The walk of partial sums
Fourth powers in the quarters of (0, p)
N0, N1, N2, N3
A, B
k, t from counts
Lesson 09 · The quartic value β · 55–70 minutes
What β Remembers
Lesson 08 defined β = k + it by weighting every n with χ4(n). How much of that weighted sum is really needed? Trace and norm keep k and |t| but lose the sign of t, which only the pin supplies. On the other hand, far less than the full sum is enough: the character summed over just the first half of 0 to p already determines β, and that half-sum can be read off from where the fourth powers fall in the four quarters of the interval.
Where this lesson is headed
∑0<n<p/2 χ4(n) = (χ̄4(2) − 2)·β = −(2 + i)β, A + Bi = −(2 + i)(k + it)
In words: the half-interval sum of χ4 is a fixed Gaussian multiple of β, and its coordinates A, B are simple counts of fourth powers in the quarters of (0, p). So k = −(2A + B)/5 and t = (A − 2B)/5 can be computed from the fourth powers alone. Tr β = 2k and N β = k2 + t2 see only k and |t|.
p: a prime in P*
σ = ∑0<n<p/2 χ4(n): the half-interval sum
Qr = (rp/4, (r + 1)p/4), r = 0, 1, 2, 3: the four quarters of (0, p)
Nr: the number of fourth powers h ∈ H (taken in 1 to p − 1) lying in Qr
A = (N0 + N1) − (N2 + N3), B = (N0 + N2) − (N1 + N3)
01. How much do you need to know?
Start with p = 13, where β = −1 − i. The left column asks what trace and norm know. The walk adds χ4(1) + χ4(2) + ⋯ step by step: the sum over the whole interval returns to 0, but the halfway point sits at 1 + 3i, which is −(2 + i)β. The number line places the fourth powers in the four quarters of (0, p).
Guided experiments
Do these in order. "Set this up" chooses p.
Why "Galois-invariant"?
Replacing i by −i everywhere is a symmetry of the Gaussian numbers (Lesson 04): it respects sums and products. It sends χ4 to χ̄4 and β to β̄. Anything defined without choosing between i and −i, like the trace β + β̄ and the norm ββ̄, is unchanged by this swap. Such quantities are called invariant under the swap (the swap is the nontrivial element of a Galois group).
So any formula that only uses the unit group mod p "symmetrically" can at best find β up to conjugation. To know the sign of t you must use something that breaks the symmetry, such as the value χ4(2) = i. The half-sum formula uses exactly that value, through χ̄4(2) = −i.
02. The half-sum and the quarter counts
Each step has the general argument on the left and your p from Stage 01 on the right.
1
Trace and norm see k and |t|
Tr β = β + β̄ = 2k and N β = ββ̄ = k2 + t2, and both are the same for β̄ = k − it. From them one recovers k = Tr β/2 and t2 = N β − k2, so
t = ±√(N β − k2)
and no symmetric quantity can do better, since the swap i ↔ −i exchanges the two answers.
2
Two interval sums, and a relation
Over the whole interval, ∑0<n<p χ4(n) = 0 (Lesson 06). For the half-sum σ and T = ∑0<m<p/2 χ(m)m, pair m with p − m and use χ(p − m) = −χ(m) (χ odd):
pβ = ∑n<p χ(n)n = ∑m<p/2 [χ(m)m − χ(m)(p − m)] = 2T − pσ
3
Split by parity instead
Every n in 1 to p − 1 is either even, n = 2m, or odd, n = p − 2m, for exactly one m in 1 to (p − 1)/2. With χ(2m) = χ(2)χ(m) and χ(p − 2m) = −χ(2)χ(m):
pβ = 2χ(2)T − χ(2)(pσ − 2T) = 4χ(2)T − χ(2)pσ
4
The half-sum formula
Substitute 2T = pβ + pσ from step 2 into step 3: pβ = 2χ(2)(pβ + pσ) − χ(2)pσ, so χ(2)pσ = pβ(1 − 2χ(2)). Multiply by χ̄(2) (with χ(2)χ̄(2) = 1):
σ = (χ̄(2) − 2)·β
For the pinned character χ̄4(2) = −i, so σ = −(2 + i)β. The formula holds for every odd character mod p; the pin enters only through χ̄(2).
5
Counting σ coset by coset
Group the first half by coset: σ = ∑j ij·#(Cj ∩ (0, p/2)). Now express each count through the fourth powers:
C0 = H: #(H ∩ first half) = N0 + N1.
C2 = −H: n ∈ C2 is in the first half iff p − n ∈ H is in the second half: N2 + N3.
C1 = 2H: 2h mod p lies in the first half iff h ∈ Q0 (then 2h < p/2) or h ∈ Q2 (then 2h − p < p/2): N0 + N2.
C3 = −2H: by the same argument, N1 + N3.
6
So σ = A + Bi
Adding with weights 1, i, −1, −i:
σ = [(N0 + N1) − (N2 + N3)] + [(N0 + N2) − (N1 + N3)]i = A + Bi
The half-sum is nothing but a count of where the fourth powers sit.
7
β from the fourth powers alone
Combine steps 4 and 6: A + Bi = −(2 + i)(k + it) = (−2k + t) + (−k − 2t)i. Divide by −(2 + i), using 1/(2 + i) = (2 − i)/5:
k = −(2A + B)/5, t = (A − 2B)/5
In particular 5 divides 2A + B and A − 2B, a hidden congruence on how fourth powers are spread. Nothing here needs a primitive root, a character value table, or the weights n.
03. β from the fourth powers
Assemble the argument, then test it on every prime of P* below 1000.
Theorem. Let p ∈ P* and β = k + it. Then Tr β = 2k and N β = k2 + t2 are the same for β and β̄. For every odd character χ mod p, ∑0<n<p/2 χ(n) = (χ̄(2) − 2)B1,χ. With the quarter counts Nr of fourth powers and A, B as defined, A + Bi = −(2 + i)β, so k = −(2A + B)/5 and t = (A − 2B)/5.
Complete proof
Invariants. Tr β = 2k and N β = k2 + t2 are fixed by conjugation, which exchanges β and β̄, so they determine k and t2 but not the sign of t.
Half-sum. Let χ be odd, σ = ∑m<p/2 χ(m), T = ∑m<p/2 χ(m)m, S = pB1,χ. Pairing m with p − m gives S = 2T − pσ. Writing each n as 2m or p − 2m gives S = 4χ(2)T − χ(2)pσ. Eliminating T: S = 2χ(2)S + χ(2)pσ, so pσ = (χ̄(2) − 2)S, i.e. σ = (χ̄(2) − 2)B1,χ.
Quarters. For χ4, C0 = H, C1 = 2H, C2 = −H, C3 = −2H. Counting members below p/2 gives N0 + N1, N0 + N2, N2 + N3, N1 + N3 respectively, so σ = A + Bi. Hence A + Bi = −(2 + i)(k + it), which solves to k = −(2A + B)/5, t = (A − 2B)/5. ∎
Computer check
For each of the 42 primes of P* below 1000: compares Tr and N of β and β̄; checks σ = (χ̄(2) − 2)B1,χ for every odd character mod p (not only χ4); and checks that the quarter counts give k and t exactly.
Not run yet.
Exit problem
Use p = 37, where β = −1 + i.