Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-166/3/d/solution

The determinant in the hint is independent of and equals the Wronskian
It is nonzero because are linearly independent. Also , and coefficient convolution gives
for a constant depending only on .
Fix , and suppose has multiplicity at . In
the two terms vanish to orders at least and , so has multiplicity at least at . The primitive polynomial therefore divides in by Gauss lemma for polynomials. Comparing leading coefficients gives
If , this implies and hence . Choosing larger than this bound proves that , uniformly in .
Solved by gpt-5.6-sol high.

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