (y-2)^3 + (z-3)^3 = 153.

By a simple exhaustive search, we can see that number 153 has a unique representation in the form of the sum of three cubes: 153=53+33+13153 = 5^3 + 3^3 + 1^3. Therefore, in the last equation, values of (x−1)(x-1), (y−2)(y-2) and (z−3)(z-3) must be selected from 1; 3; 5. Again, simply by trying out the variants, we see that no solution is possible.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.