Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 27 / 1 / c / Solution

For coefficients $b_n$ supported on an interval of length $H$, write $B=\sum b_n$ and $B_p(a)={\sum }_{n\equiv a(\mathrm{mod}p)}{b}_n$. The variance form of the large sieve statesThe constant is absolute. One may take the explicit right side $(Q^2+2\pi H)\sum |b_n{|}^2$, by orthogonality of roots of unity and the exponential-sum large sieve proved in Question 2.

Take $H=\lfloor N^2\rfloor$, $Q=N$, and $b_n$ the indicator function of the $N^{\epsilon }$-smooth numbers up to $N^2$. Applying the given smooth-number density with parameter $\epsilon /2$ giveswith a harmless adjustment of the constant for integer endpoints. If an odd prime $p\le N$ has least quadratic nonresidue $n(p)>N^{\epsilon }$, then $p>N^{\epsilon }$: a quadratic nonresidue always occurs among $1,\dots ,p-1$. Every prime factor of every selected smooth number is thus a nonzero quadratic residue modulo $p$. By the multiplicativity of the Legendre symbol, every selected number is a nonzero quadratic residue modulo $p$.

There are $(p-1)/2$ nonzero quadratic nonresidue classes, and $B_p(a)=0$ on all of them. Their contribution to the variance is at leastIf $R$ denotes the number of exceptional primes, the variance form of the large sieve, with $\sum |b_n{|}^2=B$, yields $R{B}^2\ll N^{2}B$. ConsequentlyThis is the bounded exceptional primes for least quadratic nonresidues argument. Using an interval of length $N^2$ is what matches the $Q^2$ term; an interval of length $N$ would not give a bounded exceptional set.