Modeling Chebyshev s Bias in the Gaussian Primes as a Random Walk

Modeling Chebyshev s Bias in the Gaussian Primes as a Random Walk Daniel J. Hutama July 18, 2016 Abstract One asect of Chebyshev s bias is the henomenon that a rime number,, modulo another rime number,, exerimentally seems to be slightly more likely to be a nonuadratic residue than a uadratic residue. We thought it would be interesting to model this residue bias as a random walk using Legendre symbol values as stes. Such a model would allow us to easily visualize the bias. In addition, we would be able to extend our model to other number fields. In this reort, we first outline underlying theory and some motivations for our research. In the second section, we resent our findings in the rational rime numbers. We found evidence that Chebyshev s bias, if modeled as a Legendre symbol ( ) walk, may be somewhat reduced by only allowing to iterate over rimes with nonuadratic residue (mod ). In the final section, we extend our Legendre symbol walks to the Gaussian rimes and resent our main findings. Let π 1 + βi [ and] π 2 [ β + ] i. We a+bi a+bi observed strong (±) correlations between Gaussian Legendre symbol walks for π 1 and π 2 where N(π 1) N(π 2) and a+bi iterates over Gaussian rimes in the first uadrant. We attemt an exlanation of why, for some norms, the lots for π 1 and π 2 have strong ositive correlation, while, for other norms, the lots have strong negative correlation. We hoe to have written in a way that makes our observations accessible to readers without rior formal training in number theory. 1 Introduction 1.1 Prime Numbers Definition 1. A rime number is any integer > 1 whose divisors are only 1 and itself. A comosite number is any integer that is not a rime number or the unit number, 1. One of the first mathematicians to study the rimes was Eratosthenes, to whom is attributed an algorithm to find all rimes less than or eual to a certain value. The Sieve of Eratosthenes starts by marking all multiles of 2 as comosite, then roceeding to multiles of 3, 5, 7 and so on u to x. For examle, after all even numbers u to (and including) 30 have been marked as comosite, we have: 2, 3,, 5, 6, 7, 8, 9, 10, 11, 12, 13, 1, 15, 16, 17, 18, 19, 20, 21, 22, 23, 2, 25, 26, 27, 28, 29, 30 Next, we mark comosite all multiles of 3 not already marked: 2, 3,, 5, 6, 7, 8, 9, 10, 11, 12, 13, 1, 15, 16, 17, 18, 19, 20, 21, 22, 23, 2, 25, 26, 27, 28, 29, 30 Next, we continue to multiles of 5 and roceed as before, continuing until multiles of 29: 2, 3,, 5, 6, 7, 8, 9, 10, 11, 12, 13, 1, 15, 16, 17, 18, 19, 20, 21, 22, 23, 2, 25, 26, 27, 28, 29, 30 The remaining values form the set {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}, which are the rime numbers less than or eual to 30; i.e. the set of numbers less than or eual to 30 whose divisors are only 1 and itself. 1

Proosition 1. (Fundamental Theorem of Arithmetic) Every integer has a uniue rime factorization. In other words, every integer can we exressed in a uniue way as an infinite roduct of owers of rimes: n 2 1 3 2 5 3 7 i i (1) where rimes, and a finite number of i are ositive integers with the rest being zero. For examle, we can write 10 2 1 3 0 5 1 7 0 11 0. Proosition 2. (Euclid s Theorem) There are infinitely many rime numbers. There are many well-known roofs of Euclid s theorem. Euler s roof is as follows: Let denote rime numbers and P denote the set of all rime numbers. Then, 1 1 2 1 3 1 5 1 7 1 2 1 3 2 5 3 P 0 0 0 0 1, 2, 3,... 0 0 However, by (1), we know that every integer can be written uniuely as a roduct of rimes. Thus, we can rewrite our euation as: 1 1 2 1 3 2 5 3 1 (2) n n P 0 1, 2, 3,... 0 We then recognize the right hand side of (2) as the harmonic series. Because of the divergence of the harmonic series, we know our roduct must be infinite as well. Since each term of our roduct is a finite number, there must be an infinite number of terms for the roduct to be infinite. Euler also roved a stronger version of the divergence of the harmonic series, in which he shows the sum of recirocals of rimes also diverges [1]. We will use this fact in a later roof. 1 (3) P 1.2 Arithmetic Progressions The Sieve of Eratosthenes is effective because of the simlicity of identifying multiles of a number. For examle, it is easy to identify all numbers of the form 3n (which is the set {3, 6, 9, 12, 15, 18,...} for n 1) as multiles of 3, and subseuently mark them as comosite (with the excetion of the first element). However, what haens if we change the starting value of the set, while keeing the distance between elements the same? Definition 2. We call a seuence of numbers with constant difference between terms an arithmetic rogression. For examle, consider all numbers of the form 3n+2 and 3n+1, which reresent the sets {2, 5, 8, 11, 1, 17...} and {1,, 7, 10, 13, 16,...} resectively. Both sets of numbers are arithmetic rogressions with a difference of 3. The reader might then inuire: Between 3n + 2 and 3n + 1, which arithmetic rogression contains more rimes u to a value x? In other words, if we consider the count of rimes in each rogression as a race, which team is in the lead at a given x? Can we extend Euclid s Theorem to rimes in arithmetic rogressions? In other words, do arithmetic rogressions contain infinitely many rimes? What is the distribution of rimes in these rogressions? To answer these uestions, we must first introduce a few tools to give our analysis some sohistication. 2

1.3 Euclidean Algorithm, Euler s Totient Function, and Modulo Definition 3. An integer a 0 divides another integer b if there exists another integer c, such that b ac. We denote that a divides b with a b. Definition. Pick two integers a and b. An integer c such that c a and c b is said to be a common divisor of a and b. If there exists another integer d c that also divides a and b, we say that d is the greatest common divisor of a and b. We denote this by gcd(a, b) d. Proosition 3. Let a and b be integers. The Euclidean Algorithm allows us to comute the greatest common divisor of a and b; i.e. it allows us to find the largest number that divides both a and b, leaving no remainder. The algorithm is as follows: a b 0 + r 0 for 0 < r 0 < b b r 0 1 + r 1 for 0 < r 1 < r 0 r r 1 2 + r 2 for 0 < r 2 < r 1... r k 1 r k k+1 + r k+1 for 0 < r k+1 < r k r k r k+1 k+2 + 0 Then gcd(a, b) r k+1. For examle, to find gcd(6188, 709), we aly the Euclidean Algorithm as follows: 6188 709 1 + 179 709 179 3 + 272 179 272 5 + 119 272 119 2 + 3 119 3 3 + 17 3 17 2 17 gcd(6188, 709) Definition 5. a and b are said to be relatively rime, or corime if gcd(a, b) 1. Two rime numbers, and, will always be corime to each other. A comosite number, a, will be corime to rime number,, if and only if a is not a multile of. Definition 6. Euler s totient function, denoted φ(n), counts the number of totatives of n, i.e. the number of (ositive) integers u to n that are corime to n. For examle, φ(10) #{1, 3, 7, 9}. In this examle, the numbers 1, 3, 7, and 9, are the totatives of 10. For a rime number, φ() #{1, 2,..., 1} 1 since all integers < are also corime to. Definition 7. We say that a is congruent to r modulo b if b a r. We write this relation as a r (mod b) In other words, we say that a r (mod b) if r is the remainder when a is divided by b. For examle, when 9 is divided by 7, the remainder is 2. In other words, 9 2 (mod 7). This concet allows us to conveniently refer to arithmetic rogressions by their congruences modulo a. For instance, we can refer to the rogression n + 3 as the set of all integers congruent to 3 (mod ). Furthermore, we can refer to all rimes in the rogression n + 3 as the set of rimes congruent to 3 (mod ). Corollary. Let Z denote the set of all integers. The modulo oeration allows us to define a uotient ring, Z/nZ, which is the ring of integers modulo n. For examle, the set of all integers modulo 6 reeats as {..., 1, 2, 3,, 5, 0, 1, 2, 3,, 5,...}. The uniue elements of this set are {0, 1, 2, 3,, 5}, which is the ring Z/6Z. We say that an element u in Z/nZ is a unit 3

