An elementary investigation of the p = 4k – 1 asymmetry theorem for quadratic residues

An elementary investigation of the p = 4k – 1 asymmetry theorem for quadratic residues by Jim Adams

Sources Part I: Mathematics – Innovation in mathematics, chapter 8 in www.jimhadams. org. Part II: Mathematics – Innovation in mathematics, chapter 9 in www.jimhadams.org.

Part I The problem • If a prime p = 4k – 1, there are more quadratic residues in the interval [1, 2k – 1] than in [2k, 4k – 2]. • All previous proofs used Dirichlet’s class-number formula. • Is there a proof by elementary methods?

Definition of the disparity • The disparity is the number of quadratic residues in the interval [1, 2k – 1] minus those in [2k, 4k – 2].

Sophisticated methods • Herman Weyl (1940) used transcendental methods. • Class number H for quadratic forms. • For primes p = 4k – 1, but not q = 4k + 1, we must consider negative discriminants. • He showed that, for p  7 (mod 8), the disparity is equal to H. For p  3 (mod 8) it equals 3H. • Quote: “A non-transcendental derivation of these wondrous results is unknown.”

Basic definitions • We call n2 a square or a perfect square. • A quadratic residue, b, is then a square reduced (mod p), so n2 = ap + b, where b < p. • Natural numbers here are in lower case.

Basic definitions • A row is the corresponding interval not reduced (mod p), so the first row is [0, p – 1] and the second row is [p, 2p – 1], etc. • We specify that [0] is at column 0. p – 1 0 1 0 1 (p – 1)/2 p – 1 low row number high row number • Each traversal of the clock on the left with prime p = 4k – 1 hours is pictured as transformed into a row on the right, and correspondingly so are the quadratic residues belonging to it. • The theorem to be proved states there are more quadratic residues on the right hand side of the clock, or equivalently more in total on the left hand side of the rows.

Part I results Standard results • There are (p – 1)/2 quadratic residues  0 (mod p) occupying (p – 1)/2 separate columns. • If y is a quadratic residue  0 (mod p), p – y is not, and vice versa.

Crossing out method To find if n is a quadratic residue • Put X in column 0 • Put X in column 1 (no spaces from previous) • Put X in column 4 (two spaces from previous) • ... • Put X in column z (increased by two spaces from previous) continue to other rows if necessary • If column n is crossed out, it is a residue.

Number of rows before a residue repeats • Rows before a repetition = (p + 1)/4. • The proof uses the maximum perfect square converted to a distinct residue is [(p – 1)/2]2. • All residues occur on or before this row.

Row T • T is the row up to which the difference between the next perfect square is < (p – 1)/2. • T = (p + 9)/16 . •  is the floor function.

Row region up to row T p = 1031 Rows increase downwards. Columns contain residues.

First row • The disparity is non-negative and positive for p  7. • The disparity >[(2) – 1](p – 1) – 1.

Any row This includes the trajectory region, where rows are > T and < (p + 1)/4. • The lowest disparity is -1. • The proof uses 2[a + (b/2)] > (a + b) + (a). • Implies for an even number of quadratic residues in a row the disparity is non-negative.

The disparity for a row, r 2[rp – (p/2)] – (rp) – [(r – 1)p]

Disparities for low row numbers • The disparity is always > 0 for p > 32(2r – 1)3. • The proof uses 2X>X + A – ½ + X – A – ½ and the binomial theorem. • The disparity for row 2 is always > 0. • All primes with negative disparity in row r up to r = 5 have been determined by computer program.

Disparities for rows approaching row T • Except for p = 67, the disparity is non-negative up to row y above row T. • y = 0 is at row T. • y2 < 2. • p = 4k – 1, k = 4 +  and  = 0, 1, 2 or 3. • The proof uses enumeration of all cases.

The total disparity For row and trajectory regions this is • 2Σ[r = 1 to (p + 1)/4][[rp – (p/2)] – (rp)] + (p – 1)/2. • The proof uses previous results. • The total disparity is odd. The objective is now • To prove this positive. • To obtain an estimate of its value.

Trajectories • The trajectory region is situated after row T and up to row (p + 1)/4. • Trajectories ascend from the bottom row. • The first trajectory is labelled as m = 0 and the next is at m = 1, etc. • The m = 0 trajectory starts at column (p + 1)/4. • Here the vth trajectory residue starting at 1 is at column (p + 1)/4 + v(v – 1) (mod p). • When a trajectory meets the right hand edge, the next trajectory continues upwards from the same row, starting out switched to the left.

