Hanna's Series A(x - A(x)) = x^2 + x A(x) Is Rational Modulo Eight

OEIS A392203, by Paul D. Hanna, Jan 03 2026:

G.f. A(x) satisfies A(x - A(x)) = x^2 + x*A(x).

The comment carrying the question:

It appears that a(2n) == 1 (mod 8) and a(2n+1) == 3 (mod 8) for n >= 1.

The frozen theorem D5/S1/Recurrence/Invariants/HannaShiftSquareModEight.result settles the comment for every integer solution of order at least two, and shows that such a solution exists.

Issue 9678 records the screen carried out before the work. The A-number appears nowhere under Problems/, D5/, Library/ or Blueprint/, nor in the screening records, nor in google-deepmind/formal-conjectures. The entry records no proof and no reference; its cross-reference A276370 states no congruence for this sequence, and a catalogue search by the terms returns only this entry. This is a bounded negative finding.

One, contraction. For a series f of order at least two put T(f) = x^2 + x f - (f(x - f) - f); the equation is T(f) = f. If f and g have order at least two and agree below degree d, then T(f) and T(g) agree below degree d + 1: the difference x(f - g) gains a factor x; f(x - f) - f(x - g) is a sum of terms c_k((x-f)^k - (x-g)^k) with k ≥ 2, each divisible by (f - g) times a series without constant term; and writing f - g = x^d h, the difference (f - g)(x - g) - (f - g) = x^d((1 - g/x)^d h(x - g) - h) has no term of degree d.

Two, existence. Iterating T from zero over the integers, the n-th coefficient stabilises from the (n + 1)-th iterate on; the series of stabilised coefficients is a fixed point of T.

Three, uniqueness modulo eight. Over ZMod 8, two fixed points of order at least two agree below every degree by induction on step one, so they are equal.

Four, the rational solution. Let S be the series over ZMod 8 with coefficient 0 below degree two, 1 at even and 3 at odd degrees, so that S(1 - x^2) = x^2 + 3x^3. With D = 1 - x^2 and N = x - x^2 - 4x^3 one has (x - S) D = N, and multiplying S(x - S)(1 - (x - S)^2) = (x - S)^2 + 3(x - S)^3 by D^3 turns the equation S(x - S) = x^2 + x S into N^2 D + 3 N^3 = (x^2 D + x^3 + 3x^4)(D^2 - N^2), a polynomial identity that holds modulo eight because the two integer sides differ by a polynomial with every coefficient divisible by eight; D and D^2 - N^2 have constant term one and so are units. Hence S is a fixed point modulo eight.

Five. The reduction modulo eight of any integer solution is a fixed point there, hence equals S by step three, and reading off coefficients gives the comment.

The statement would fail if some even-index coefficient were not congruent to one, or some odd-index coefficient not congruent to three, modulo eight. It does not extend to modulus sixteen: a(6) = 1209 is congruent to 9, not 1, modulo sixteen. Through degree sixty the even-index coefficients modulo sixteen take the values 1 and 9 and the odd-index ones 3 and 11, repeating with period six from degree four; that pattern is observed on the computed prefix only and is not proved here.

Solving the equation coefficient by coefficient reproduces the entry's first eleven terms, and every coefficient up to degree sixty satisfies the congruences. The polynomial identity of step four was checked symbolically: every coefficient of the difference of its two integer sides is divisible by eight.

theorem; Tier 1 named external open question from 2026, preregistered in issue 9678 before the work. The computational use is none: the delivered statement quantifies over every solution and every index.

The literature screen is bounded. The entry was read in full and records no proof and no reference to one. Citation indices, printed sources and the resolved set of arXiv:2608.11941 were not exhaustively compared, so no worldwide priority claim is made.