Chebyshev's bias

Let (x; n, m) denote the number of primes of the form nk + m up to x. By the prime number theorem (extended to arithmetic progression), :\pi(x;4,1)\sim\pi(x;4,3)\sim \frac{1}{2}\frac{x}{\log x}. That is, half of the primes are of the form 4k + 1, and half of the form 4k + 3. A reasonable guess would be that (x; 4, 1) > (x; 4, 3) and (x; 4, 1) < (x; 4, 3) each also occur 50% of the time. This, however, is not supported by numerical evidence — in fact, (x; 4, 3) > (x; 4, 1) occurs much more frequently. For example, this inequality holds for all primes x < 26833 except 5, 17, 41 and 461, for which (x; 4, 1) = (x; 4, 3). The first x such that (x; 4, 1) > (x; 4, 3) is 26861, that is, (x; 4, 3) ≥ (x; 4, 1) for all x < 26861. In general, if 0 < a, b < n are integers, gcd(a, n) = gcd(b, n) = 1, a is a quadratic residue mod n, b is a quadratic nonresidue mod n, then (x; n, b) > (x; n, a) occurs more often than not. This has been proved only by assuming strong forms of the Riemann hypothesis. The stronger conjecture of Knapowski and Turán, that the density of the numbers x for which (x; 4, 3) > (x; 4, 1) holds is 1 (that is, it holds for almost all x), turned out to be false. They, however, do have a logarithmic density, which is approximately 0.9959.... ==Generalizations==

This is for k = −4 to find the smallest prime p such that \sum_{q \le p,\ q\ \text{is prime}}\left(\frac{k}{q}\right)>0 (where \left(\frac{m}{n}\right) is the Kronecker symbol), however, for a given nonzero integer k (not only k = −4), we can also find the smallest prime p satisfying this condition. By the prime number theorem, for every nonzero integer k, there are infinitely many primes p satisfying this condition. For positive integers k = 1, 2, 3, ..., the smallest primes p are :2, 11100143, 61981, 3, 2082927221, 5, 2, 11100143, 2, 3, 577, 61463, 2083, 11, 2, 3, 2, 11100121, 5, 2082927199, 1217, 3, 2, 5, 2, 17, 61981, 3, 719, 7, 2, 11100143, 2, 3, 23, 5, 11, 31, 2, 3, 2, 13, 17, 7, 2082927199, 3, 2, 61463, 2, 11100121, 7, 3, 17, 5, 2, 11, 2, 3, 31, 7, 5, 41, 2, 3, ... For negative integers k = −1, −2, −3, ..., the smallest primes p are :2, 3, 608981813029, 26861, 7, 5, 2, 3, 2, 11, 5, 608981813017, 19, 3, 2, 26861, 2, 643, 11, 3, 11, 31, 2, 5, 2, 3, 608981813029, 48731, 5, 13, 2, 3, 2, 7, 11, 5, 199, 3, 2, 11, 2, 29, 53, 3, 109, 41, 2, 608981813017, 2, 3, 13, 17, 23, 5, 2, 3, 2, 1019, 5, 263, 11, 3, 2, 26861, ... For fundamental discriminants of real quadratic fields (k = 1, 5, 8, 12, 13, 17, 21, 24, 28, 29, 33, 37, 40, 41, 44, 53, 56, 57, 60, 61, ... ), the smallest primes p are :2, 2082927221, 11100143, 61463, 2083, 2, 1217, 5, 3, 719, 2, 11, 3, 2, 7, 17, 11, 2, 7, 5, 2, 13, 2, 3, 23, 7, 3, 2, 13, 19, 2, 23, 17, 2, 5, 2, 7, 3, 2, 13, 3, 2, 19, 7, 2, 31, 31, 5, 17, 2, 13, 13, 3, 47, 2, 5, 3, 2, 37, 2, 47, 2, 5, 7, 2, 43, 2, 3, 11, 5, 3, 2, 29, ... For fundamental discriminants of imaginary quadratic fields (k = −3, −4, −7, −8, −11, −15, −19, −20, −23, −24, −31, −35, −39, −40, −43, −47, −51, −52, −55, −56, −59, ... ), the smallest primes p are :608981813029, 26861, 2, 3, 5, 2, 11, 3, 2, 5, 2, 11, 2, 11, 53, 2, 13, 17, 2, 3, 5, 163, 3, 2, 2, 11, 5, 2, 31, 31, 2, 2, 3, 23, 2, 41, 3, 2, 13, 47, 2, 5, 19, 7, 11, 2, 191, 2, 3, 19, 2, 15073, 3, 2, 29, 5, 2, 41, 109, 2, 11, 2, 31, 59, 3, 2, 19, 2, 11, 53, 2, 1019, 137, ... For every (positive or negative) nonsquare integer k, there are more primes p with \left(\frac{k}{p}\right)=-1 than with \left(\frac{k}{p}\right)=1 (up to the same limit) more often than not. ==Extension to higher power residue==

Let m and n be integers such that m ≥ 0, n > 0, gcd(m, n) = 1, define a function f(m,n)=\sum_{p \text{ is prime, } p\,\mid\,\varphi(n),\ x^p \,\equiv\, m \pmod n \text{ has a solution }} \left(\frac{1}{p}\right), where \varphi is the Euler's totient function. For example, f(1, 5) = f(4, 5) = 1/2, f(2, 5) = f(3, 5) = 0, f(1, 6) = 1/2, f(5, 6) = 0, f(1, 7) = 5/6, f(2, 7) = f(4, 7) = 1/2, f(3, 7) = f(5, 7) = 0, f(6, 7) = 1/3, f(1, 8) = 1/2, f(3, 8) = f(5, 8) = f(7, 8) = 0, f(1, 9) = 5/6, f(2, 9) = f(5, 9) = 0, f(4, 9) = f(7, 9) = 1/2, f(8, 9) = 1/3. It is conjectured that if 0  f(b, n), then (x; n, b) > (x; n, a) occurs more often than not. ==References==