Trajectories • Trajectories are segments of parabolas. • The disparity for a trajectory is the number of its residues to the left of column (p – 1)/2 minus those to its right. • The value of the disparity if m  0 is 2[(m + ¼)p – ½] + ½ – [(m + ¾)p – 1] + ½ – [(m – ¼)p] + ½. • The lowest disparity for a whole trajectory is -1. • The m = 0 trajectory has non-negative disparity. • There is a bijection between disparity expressions for trajectories and those for rows.

Comparing q = 4k + 1 with prime p = 4k – 1 • For prime q = 4k + 1, there are (q – 1)/4 quadratic residues in each of the left and right hand parts – so the disparity is zero. • The number of rows is (q – 1)/4. The last square [(q – 1)/2]2 at (q – 1)/4 from the right edge is in the right hand part. • The disparity intervals are [1, (q – 1)/2] and [(q + 1)/2, q – 1]. • When q is not prime but still 4k + 1 the occupancy theorem fails, so multiply occupied residues are at different squares. • But the expressions involving floor functions and square roots we use ignore this. • We compare the zero disparity for q with the disparity for p.

The proof of positive disparity • Number of q residues on the left hand side is k. • Number of p residues on the left hand side should be > k. • We compare these disparity sums from 1 to k – 1. • Subtracting q residues from p residues gives > 0. • Row zero for p is > 0 on the left hand side. • Row k for q has the residue on the right. • Adding these together gives positive disparity.

Part II Parabolas for rows • Residues are given by squares n2 (mod p). • By the Euclidean algorithm for j < 0 < r, where h and j are unique n = hr + j. • Choose h free, so there are multiple representations of n. • To form the column for the first row, from n2 subtract nothing, and for the rth row subtract (r – 1)p. • The residue is then in column G = (hr + j)2 – (r – 1)p. • This is a parabola.

Stratified parabolas • G = (hr + j)2 – (r – 1)p. • Retain the row as a whole number, but h as fractional. • If the denominator of h does not divide r, then j is fractional. • These are defined as stratified parabolas. • The constant denominator of h over all rows where it operates is the number of stratums, or strata.

The minimum value of G, Gmin • The minimum value of G occurs when dG/dr = 0. • This means 2h2r + 2hj – p = 0. • The value of r for Gmin is a rational number rmin = (p – 2hj)/ 2h2. • This gives the minimum value of the parabola G as Gmin = -(p2/4h2) + [(j/h) + 1].

The interpretation of h • For unstratified Gmin, the difference in its values between j and (j + δ) at constant rmin is pδ/h < p, reducing to δ < h. • A sequence of δ intervals contains (δ + 1) end points for the intervals, so the maximum number of residues is Mmax = h.

The slope of Gmin over all parabolas for the row • jstartis the value of j for the leftmost Gmin parabola. • The increment of rmin at jstartto rmin at (jstart + Mmax – 1) is (- Mmax + 1)/h = (1/h) – 1. • Row numbers increase going downwards, so this negative slope is pictured as an ascending set of parabolas.

The determination of j For unstratified parabolas • j = jstart+ δ. • jstart= p/4h– h + 1. The proof uses • The leftmost value of Gmin is < p/h, where the row length is p and there are h parabolas with increasing spacing between them from left to right. • This means jstart< 1 + (p/4h) – h. • The leftmost value of Gmin is > 1. • This means jstart> (p/4h) – h. This implies • Gmin = (p/h)[1 – (p/4h) +p/4h + δ].

A maximum suitable value of h, hmax • With increasing row number (going down in the diagram), h decreases. • The maximum value of h we want is defined as occurring when the differences for rmin at h and rmin at (h + 1) is about 1. • An approximate calculation shows (1 + 2h)3 4p. • Putting hmax = h + 1 gives as a definition from the approximate value hmax = [(p + 3)/2]1/3 .

Ambits, gaps, bands and fragments

Unstratified ambits and gaps • Unstratified parabolas are parameterised by h, and their instances are given by δ = 0 near the left edge then successively to δ = (h – 1) near the right edge. • An ambit for an nth parabola from the right edge is the range of rows within it, intersecting with edge column (p – 1). • If two ambits intersect, eliminate the bottom row, so they fit together. • A gap outside of an ambit for an nth parabola is the external range of rows intersecting with the nth parabola gap for an (h + 1) or (h – 1) parabola.