in the ring if there exists a multilicative element v, such that uv vu 1. We denote the grou of units as (Z/nZ). The grou (Z/nZ) has φ(n) elements, which are the totatives of n. For examle, for the ring Z/6Z, the grou of units, (Z/6Z) is given by the totatives of 6: {1,5}. We notice that 1 and 5 are both units in Z/6Z since 1 1 (mod 6) and 5 5 1 (mod 6). Thus for a rime number, the grou (Z/Z) has 1 φ() elements. 1. The Prime Number Theorem and Dirichlet s Theorem on Arithmetic Progressions Let π(x) denote the number of rimes u to x. Proosition. Gauss s Prime Number Theorem (PNT), which Hadamard and Vallèe-Poussin roved indeendently in 1896, states that π(x) behaves asymtotically to x/ log(x) 1 Put another way: lim x π(x) x/ log(x) 1 () Thus for an arbitrarily large value of x, one can exect π(x) to be close to x/ log(x), with some error term. One might next wonder about aroximating the count of rimes within an arithmetic rogression. One way of intuitively aroaching this roblem is by viewing the set of all ositive integers as a union of arithmetic rogressions. For examle, if we consider the arithmetic rogressions with a difference of 3 between elements in each set, we have the three rogressions: {3n + 1 for n N 0 } {1,, 7, 10, 13, 16,...} {3n + 2 for n N 0 } {2, 5, 8, 11, 1, 17,...} {3n for n N 1 } {3, 6, 9, 12, 15, 18,...} Combining these three sets will yield the set of all ositive integers. Since each element in the third set is a multile of 3, and thus a comosite number, we can ignore this set and only consider the first two. We can then exect the rimes to be slit aroximately eually between 3n + 1 and 3n + 2. Similarly, for a difference of between elements in each set, rimes would be slit aroximately evenly between n + 1 and n + 3. Thus alying our intuition to (), we arrive at: Theorem 1. (Dirichlet s Theorem on Arithmetic Progressions) If gcd(a, b) 1, there are infinitely many rimes congruent to b modulo a. In addition, for rogressions of the form an + b, the rimes will be slit among φ(a) different rogressions. In other words, the roortion of rimes in a rogression with increment a is 1 φ(a). lim x π(x; a, b) x/(φ(a) log(x)) 1 (5) For examle, the rogression 5n + 1 holds one-fourth of rimes (φ(5) ), and we write: lim x π(x; 5, 1) x/(φ(5) log(x)) 1 1 log(x) here is actually the natural log of x, but we wish to use the same notation as in our references

The comlete roof of Dirichlet s Theorem is uite lengthy, but excellently shown by Pete L. Clark [2] and Austin Tran [3]. Here, we only briefly introduce imortant concets from analytic number theory and highlight crucial oints of the roof as shown by Clark and Tran. For readers not familiar with analytic number theory, this section may be articularly difficult. Nevertheless, we encourage the reader on. Definition 8. A Dirichlet Character modulo a is a function χ on the units of Z/aZ that has the following roerties: χ is eriodic modulo a, i.e. χ(b) χ(b + a) for b N. χ is multilicative, i.e. χ(b) χ(c) χ(bc). χ(1) 1. χ(b) 0 if and only if gcd(a, b) 1. We say that a character is rincial if its value is 1 for all arguments corime to its modulus, and 0 otherwise. We denote the rincial character modulo a as χ 0. Note that the rincial character still deends on a. Examle. Consider the Dirichlet characters modulo 3. We have χ(1) 1 and χ(3) 0 by roerties stated above. Using the multilicativity and eriodicity of χ we note that (χ(2)) 2 χ(2) χ(2) χ(1) 1. This imlies that (χ(2)) 2 χ(2) ±1. If χ(2) 1, then χ χ 0 is a rincial character by definition. On the other hand, we use χ 1 to denote the character for when χ(2) 1. We note that χ 1 also satisfies all necessary roerties to be a Dirichlet character, but is not a rincial character. Proosition 5. Let X(a) denote the set of all Dirichlet Characters modulo a. X(a) is a grou with multilication and an identity element given by the rincial character χ 0 modulo a. In addition, the following orthogonality relation holds (orthogonality of characters): χ (mod a) { 1 if b 1 (mod a), 0 otherwise (A roof of the orthogonality of characters is nicely shown by A. Tran in [3]). Corollary. The values of a character χ are either 0 or the φ(a) th roots of unity. Recall that if χ(b) 0, then gcd(a, b) 1. If order of the grou is φ(a), then χ(b) φ(a) is rincial, so χ(b) φ(a) 1. Thus, χ(b) e 2πiν φ(a) for ν N. Definition 9. A Dirichlet L-series is a function of the form: L(χ, s) where s is a comlex variable with Re(s)> 1. Proosition 6. The Dirichlet L-function can be also exressed as an Euler roduct as follows (A roof can be found in []): n1 χ(k) n s L(χ, s) ( 1 χ() ) 1 s (6) We introduce an intermediate theorem necessary for the roof of theorem 1: Theorem 2. Dirichlet s Non-vanishing Theorem states that L(χ, 1) 0 if χ is not a rincial character. Here, we will only highlight crucial sections of the roof of Dirichlet s non-vanishing theorem (as shown by J.P. Serre). A more comlete roof of Theorem 2 can be found in [5]. 5

