Download An elementary proof of a formula on 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,
chapters 8 and 10 in www.jimhadams.com.
Part II: Mathematics – Innovation in mathematics,
chapter 9 in www.jimhadams.com.
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.
The average square
• The average of the squares in arithmetic (mod p), p is prime, is
12 + 22 + ... + [(p – 1)/2]2 = [(p – 1)/2]p[(p + 1)/2]/6.
• There are (p – 1)/2 such squares, so that their arithmetic
average is p(p + 1)/12.
• We can find the average square column for the average row.
There are (p + 1)/4 rows and the extent of columns of the
average square is p(p + 1)/12.
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
• jstart is the value of j for the leftmost Gmin parabola.
• The increment of rmin at jstart to 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.
Prospects
• Formulas derived using floor functions of square roots may not estimate
computably the distance distribution of residues from the rightmost edge,
but parabola techniques may. 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.
• 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.