Unstratified fragments, and bands • A fragment is a parabola given by δ < 0. • Fragments may be thought of as continuations of parabolas intersecting the right edge ‘wrapped round’ to continue from the left edge. • This continuation is one row lower in the diagram than its intersection with the right edge. • The top of a band is the intersection row of fragments given by h and (h + 1). • This intersection is called a join. • If there are no fragments, the intersection is given instead by the join of rightmost parabolas. • The bottom of a band is the corresponding join given by parabolas h and (h – 1), minus a row, so the bands fit together.

Trajectories and trajectory parabolas p = 1031. Trajectories m = 0 to m = 5 are at the bottom. Trajectory parabolas are at the top.

Trajectories and trajectory parabolas • The vth quadratic residue starting from v = 1 at the bottom row is at trajectory column D = v(v – 1) + (p + 1)/4 (mod p). • Trajectories occupy the region from row T = (p + 9)/16 + 1 to U = (p + 1)/4. • The same residues trace out trajectory parabolas. • These trajectory parabolas are described by the same parabola formula as for rows G = (hr + j)2 – (r – 1)p.

Trajectory parabolas are stratified • e is the number of trajectory parabola continuations cutting across a row. • Say there are g trajectory parabola rows with a single residue and d rows with a pair of residues, so e = 2d + g. • We will represent a v which increases, at a row r which decreases by v = (-hr + f). • For a vth residue, a trajectory parabola’s nth residue along this parabola is at column v + (n – 1)e. • For the increment v  v + (n – 1)e, the row of the trajectory parabola for that quadratic residue decrements by (d + g) = (e – d) rows under the mapping v  {-h[r – (n – 1)(e – d)] + f} = {v + h(n – 1)(e – d)}, so we identify (n – 1)e and h(n – 1)(e – d): h = e/(e – d). • Thus as previously defined, the trajectory parabolas correspond to stratified parabolas for the row equations.

Some stratified trajectory parabolas p = 1031 h = 5/3 Residues in pairs for a row are connected by a horizontal line.

Ambit and fragment joins • When ambits are joined, there are no fragments. • The upper ambit join is at row rjoin = p/4h(h + 1) or p/4h(h + 1). • The proof uses (where jh is the rightmost j parameter associated with h) G = [hr + jh]2 – p(r – 1) = [(h + 1)r + jh+1]2 – p(r – 1) and jh = p/4h. • The join at fragments is one row greater than the formula for the join between ambits (because of ‘wrap round’).

Parabolas with two strata • Stratified parameters are subscripted by s. • Say Gs = (hsr + js)2 – p(r – 1) and hs is a multiple of ½, hs = h – ½. • Then js = p/4hs – hs + 1 + δ – ½νs. • νs = 0 or 1 is the stratification number. • The fragment join of parabolas given by hs and h then satisfies at νs = 1 rjoin = 2[p/4hs – p/4h], being one more at νs = 0.

Ambits ignoring floor functions • The rightmost ambit is Aedge = 2{(p/h)[(p/4h) – p/4h – 1 + h – δ] – 1}½/h. • For odd h, the ambit when it exists straddling the mid column (p – 1)/2 is Amid = {(p – 2hj)2 – 4h2[j2 + (p + 1)/2]}½/h2. • Amid < Aedge.

The disparity within a band • Single fragments are on the left. • Ambit K = fragment gap F displaced downwards one row. • The band is Γ. • The disparity is (Γ – F) + I + J – (Γ – I) – (Γ – J) – K = 2I + 2J – 2K – Γ.

Multiple fragments (high p) h = 3

Multiple fragments (high p) • Here fragments traverse the entire range of columns. • There are h parabolas, so fragments are stratified in h trajectory sets. • The original fragment stratum returns to itself cyclically at the (h + 1)th trajectory set. • These fragments can nest so that they define parabolas, and these parabolas define further fragment trajectories, etc.

Interspersion • Between contiguous possibly stratified h parameters hs and ht, define a stratified parameter ½(hs + ht). • This is called interspersion. • This process can nest, as for multiple fragments. • Interspersed parabolas cover the entire row region. • A suitable interspersion depth has been calculated.

Further prospects • Formulas derived using parabola techniques may estimate computably the distance distribution of residues from the rightmost edge. This distribution is implicit in proving positive total disparity. • Parabolas for h = 3 have positive disparity. • Various rule of thumb hypotheses have been formulated. • It may be possible to prove the interspersed region disparity for h = 3 plus the top row disparity exceeds the other odd h disparities. • Conjecture: The max. disparity for an h band including fragments is -1. • Parabolas for q = 4k + 1 could be compared with those for prime p. • A project is to enumerate the total disparity by these means and compare it with the class number, H, thereby proving there is no tenth discriminant of form -p.

An elementary investigation of the p = 4k – 1 asymmetry theorem for quadratic residues