Let a be a fixed integer 1. If m, we denote the image of in (Z/aZ) by. In addition, we use f() to denote the order of in (Z/aZ) ; i.e. f() is the smallest integer f such that f 1 (mod a). We let g() φ(a) f(). This is the order of the uotient of (Z/aZ) by the subgrou () generated by. Lemma 1. For a, we have the identity: χ X(a) (1 χ()t ) (1 T f() ) g() For the derivation of lemma 1, we let µ f() denote the set of f() th roots of unity. We then have the identity: (1 wt ) 1 T f() (7) w µ f() For all w µ f(), there exists g() characters χ X(a) such that χ() w. This fact, together with (7), brings us to lemma 1. We now define a function ζ a (s) as follows: ζ a (s) : χ X(a) L(χ, s) We continue by relacing each L(χ, s) in the roduct by its roduct exansion as in (6), and then alying lemma 1 with T s. Proosition 7. We can then reresent the roduct exansion of ζ a (s) as follows: ζ a (s) 1 ( a 1 1 f()s We note that this is a Dirichlet series with ositive integral coefficients converging in the half lane Re(s) > 1. We now wish to show (a) that ζ a (s) has a simle ole at s 1 and (b) that L(χ, 1) 0 for all χ χ 0. The fact that L(1, s) has a simle ole at s 1 imlies the same for ζ a (s). Thus, showing (b) would imly (a). Suose for contradiction that L(χ, 1) 0 for χ χ 0. Then ζ a (s) would be holomorhic at s 1, and also for all s with Re(s) > 0. Since by roosition 7, ζ a (s) is a Dirichlet series with ositive coefficients, the series would converge for all s in that domain. However, this cannot be true. We show this by exanding the th factor of ζ a (s) as follows: ) g() 1 (1 f()s ) g() (1 + f()s + 2f()s + 3f()s +...) We then ignore crossterms with negative contribution to arrive at an uer bound: 1 + 1 φ(a)s + 1 2φ(a)s + 1 3φ(a)s +... Multilying over, it follows that ζ a (s) has all its coefficients greater than the series: Evaluating euation (8) at s 1 series: φ(a) n gcd(a,n)1 1 n φ(a)s (8), we finish the roof of theorem 2 by arriving at the following divergent n gcd(a,n)1 1 n. 6

We now roceed with the roof of Dirichlet s Theorem. Proof of Theorem 1. Let X(a) denote the grou of Dirichlet characters modulo a. We then fix gcd(a, b) 1 as stated in Dirichlet s Theorem. In addition, we let Ψ denote the set of rime numbers b (mod a). Our goal is to show that Ψ is an infinite set. We wish to consider a function similar to the one in (2). We define: P b (s) : Ψ 1 s (9) In articular, we wish to show that the function P b (s) aroaches as s aroaches 1. This would imly infinitely many elements in Ψ. We also define θ b to be the characteristic function of the congruence class b (mod a). In other words: { 1 if n b (mod a), θ b (n) 0 otherwise Note that θ b is eriodic modulo a and is 0 when gcd(n, a) > 1. Using this characteristic function, we wish to exress P b (s) as a sum over all rimes: Lemma 2. For all n Z, we have: θ b P b (s) P χ X(a) θ b () s χ(b 1 ) φ(a) χ(n) Proof of Lemma 2. Using the multilicative roerty of the Dirichlet character: θ b 1 χ(b 1 n) φ(a) χ X(a) By our orthogonality relation, the summation term becomes φ(a) if b 1 n 1 (i.e. if n b (mod a)) and zero otherwise. The result is exactly θ b. Alying Lemma 2 to (9), we arrive at: P b (s) χ X(a) χ(b 1 ) φ(a) χ() s (10) We recognize the second summation term as reminiscent of the Dirichlet series we defined earlier. We will come back to this euation later. Consider the convergent Taylor series exansion of log(1 z) for z < 1 log(1 z) z n n In addition, consider the Euler roduct reresentation of our Dirichlet series in (6). Alying logarithms, we get: log(l(χ, s)) ( log 1 χ() ) s (12) n1 (11) 7

Combining (11) and (12), we have: log(l(χ, s)) n 1 n ( ) n χ() (13) s The right side of (13) is absolutely convergent for Re(s) > 1, and is therefore an analytic function on that half lane. We now denote the right hand side of (13) as l(χ, s). Lemma 3. In the half lane with Re(s)> 1, e l(χ,s) L(χ, s). The roof of Lemma 2 is shown in [3]. We now slit l(χ, s) into two arts. The first art will be for the sums when n 1, and the second art will be for the sums when n > 1. We denote these as I (χ, s) and R(χ, s) resectively. Symbolically, I (χ, s) We now note that we can write P b (s) from (10) as: l(χ, s) I (χ, s) + R(χ, s) P b (s) χ() s, R(s, χ) χ() n n ns n 2 χ X(a) χ(b 1 ) I (χ, s) (1) φ(a) Lemma. R(χ, s) is bounded when s 1 (Recall, that we wish to show that P b (s) as s 1). This can be shown by comaring R(χ, s) to the well-known Basel roblem: R(χ, 1) 1 n n n 2 n 2 1 n 2 n 1 n 2 2π2 6 Since we know that R(χ, 1) is bounded, we can ignore it as it will not hel us in showing that P b (s) diverges as s 1. We now wish to slit our summation from (1) into an exression with only rincial characters, and a sum over non-rincial characters. Recall that a rincial character χ 0 (n) 1 for gcd(n, a) 1, and 0 otherwise. P b (s) χ X(a) χ(b 1 ) I (χ, s) φ(a) χ 0(b 1 ) I (χ 0, s) + χ(b 1 ) I (χ, s) φ(a) φ(a) χ χ 0 P b (s) 1 1 φ(a) s + l(χ, s) (15) a χ χ 0 We know that a will have a finite number of rime divisors. This fact, together with euation (3), tells us that the first term in (15) is unbounded. All that remains is to show that the second summation in (15) is bounded as s 1. Doing so will show that the rimes (mod a) will fall into one of the φ(a) congruence classes as claimed in theorem 1. To do this, we must use Dirichlet s non-vanishing theorem (theorem 2). Recall that L(χ, 1) 0 if χ is not a rincial character. Thus: L(χ, s) lim s 1 L(χ, s) lim s 1 e l(χ,s) 8

