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If, for every $\eta >0$, the smooth numbers up to $T$ with all prime factors at most $T^{\eta }$ have density at least $\kappa (\eta )>0$, then for fixed $\epsilon >0$ there are only $O_{\epsilon }(1)$ primes $p\le N$ with $n(p)>N^{\epsilon }$. Apply the variance form of the large sieve to the indicator function of the $N^{\epsilon }$-smooth numbers up to $N^2$. For every exceptional prime, all these numbers are nonzero quadratic residues, so half its nonzero residue classes are empty. Their variance contribution is at least one third of the square of the total count. The large sieve bounds the sum of contributions by a constant times $N^2$ times that count. The density assumption with $\eta =\epsilon /2$ finishes the argument.

For a prime $p\equiv 3(\mathrm{mod}8)$, apply two-isogeny descent to the congruent number elliptic curve $y^2=x^3-p^{2}x$ and its companion $y^2=x^3+4p^{2}x$. The first square-class image is exactly $\{1,-1,p,-p\}$, already supplied by 2-torsion. The second has only positive candidates $1,2,p,2p$. The quartic covering in a two-isogeny descent equations for $p$ and $2p$ are impossible modulo $p$, because $-1$ is not a quadratic residue. For $2$ the equation is $W^2=2U^4+2p^2{V}^4$. Coprimality forces $U,V$ odd and $W=2Z$; modulo sixteen this gives $2Z^2=1+9=10$, which is impossible. Thus the second image is trivial and the square-class index formula for two-isogeny descent gives rank zero. The same prime-reduction argument bounds its rational torsion by four, and it already has four rational 2-torsion points, so it has no rational point corresponding to a right triangle of area $p$.

If a squarefree positive $d$ has a prime divisor $p\equiv 3(\mathrm{mod}4)$, a unit of norm $-1$ in $\mathbb{Q}(\sqrt{d})$ would give $a^2-d{b}^2=-4$ for integers $a,b$. Modulo $p$ this makes $-1$ a quadratic residue, which is impossible. The obstruction is sufficient, not a full criterion for negative-norm units.

Let $\chi$ be the quadratic character, equivalently the Legendre symbol. Zero is included among the square residue classes. For every integer $n$ its square-class indicator isFor a nonzero quadratic residue the right side is one, for a nonresidue it is zero, and for a multiple of $q$ it is one. Summing over the interval, the character contribution is $O(\sqrt{q}\log q)$ by the previous part, while the number of multiples of $q$ is $N/q+O(1)$. Thus the required count isThis counts the integers in the interval whose residue classes are squares. When the interval exceeds one period, repeated appearances are counted; the displayed main term could not describe a count of distinct residue classes for arbitrary $N$.

For the odd prime $p$, the Fermat-Euler theorem gives $x^{p-1}\equiv 1(\mathrm{mod}p)$. Therefore the residue class $z=x^{(p-1)/2}$ obeys $(z-1)(z+1)\equiv 0(\mathrm{mod}p)$. Since a prime number dividing a product divides one of its factors,If $x$ is a quadratic residue, choose $y$ with $y^2\equiv x(\mathrm{mod}p)$. Then $y$ is nonzero modulo $p$, and Fermat's little theorem givesThis proves the quadratic residue direction of Euler's criterion.

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.