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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-3/10a/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 3 / 10A / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

Write

p = \sum_{j = 0}^{n} a_{j} x^{n - j} y^{j}

. Equating the coefficient of

x^{n - j - 2} y^{j}

\nabla^{2} p

gives

(n - j) (n - j - 1) a_{j} + (j + 2) (j + 1) a_{j + 2} = 0.

(1)

Thus all even coefficients are determined by $a_{0}$ and all odd coefficients by $a_{1}$ , with no further constraints. These two choices give two independent harmonic homogeneous polynomials.

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