Since logarithms of an analytic function differ only by multiles of 2πi, l(χ, s) log L(χ, s) always remains bounded as s 1. As a result, the contribution to P b (s) from non-rincial Dirichlet characters remains bounded, while the contribution from rincial characters is unbounded. P b (s) itself is then unbounded as s 1. In conclusion, we have: 1 s lim P b(s) s 1 Ψ Thus, there must be infinitely many elements in Ψ, i.e. there are infinitely many rimes congruent to b modulo a for gcd(a, b) 1. 1.5 Chebyshev s Bias, Quadratic Residue, and the Legendre Symbol As uite thoroughly shown by A. Granville and G. Martin in their aer, Prime Number Races [6], when we race rogressions, some rogressions hold the lead for an overwhelming majority of the time. For examle, in the mod race of n + 1 against n + 3, the bias is as much as 99.59% in favor of the n + 3 team! This bias, first observed by Chebyshev in 1853, is attributed to rimes in the n + 1 rogression being uadratic residues modulo. As noted by Terry Tao [7]:...Chebyshev bias asserts, roughly seaking, that a randomly selected rime of a large magnitude x will tyically (though not always) be slightly more likely to be a uadratic non-residue modulo than a uadratic residue, but the bias is small (the difference in robabilities is only about O( 1 x ) for tyical choices of x) Definition 10. Let be an odd rime number 2. We say that a number a is a uadratic residue (QR) modulo if there exists an element x in the set of totatives of, such that x 2 a (mod ). (Note: does not necessarily need to be rime for the definition of uadratic residues. However, as we will see later, the modulus must be rime for our Legendre symbol model to work. Thus, we restrict our study to only rime moduli). For examle, let us consider the set of totatives of 7, which is the set {1,2,3,,5,6}: Table 1: Quadratic Residues (mod 7) x x 2 x 2 (mod 7) Conclusion 1 1 1 1 is a QR (mod 7) 2 is a QR (mod 7) 3 9 2 2 is a QR (mod 7) 16 2 2 is a QR (mod 7) 5 25 is a QR (mod 7) 6 36 1 1 is a QR (mod 7) In this examle, 1 is a uadratic residue since both 1 2 and 6 2 are congruent to 1 (mod 7). In addition, is a uadratic residue since 2 2 and 5 2 are congruent to (mod 7), and 2 is a uadratic residue since 3 2 and 2 are congruent to 2 (mod 7). Note the symmetry of uadratic residues when ordered by x. We now might like a convenient notation to uantify the notion of uadratic residues. 2 Restriction is such that the Legendre symbol will be defined for any. 9

Definition 11. The Legendre symbol searates an integer a into three classes, deending on its residue modulo an odd rime. ( ) a 1 if a is a uadratic residue (mod ), 1 if a is a nonuadratic residue (mod ), 0 if a 0 (mod ) Note: the Legendre symbol is only defined for being an odd rime number. If a is a rime number, the Legendre symbol will never be 0 (since two different rime numbers will be corime) We know by Theorem 1 that the residues of a (mod ) are then eually distributed among congruence classes in {1, 2, 3..., 1}. Continuing with our definition, we introduce several roerties of the Legendre symbol: The Legendre symbol is eriodic on its to argument modulo. In other words, if a b (mod ), then ( ) ( ) a b The Legendre symbol is multilicative on its to argument, i.e. ( ) ( ) ( ) a b ab The roduct of two suares is a suare. The roduct of two nonsuares is a suare. The roduct of a suare and a nonsuare is a nonsuare. This can be exressed as follows: Two suares : 1 1 1 Two nonsuares : 1 1 1 Suare and nonsuare : 1 1 1 The Legendre symbol can also be defined euivalently using Euler s criterion as: ( ) a a ( 1)/2 (mod ) Proosition 8. (Law of Quadratic Recirocity) For and odd rime numbers: ( ) ( 1) 1 1 2 2 ( ) The Law of Quadratic Recirocity [8] has several sulements for different values of a. introduce the first two sulements without roof. For x in the set of totatives of : 1. x 2 1 (mod ) is solvable if and only if 1 (mod ). 2. x 2 2 (mod ) is solvable if and only if ±1 (mod 8). These sulements can be exressed euivalently as follows: 1. ( ) { 1 ( 1) 1 1 if 1 (mod ), 2 1 if 3 (mod ) Here, we only 2. ( ) { 2 ( 1) 2 1 1 if 1, 7 (mod 8), 8 1 if 3, 5 (mod 8) 10

Continuing with our examle for a in (Z/7Z), we have: Table 2: Legendre Symbols (mod 7) a 1 2 3 5 6 ( a 7 ) 1 1 1 1 1 1 Proosition 9. In general, Chebyshev s bias suggests that, in a race between n + β 1 and n + β 2, the rogression in which β i is a nonuadratic residue (mod ) will likely contain more rimes u to x. For instance, when racing 1 (mod 3) against 2 (mod 3), we observe that 2 (mod 3) almost always has more rimes u to x. Indeed, 1 is a uadratic residue (mod 3), and 2 is a nonuadratic residue (mod 3). Table 3: Count of Primes in the mod 3 Race x Primes in 3n + 1 u to x Primes in 3n + 2 u to x 10 1 1 2 10 2 11 13 10 3 80 87 10 611 617 10 5 78 807 10 6 39231 39266 Desite the aarent domination by the 2 (mod 3) team, a theorem from J.E. Littlewood (191) asserts that there are infinitely many values of x for which the 1 (mod 3) team is in the lead (of course, this theorem alies to races in other moduli as well). In fact, the first value for which this occurs is at 608, 981, 813, 029 (discovered by Bays and Hudson in 1976). In 1962, Knaowski and Turán conjectured that if we randomly ick an arbitrarily large value of x, then there will almost certainly be more rimes of the form 3n+2 than 3n+1 u to x. However, the Knaowski- Turán conjecture was later disroved by Kaczorowski and Sarnak, each working indeendently. In fact if we let ν denote the number of values of x( X) for which there are more rimes of the form 3n + 2, the roortion ν X does not tend to any limit as X, but instead fluctuates. This oens the uestion of: what haens if we go out far enough? Will the race be unbiased if we set X sufficiently far away from 0? That is, is Chebyshev s bias only aarent for small values of X? In 199, while working with the mod race, Rubinstein and Sarnak introduced the logarithmic measure to find the ercentage of time a certain team is in the lead [9]. Instead of counting 1 for each x( X) where there are more rimes of the form n + 3 than of the form n + 1, Rubinstein and Sarnak count 1 x. Instead of ν, the sum is then aroximately ln X. They then scale this with the exact value of ln X to find the aroximate roortion of time the n + 3 team is in the lead: 1 ln X ln X > 1 ln X x X 1 0.9959... x where x in the summation is only over values where there are more rimes of the form n + 3 than of the form n + 1. For the mod 3 race, we have: 1 ln X x X 1 0.9990... x Using the logarithmic measure, we see that the 3n + 2 team is in the lead 99.9% of the time! 11

