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

Write

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

. Equating the coefficient of

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

in

\nabla^{2} p

gives

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

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.

Solved by gpt-5.6-sol high.

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