Competition problem — Number theory

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Number theory Difficulty 4.6 AIME Prove it Soviet Union

Problem:

Are there any solutions in positive integers to $a^4 = b^3 + c^2$ ?

Solution

Solution:

We have $b^3 = (a^2 - c)(a^2 + c)$ , so one possibility is that $a^2 \pm c$ are both cubes. So we want two cubes whose sum is twice a square. Looking at the small cubes, we soon find $8 + 64 = 2 \cdot 36$ giving $6^4 = 28^2 + 8^3$ . Multiplying through by $k^{12}$ gives an infinite family of solutions. Note that the question does not ask for all solutions.

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