1.6 The Gaussian Primes Definition 12. A Gaussian integer is a comlex number whose real and imaginary arts are both integers. The Gaussian integers form an integral domain, which we denote with Z[i]. In other words, for i 2 1, we have: Z[i] {a + bi a, b Z}. The units of Z[i] are ±i and ±1. In addition, we say that two elements, µ and ν are associated if µ uν for u being a unit in Z[i]. Because of the four units, Gaussian rimes (along with their comlex conjugates) have an eightfold symmetry in the comlex lane (figure 1). For convenience, we often write rimes in lace of rimes uniue u to associated elements. Definition 13. We say that an element in Z[i] is a Gaussian rime if it is irreducible, i.e. if its only divisors are itself and a unit in Z[i]. One might initially believe that the rimes in Z are also irreducible elements in Z[i]. However, this is not the case. In fact, there is a surrising connection between rimes in mod arithmetic rogressions in Z and the Gaussian rimes. To understand this connection, we must first introduce the concet of norm. Definition 1. The norm function takes a Gaussian integer a + bi and mas it to a strictly ositive real value. We denote the norm of a Gaussian integer as N(a + bi) (a + bi)(a + bi) (a + bi)(a bi) a 2 + b 2. In other words, the norm function takes a Gaussian integer and multilies it by its comlex conjugate. One can geometrically understand the norm as the suared distance from the origin. Let γ β. The norm function is multilicative; i.e. for γ,, β elements in Z[i], N(γ) N(β) N()N(β) We also note that the norm of any unit is 1. For examle, if i 0 + 1i, then N() 0 2 + 1 2 1. In addition, we note that if an integer can be written as a sum of two suares, we can reduce it to two elements with smaller norms. For examle, we note that 5 2 2 + 1 2 (2 + i) (2 i) (2 + i) (2 + i) N(2 + i). Thus, if a rime (in Z) can be written as a sum of suares, we know it is not a rime element in Z[i]. Proosition 10. If an odd rime is a sum of suares, it is congruent to 1 (mod ) and not a rime element in Z[i]. Suose a 2 + b 2. Since is odd, exactly one of a or b must be odd, and the other even. For the roof, we let a be odd. Let a 2m + 1 and let b 2n. Then we have: a 2 + b 2 (2m + 1) 2 + (2n) 2 m 2 + m + 1 + n 2 1 (mod ) Thus if 1 (mod ), reresents the norm of two rimes in Z[i]. For examle, 13 1 (mod ) and 13 N(π 1 ) N(π 2 ), where π 1 2 + 3i and π 2 3 + 2i. We note that π 2 i π 1. (Here, we also note that counting rimes in one uadrant is the same as counting rimes uniue u to associated elements). Proosition 11. If an odd rime is congruent to 3 (mod ), then is a rime element in Z[i]. For the roof, suose for contradiction that we can factor into (a + bi) (c + di). Using the multilicative roerty of the norm function, we have: N() N(a + bi) N(c + di) 2 (a 2 + b 2 ) (c 2 + d 2 ) Since is rime, 2 can only be either 1 2 or. Since we do not want a unit as a factor, we let (a 2 + b 2 ) and (c 2 + d 2 ). However, by roosition 10, we know that a solution would imly that is a sum of suares; i.e. 1 (mod ). Thus, 3 (mod ) cannot be factorized; i.e. is a Gaussian rime. 12

We now have enough information to classify a Gaussian rime into one of three general cases. Let u be a unit in Z[i]. Then: u(1 + i) Since 2 N(1 + i) u(a + bi) a 2 + b 2 1 (mod ) u() 3 (mod ) Figure 1: Plot of Gaussian rimes with norm 103 2 Thus, we can see that rimes in Z with uadratic residue (modulo ) are not rimes in Z[i]. Instead, they reresent the norms of two searate Gaussian rimes. We can use this to derive an euation for the exact count of Gaussian rimes (uniue u to associated elements) within a certain norm. Let π G (x) reresent the count of Gaussian rimes u to norm x, then: π G (x) 2π(x;, 1) + π( x;, 3) + 1 The extra count is to include the Gaussian rime at 1 + i, which has norm 2. In addition, we can extend our rime number theorem in the rational integers () to a rime number theorem in the Gaussian integers by a modification of Dirichlet s Theorem (5). Moreover, we note the infinitude of Gaussian rimes by their intimate connection with Dirichlet s Theorem for rimes in mod rogressions. π G (x) 2x x φ() log(x) + φ() log( x) The first term reresents the aroximation of rimes congruent to 1 (mod ), which are the norms of two rimes in Z[i]. The second term reresents the aroximation of rimes congruent to 3 (mod ), which have a norm of 2 for n + 3. More recisely, we have: lim x π G (x) 2x φ() log(x) + 13 x φ() log( x) 1 (16)

The following code can be used in Sage to generate lots of Gaussian rimes within a secified norm 3 : def gi_of_norm(max_norm): Gaussian_rimes {} Gaussian_integers {} Gaussian_integers[0] [(0,0)] for x in range(1, ceil(srt(max_norm))): for y in range(0, ceil(srt(max_norm - x^2))): N x^2 + y^2 if Gaussian_integers.has_key(N): Gaussian_integers[N].aend((x,y)) else: Gaussian_integers[N] [(x,y)] if(y 0 and is_rime(x) and x%3): have_rime True elif is_rime(n) and N%1 or N2: have_rime True else: have_rime False if have_rime: if Gaussian_rimes.has_key(N): Gaussian_rimes[N].aend((x,y)) else: Gaussian_rimes[N] [(x,y)] return Gaussian_rimes,Gaussian_integers def all_gaussian_rimes_u_to_norm(n): gis gi_of_norm(n)[0] return flatten([uni([(x,y), (-y,x), (y,-x), (-x,-y)]) for x,y in flatten(gis.values(), max_level1)], max_level1) N10609 + 1 ### Declare norm here (in lace of 10609) Pscatter_lot(all_gaussian_rimes_u_to_norm(N), markersizerr(1000)/(n/50)) P.show(asect_ratio1, figsize13, svgfalse, axes False) 2 Findings in the Rational Primes 2.1 Bias in the Legendre Symbols of Primes Modulo Another Prime One henomenon we wished to study in detail was Chebyshev s bias, secifically in regards to a randomly selected rime being more likely to have nonuadratic residue modulo some other rime. We aroached this by first attemting to model the bias as a random walk using Legendre symbol values as stes. Let and be two randomly selected rime numbers. Then, according to Chebyshev s bias, ( ) has a slightly less than half robability of being a uadratic residue (i.e. returning a 1). If we fix and let iterate through all rimes, we get a seuence of 1s and ( 1)s (with the excetion of when, in which case we have 0). If modeling as a random walk, the summation of our seuence should not wander far from y t, where t denotes the index of the rime number. Indeed, this is the case with all observed values of u to the final value of (we tested for rimes < 1000 and for iterating over rimes < 10, 000, 000). However, there is a noticeable bias in the summation. Most of the time, the summation of Legendre symbol values is negative, suorting the claim that there are slightly more nonuadratic residues. 3 We also created a video animation of Gaussian rime lots with norms from 10 1 to 10 7 : htts://youtu.be/jrbcmxglvju 1

Figure 2: Legendre Symbol walk for 97 We wished to model the average behavior of our Legendre symbol walks. To do this, we recorded the ratio of uadratic residues in each of our walks for fixed as we increase the range of rimes over which iterates. For examle, when 97 and iterates over all rimes less than 1000, the ratio is 0.698795. When we allow to iterate over all rimes less than 10, 000, 000, the ratio of uadratic residues increases to 0.997826. We then lotted the average ratio for 167 values of ( {3 all rimes < 1000}). In addition, we lotted the within- standard deviation of our ratio for each range of iterated. Since most rimes have nonuadratic residue modulo another rime, the average ratio seems to converge to 0.50 from below as we increase the -range. Figure 3: Plot of the Average Ratios. Horizontal axis denotes log(x), where x is the range over which iterates. Vertical bars reresent 1 standard deviation ( ) We reeated our exeriments with for fixed and varying and arrived at similar results. For 1 ( ) ( ) (mod ), we know from uadratic recirocity that, so the contribution is the same (see theorem ) ) 10 in [10]). For 3 (mod ),. However, Chebyshev s bias still exists (i.e. there are slightly fewer +1s than -1s). As a result, the average behavior is similar. ( ( 15

