Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-27/1/b/solution
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 27 1 b Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-06
The remainders in the sieve distribution are defined byMultiply the pointwise inequality from part (a) by and sum. Finite rearrangement givesTo identify the main quadratic form, put . For squarefree integers , multiplicativity givesThe second identity is also valid if or is not squarefree: both sides then vanish. Therefore the Selberg sieve diagonalization givesThis proves . Notice that the hypotheses do not force to vanish on non-squarefree integers; those coefficients simply make no contribution to this main quadratic form because .
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