2.2 Bias in the Legendre Symbols of consecutive Primes Our next exeriment in the rational rimes was to examine the ratio of consecutive uadratic or nonuadratic residues for rimes modulo a fixed rime. I.e. we wished to model the behavior of the ratio of 1, 1s or 1, 1s. Since the robability of (mod ) being is uadratic residue is very slightly less than 0.5, we should exect exect our average ratio to converge to ( 1 n 1 2) from below, where n denotes the length of the consecutive chain. For examle, for the ratio of three consecutive uadratic or nonuadratic residues, we exect to obtain aroximately: ( 1 2 )3 + ( 1 2 )3 ( 1 2 )3 1. (The first term in the summation reresents the robability of 3 consecutive uadratic residues, and the second term reresents the robability of 3 consecutive nonuadratic residues). However, in a very recent aer (March, 2016), R. Lemke Oliver and K. Soundararajan [11], note that there is a much stronger bias in the residue of consecutive rimes than exected. We set out to model this (stronger) bias with our Legendre symbol walk. We reeated our average ratio exeriment as in section 2.1. However, we instead searched for 2, 3, and consecutive residues having the same sign. We notice that the average ratios converge to their exected values uite slowly, suorting R. Lemke Oliver and K. Soundararajan s recent discovery. Figure : From Left to Right: Two Consecutive, Three Consecutive, Four Consecutive 2.3 Bias in the Legendre Symbols Modulo Primes in the Mod Races We reeated our Legendre symbol walk with fixed, but for varying only over rimes congruent to 1 (mod ), and again with rimes congruent to 3 (mod ). We observed Chebyshev s bias in both cases (on average). However, when varied over rimes congruent to 1 (mod ), we noticed a much stronger bias. For examle, if we consider the walks for 97, the walk for 1 (mod ) seems to lie mostly below the t-axis. On the other hand, the walk for 3 (mod ) seems to lie mostly above the t-axis. Figure 5: From Left to Right: iterating over all, 1 (mod ), 3 (mod ). Iteration range for in all lots is 10, 000, 000. 16

We wished to check if this attern exists on average. For 1 (mod ), the average converges to 0.50 more slowly than the average for 3 (mod ). It seems that only allowing to iterate over rimes with nonuadratic residue (mod ) removes, or at least diminishes, some art of Chebyshev s bias. We noticed a similar, but less distinct (see section 2.2 and [7]), attern while testing for consecutive residues being the same. Figure 6: Average ratios of uadratic residues for Left: 1 (mod ). Right: 3 (mod ). The following simle code can be used in Sage to generate a lot for Legendre symbol walks of #declares maximum -iteration range maxn10^7 #P must be an odd rime for legendre_symbol(,p) to be defined P 97 rimes rime_range(3, maxn) m{1:[], 3:[]} m[1] [ for in rimes if % 1] m[3] [ for in rimes if % 3] #relace "3" with "1" to model walk with uadratic residues (mod ) lp [legendre_symbol(, P) for in m[3]] rint "Legendre symbol walk for P{} and iterating over rimes less than {}".format(p,maxn) ( ) sum_lp TimeSeries(lP).sums() #relace "3" with "1" to model walk with uadratic residues (mod ) sum_lp.lot()+lot([srt(x),-srt(x)],(x,0,len(m[3]))) ( ) : 17

3 Findings in the Gaussian Primes Chebyshev s bias in the rational rimes has been well-documented. However, there has been comaratively less exerimental research on such a bias in the Gaussian rimes. In this section, we extend our model of Legendre symbol walks to the Gaussian rimes to see if a similar bias occurs. To do this, we must first introduce a way to ma a Gaussian integer to its residue in the rational integers modulo a Gaussian rime. Proosition 12. A ma that sends a Gaussian rime a + bi to a residue r (mod π), where π + βi, is an isomorhism of rings between Z[i]/πZ[i] and Z/Z, where N(π). In articular, if π is an irreducible element in Z[i], then the residue class ring Z[i]/πZ[i] is a finite field with N(π) elements. A rigorous roof of roosition 12 can be found in [12] as Theorem 12. We first start with a soft roof as motivation for calculating a residue before showing a more rigorous roof. For two rimes and, the Euclidean algorithm shows that the gcd(, ) 1. This fact allows us to easily calculate the residue of (mod ). Let and be rime numbers with >. Let n and r be integers: n + r r n r 0 (mod ) r (mod ) Where r is a element from (Z/Z) ; i.e. r is an element from the set of totatives of. We can extend this algorithm to the Gaussian rimes. Let a + bi and π + βi denote Gaussian rimes with N(a + bi) > N( + βi) N(π). We can then write: We then grou the real and imaginary terms: a + bi π(φ + iψ) + r a + bi ( + βi)(φ + iψ) + r a + bi φ + iψ + βiφ βψ + r a φ βψ + r b ψ + βφ Use the imaginary comonent to solve for ψ, then solve for a: ψ b βφ ( b βφ a φ β a φ bβ + β2 φ Rearrange, multily both sides by, and solve for r: a + bβ + r φ + β2 φ a + bβ r φ( 2 + β 2 ) ) + r + r a + bβ r 0 (mod 2 + β 2 ) a + bβ r (mod 2 + β 2 ) r a + 1 bβ (mod 2 + β 2 ) (17) where r is an element from (Z/( 2 + β 2 )Z) (Z/N(π)Z) (Z/Z) since 2 + β 2 N(π). 18

The idea is to use this residue to calculate the value of a Gaussian Legendre symbol [ ] a+bi π with the hoe of observing a bias as in the rationals. First, we must lay the groundwork by introducing several concets. (A comrehensive reference by Nancy Buck regarding Gaussian Legendre symbols, which includes the full roofs for the following roositions, can be found in [12]. Since many of the roofs are uite lengthy, we will only highlight sections relevant for our model). Definition 15. For k, l, π Z[i], let π be a Gaussian rime u(1+i) and such that k and l are not divisible by π. The Gaussian Legendre symbol has the following roerties: [ ] [ ] k l for k l (mod π) π π [ ] [ ] [ ] k l kl π π π For N(π), the second oint can be euivalently exressed as: [ ] k 1 1 2 l 2 (kl) 1 kl 2 (mod π) π In addition, we have an analog of Euler s criterion in the Gaussian Legendre symbols: [ ] k k ( 1)/2 π Theorem 3. Every Gaussian Legendre symbol can be exressed in terms of a Legendre symbol in the rational integers. [ ] k In articular, we have the following two euations for. Let k a + bi, π + βi, and N(π). Then: π [ ] ( a + bi a 2 + b 2 ) ; π 3 (mod ) (18) [ ] ( ) a + bi a + bβ ; N(π) 1 (mod ) (19) + βi Recall that if π is a rime element in Z[i], a zero imaginary art imlies that π 3 (mod ). For the roof of euation (18), we must show that there exists an element x Z[i] such that x 2 a + bi (mod ) has a solution. We set x φ + ψi so that φ 2 ψ 2 + 2φψi a + bi (mod ). Then we have the following two congruences by grouing real and imaginary terms: φ 2 ψ 2 a (mod ) 2φψ a (mod ) We then suare each congruence and add them together to get: φ + 2φ 2 ψ 2 + ψ (φ 2 + ψ 2 ) 2 a 2 + b 2 (mod ) It then suffices to check that there exists φ and ψ Z[i] such that both congruences have [ simultaneous ] a + bi solutions for the cases a 0 (mod ) and a 0 (mod ) (shown in [12]). Doing so shows that 1 ( a 2 + b 2 ) [ ] ( a + bi a 2 + b 2 ) if and only if 1. In other words, we arrive at euation (18):. We now wish to consider the more interesting case when N(π) 1 (mod ); i.e. when π + βi for, β Z\{0} and π (1 + i). Let be odd and β be even. Let k a + bi with a, b Z and gcd(π, k) 1. As above, we wish to determine if x 2 a + bi (mod π) has a solution for x Z[i]. 19

Recall that N(π) is a rime congruent to 1 (mod ). By roosition 12, we know the set of congruence class reresentatives modulo π is {0, 1, 2,..., 1}. This allows us to only consider x Z when determining if x 2 a + bi (mod π) has a solution. We start by writing our congruence as an euivalence. The congruence x 2 a + bi (mod π) is solvable if and only if there exists x, φ, ψ Z such that: x 2 a bi (φ + ψi)( + βi) x 2 a bi φ + φβi + ψi βψ We then grou the real and imaginary terms into searate euations: x 2 φ βψ b φβ + ψ Then we multily the real art by and the imaginary art by β and add: x 2 a φ 2 βψ bβ φβ 2 + βψ x 2 a bβ φ 2 + φβ 2 x 2 a bβ φ x 2 φ + a + bβ Converting back to a congruence statement modulo, we arrive at the following result: x 2 a + bβ (mod ) ( x 2 ) ( ) ( ) ( a + bβ a + 1 ) ( ) ( ) bβ r (20) ( ) All that remains is to show that 1. To do this, we use the law of uadratic recirocity as described in roosition 8: ( ) ( ( 1) ( 1)( 1)/ ) ( ) ( Since 1 (mod ), 1 0 (mod ). Thus,. In addition, recall that ) 2 + β 2, so ( ) ( ( ) ( ) ( ) β β 2 2 β β (mod ). Thus, we can write. It is then clear that regardless ( ) ( ) ) β of the value of, we have 1. In conclusion, we arrive at euation (19): [ ] a + bi + βi The Exeriment. ( ) a + bβ ( ) r While imlementing our random walk model on Sage, we decided to fix π + βi and let a + bi iterate over Gaussian rimes in the first uadrant sorted by increasing norm. In the case of a + bi a 3 (mod ), the sorting is obvious. However, when N(a + bi) 1 (mod ), there are exactly two (distinct) Gaussian rimes with norm (we have a + bi and b + ai i(a + bi), where a 2 + b 2 ). When this is the case, we sort by the size of the real comonent. (For examle, when 17 N(1 + i) and N( + i), we find the residue of 1 + i (mod π) first and then roceed to find the residue of + i (mod π)). 20

When viewed individually, the resulting lots resemble the Legendre symbol walks in section 2.1. However, we observe an interesting henomenon when comaring walks that have the same N(π 1 ) N(π 2 ) where π 1 and π 2 are fixed with a + bi iterating. We noticed for some, the lots for π 1 and π 2 have strong ositive correlation. For other, the lots for π 1 and π 2 have strong negative correlation. Figure 7: Gaussian Legendre symbol walks for 97 (strong [ ositive ] correlation) [ ] a + bi a + bi Left:. Right: + 9i 9 + i. Figure 8: Gaussian Legendre symbol walks for 29 (strong [ negative ] correlation) [ ] a + bi a + bi Left:. Right: 2 + 5i 5 + 2i. Before we attemt to (artially) exlain this henomenon, we must first introduce additional theory. Theorem. The following 3 roerties hold for the Gaussian Legendre symbol [ ] i + βi [ ] 1 + i + βi [ ] a + bi + βi ( 1) 1 (21) ( 1) (+β) 2 1 8 (22) [ ] + βi a + bi (23) 21

The roof of euation (21) is as follows: [ ] From Euler s criterion in the Gaussian Legendre symbols, we know that i ( 1)/2 i + βi We note that i ( 1)/2 can be rewritten as follows: i 1 1 2 2 i ( 1) 1. (mod + βi). Thus, we have the congruence: [ ] ( 1) 1 i + βi For a roof by contradiction, we assume that the left side the right side. Then let 1 1 (mod + βi). Converting the congruence to an euivalence, we get: We then take norms of both sides and simlify: 2 ( + βi)(φ + ψi) N( 2) N( + βi)n(φ + ψi) N(φ + ψi) This imlies that, which cannot be true since 1 (mod ). Therefore, we arrive at euation (21): [ ] i ( 1) 1. + βi For the roof of euation (22), we must consider two cases: when β 0 and when β 0. Case 1: let β 0, so 2 and 3 (mod ). Recall our relations between the Gaussian Legendre symbols and the Legendre symbols in the rational integers as shown in theorem 3. From euation (18), we have: [ ] ( ) ( ) 1 + i 1 + 1 2 Recall our second sulement of uadratic recirocity in the rational integers. We can then exress this as: ( ) 2 ( 1) 2 1 8 ( 1) (+β)2 1 8 Case 2: Let β 0, so 2 + β 2 and 3 (mod ). By euation (19), we have: [ ] ( ) 1 + i + β. + βi Since our model only uses rime elements in the first uadrant, we assume that + β > 1 (the full roof without this assumtion can be found in [12]). We continue by using the law of uadratic recirocity: ( ) + β ( 1) ( 1 2 )( +β 1 2 ) ( ) + β Since 1 (mod ), then ( ) ( 1 2 is always even. Thus, 1 ) ( ) ( ) ( ) +β 1 + β 2 2 is even. So. + β 22

Next, we multily by 2 and aly a clever series of maniulations. We note that: 2 2( 2 + β 2 ) 2 + 2β + β 2 2β ( 2 + β 2 )( 2 β 2 ) 0 ( + β) 2 + ( β) 2 2 ( + β) 2 ( β) 2 2 0 ( β) 2 2 (mod + β) 2 ( β) 2 (mod + β) Let x ( β) 2. Then there exists a solution to the congruence x 2 2 (mod + β). Then we have: ( ) ( ) ( ) x 2 x x 1 + β + β + β ( ) ( ) x 2 2 1 + β + β ( ) ( ) ( ) 2 2 1 + β + β + β ( ) ( ) ( ) [ ] 2 + β 1 + i Which imlies that. Using the second sulement to + β + β + βi uadratic recirocity, we have: [ ] ( ) 1 + i 2 ( 1) (+β) 2 1 8. + βi + β For the roof of euation (23), we must consider three cases: 1. b β 0 2. b 0 and β 0 3. b 0 and β 0. Case 1: Let b β 0. Then by euation (18): [ [ a a 2 ] [ ] [ 2 a a [ a [ ] It is then clear that 1. ] a ] 1 ] 1 Case 2: Assume b 0 and β 0. Then: [ ] ( ) ( ) ( ) ( ) a a a a + βi ( ) Recall we have already shown in theorem 3 that 1. Then we have: [ ] ( + βi 2 + β 2 ) ( a a a) 23

From uadratic recirocity, we know that ( ) a (. Thus, we have: a) [ ( ) a ( ( 1) ( 1)(a 1) Since 1 (mod ), we then see that a) a + βi ] [ + βi a Case 3: Assume both b and β are nonzero. Since a + bi and + βi are distinct odd Gaussian rimes, we have: [ ] [ ] a + bi a + bβ + βi [ ] [ ] + βi a + bβ a + bi where 2 +β 2 and a 2 +b 2. Since we are working in the first uadrant, we assume that a+bβ > 1. We then wish to erform another maniulation (the idea is similar to the roof of euation (22)). In articular, we wish to show that a certain congruence is solvable (mod a + bβ). We note that: (a + bβ) 2 + (aβ b) 2 a 2 2 + 2abβ + b 2 β 2 + a 2 β 2 2abβ + b 2 2 (a + bβ) 2 + (aβ b) 2 ]. a 2 2 + b 2 β 2 + a 2 β 2 + b 2 2 ( 2 + β 2 )(a 2 + b 2 ) (a + bβ) 2 (aβ b) 2 0 (aβ b) 2 (mod a + bβ) (aβ b) 2 We then set aβ b x. Thus we have the congruence: (mod a + bβ) x 2 (mod a + bβ). To finish the roof, we show: ( ) ( ) ( ) x 2 x x 1 a + bβ a + bβ a + bβ ( ) ( ) ( ) 1 a + bβ a + bβ a + bβ ( ) ( ) which imlies that. Since we know that and are rimes in Z that are congruent a + bβ a + bβ ( ) ( ) a + bβ a + bβ to 1 (mod ), by uadratic recirocity, we can euivalently write this as:. By [ ] [ ] a + bi + βi alying euation (19) of theorem 3, we then see that. + βi a + bi 2

We now attemt to exlain the strong (±) correlations we observed between Gaussian Legendre symbol walks with π 1 and π 2 fixed, where π 2 iπ 1 and for a + bi iterating over Gaussian rimes in the first uadrant. [ ] [ ] a + bi b + ai We first wish to establish a relationshi between and. This will allow us to find their + βi [ ] + βi [ ] [ ] [ ] a + bi b + ai b + ai a + bi combined contribution. (Recall the iteration order is one of or, + βi + βi + βi + βi based on the size of the real art). [ ] [ ] a + bi b + ai To find the conditions such that we set: + βi + βi [ ] [ ] a + bi b + ai 1 + βi + βi [ ab + a 2 i + b 2 ] i ab + βi [ ] [ i a 2 + b 2 ] + βi + βi [ ] ( 1) ( 1)/ + βi ( ) ( ) ( 1) ( 1)/ ( ) ( 1) ( 1)/ (2) ( ) ( ) if 1 1, or if 1 is odd and 1. The conditions [ ] [ ] a + bi b + ai Thus, + βi [ + βi ] a + bi for the euivalence of β + i is even and [ ] b + ai are similar. β + i Case ( ) 1: Let π 1 + βi and π 2 β + i, where N(π 1 ) N(π 2 ). Let 1 be an even integer. Suose 1. Then by our euivalence relations, we have: [ ] [ ] [ ] [ ] a + bi b + ai a + bi b + bi and + βi + βi β + i β + i [ ] [ ] [ ] [ ] a + bi b + ai b + ai a + bi Thus, whether the iteration order is or, the combined contribution is one of ±2. The same is true with or. + βi [ ] + βi [ ] + βi [ ] + βi [ ] a + bi b + ai b + ai a + bi β + i β + i β + i β + i ( ) We now consider the case when 1 is still an even integer, but 1. Then by our euivalence relations, we have: [ ] [ ] [ ] [ ] a + bi b + ai a + bi b + ai and + βi + βi β + i β + i Thus, for any norm-sorted iteration order, the combined contribution will be 0. ( ) Case 2: Now we let 1 be an odd integer. Suose 1. From our euivalence relations, we know that: [ ] [ ] [ ] [ ] a + bi b + ai a + bi b + ai and + βi + βi β + i β + i 25

Then for any norm-sorted iteration order, the combined contribution from a + bi and b + ai will be zero for both walks of π 2 and π 2. ( ) [ ] [ ] Now we consider the case when 1 is still odd, but a + bi b + ai 1. In this case,. Thus, + βi + βi for any norm-sorted iteration order, the combined contribution will be one of ±2. [ ] [ ] a + bi a + bi If we can establish the conditions for euivalence between and we will be able to fully + βi β + i exlain the strong ositive and negative correlations observed. (Note: it still remains to show what haens when a + bi iterates over Gaussian rimes a + bi a 3 (mod ). However, since rime elements of this form are much more sarse by euation (16), we can ignore them for the uroses of our exlanation). Unfortunately, we found it uite difficult to rigorously ( ) rove the euivalence conditions (in articular, because the Legendre (more recisely, Jacobi) symbol is not defined for β an even integer), so we leave it as a β conjecture. [ ] [ ] a + bi a + bi Conjecture. The euivalence between and deends only on the value of the Legendre ( ) [ ] [ + βi ] ( β ) + i [ ] [ ] ( ) a + bi a + bi a + bi a + bi symbol. In articular, if 1, and if 1. + βi β + i + βi β + i We will use the following shorthand notation for clarity and convenience: [ ] [ ] a + bi b + ai π 1a π 1b + βi + βi [ ] [ ] a + bi b + ai π 2a π 2b β + i β + i π 1 π 1a + π 1b π 2 π 2a + π 2b To summarize, we have shown (conjectured) the following relations: ( ) π 1a π 1b ( 1) ( 1)/ ( ) π 2a π 2b ( 1) ( 1)/ ( ) π 1a π 2a ( ) π 1b π 2b (25) (26) (27) (28) We can now exlain the strong (±) correlations between lots for π 1 and π 2 fixed. ( ) Consider the case when 1 is even and 1. If π 1a 1 (res. 1), then by euation (25), π 1b 1 (res. 1). Using( euation ) (27), π 2a 1 (res. 1), and by euation (26), π 2b 1 (res. 1). Thus, when 1 is even and 1, the walks for π 1 and π 2 move exactly together with combined contribution one ( ) of ±2. Consider the case when 1 is even and 1. If π 1a 1 (res. 1), then by euation (25), π 1b 1 (res. 1). Using euation (27), π 2a 1 (res. 1), and by euation (26), π 2b 1 (res. 1). Then π 1 and π 2 do not move together, but the combined contribution for that articular is 0, so there is little movement and the correlation remains close to +1. 26