One candidate, several arithmetic constraints

The idea in one paragraph

A possible solution has to be the same integer in every description. In the family studied here, one description says that an integer X is one more than a square. Another says that it is a particular coefficient in a power of a quadratic unit. That coefficient is a term of a recurrence. Elementary arithmetic excludes some recurrence indices immediately. For two other sets of indices, an explicit algebraic construction forces a sign incompatible with the square requirement. The proof comes from making these descriptions meet, not from checking larger and larger integers. The human–AI workflow has a separate role: proposing arguments, recording their assumptions and making their checks reproducible.

This guide explains the frozen release-candidate article, Shifted-square obstructions at indices divisible by 17 or 41 in a family of Lucas sequences, dated 10 September 2026. Article references below use that version’s theorem numbers. It does not extend the mathematical scope or use the earlier infographic as a proof source. Worked examples illustrate the article; the final section describes the engineering strategy separately from the mathematics.

1. Start with three consecutive integers

An integer is powerful when every prime that divides it occurs at least twice in its prime factorisation. For example,

are powerful. But 26=2\cdot13 is not: both its prime factors occur only once. Thus 25,26,27 illustrate two powerful numbers separated by a non-powerful middle, not a powerful triple.

The three-consecutive problem asks whether there can be an integer M for which M-1, M and M+1 are all powerful. This paper considers an application within the square-middle case,

Here X is the square root of the middle integer. The middle integer is X^2, not X. Every prime exponent in X^2 is even, so the middle is automatically powerful. Both neighbours must still satisfy their own powerfulness conditions.

The particular family used in Paper I has the additional requirements

for an integer y. The congruence means that X leaves remainder 2 on division by 8. This family is a restricted part of the square-middle problem, not a description of every possible square-middle triple.

The square condition gives the first description of the candidate:

The paper obtains a restriction even without assuming that the lower neighbour X^2-1 is powerful. That lower-neighbour requirement would impose a further constraint on an actual triple.

Article: §1.1 and Corollary 1.5.

2. The upper neighbour gives a second description

Every powerful positive integer can be written uniquely as

where K is squarefree, meaning that no prime occurs more than once in K, and V is a positive integer. Put into K exactly the primes whose original exponents are odd. Each such exponent is at least three, so removing the cube leaves an even exponent for V^2.

There is no requirement that K and V be coprime. For instance, 32=2^3\cdot2^2 has K=V=2.

If the upper neighbour is powerful, write

The same equality can be rearranged as

This is a negative Pell equation: an equation of the form a^2-Kb^2=-1 in integer coefficients a,b. Here its coefficients are exactly a=X and b=KV.

Introduce

Its conjugate is X-KV\sqrt K, and their product is

The norm packages the Pell equation into multiplication. An element of norm -1 has an inverse with integer coefficients in the same quadratic arithmetic, so it is a unit.

Under the paper’s hypotheses, K>1 and K\equiv5\pmod8. The paper proves that there is a least unit greater than 1,

in the order \mathbb Z[\sqrt K]. That order is the set of all expressions a+b\sqrt K with integer a,b. Its positive units are powers of \varepsilon. Consequently,

for a unique positive odd integer N. Oddness follows because \varepsilon and \eta both have norm -1: an even power would have norm +1.

This choice of order matters. Using a generator from a larger ring could change the exponent. The proof retains the integer coefficients of the original Pell equation.

Article: §6.2, especially equations (6.2)–(6.4).

3. Multiplying a unit produces the recurrence

Write the powers of the unit as

The integers X_n and Y_n are its two coefficients. In particular, X_n is not the real value of the whole expression \varepsilon^n.

The equation u^2+1=Kv^2 gives

Multiplying by \varepsilon^n and comparing the coefficients of 1 produces

This motivates the family of sequences

Here A chooses the sequence and n chooses a term within it. The first terms are

For A=2, this gives 1,2,9,38,161,682,\ldots. For the Pell unit, A=u and X_n=F_n(u).

The two descriptions of the candidate therefore meet in the exact identity

This is the central intersection. The square witness, Pell coefficient and recurrence term all concern the same X.

A worked illustration shows why each condition matters. For \varepsilon=2+\sqrt5,

So X=682 has a powerful upper neighbour and X\equiv2\pmod8. But X-1=681 is not a square: it lies strictly between 26^2=676 and 27^2=729. Satisfying the Pell-side condition is not enough; the square-side condition must hold for that very same candidate.

Article: Corollary 1.4 and §§6.1–6.2. The numerical illustration is obtained by expanding the displayed unit power.

4. What the recurrence theorem actually says

The main theorem is a statement about the whole recurrence family, before any powerful-number application is imposed:

The notation 17\mid N means that 17 divides N. The statement applies to every parameter A=2,6,10,14,\ldots, not merely the sequence at A=2.

A second theorem excludes every positive even index, whether or not it has a factor 17 or 41.

These are restrictions, not a classification of all remaining possibilities. For example, F_1(2)-1=1 is a square, so removing the index conditions would produce a false theorem.

The primes 17 and 41 have two linked roles: they are fixed recurrence indices and the degrees of two number fields used in the proof. They are not asserted to be the odd primes detecting every nonsquare locally.

Article: Theorems 1.1–1.3.

5. The first obstruction: count factors of two

A square contains each prime an even number of times. The notation v_2(m) counts the factors of 2 in a nonzero integer m. For example, 40=2^3\cdot5, so v_2(40)=3.

The even-index theorem gives an exact count:

when A\equiv2\pmod4 and N>0 is even. Its right side is odd, which is incompatible with an integer square.

The reason is a controlled doubling pattern. The recurrence has the identity

It follows by putting \alpha=A+\sqrt{A^2+1} and \beta=A-\sqrt{A^2+1}, observing that \alpha\beta=-1, and using F_n(A)=(\alpha^n+\beta^n)/2.

For odd n, the doubling identity gives

Since F_n(A)\equiv2\pmod4 at odd indices, its valuation is exactly one. The first doubling therefore gives valuation 1+2=3.

For even n, it instead gives

At even indices F_n(A)\equiv1\pmod4, so the final factor contributes exactly one further factor of 2. Each subsequent doubling increases the valuation by two.

Writing N=2^s m with m odd gives one initial valuation of 3 and s-1 further increments of 2: 3+2(s-1)=2s+1.

At A=2, the shifted terms at indices 2,4,8 are 8,160,51840, with valuations 3,5,7. The proof explains that pattern at every height; it is not an extrapolation from those examples.

Article: §2.2–§2.3 and Theorem 1.3.

6. Odd indices: what survives at the prime two

For odd indices the shifted value is odd. Every odd integer square is 1 modulo 8, whereas 5 modulo 8 is impossible. Applying the recurrence modulo 8 gives:

The surviving rows do more than pass one small test. An odd integer congruent to 1 modulo 8 is a square in \mathbb Z_2, the ring of 2-adic integers. This means that its square roots can be chosen consistently through all powers of 2. Going from modulus 8 to 16, 32 or higher powers cannot eliminate it merely as a nonsquare at 2.

That does not mean it is an integer square, or that it passes every odd-prime test. The article gives

and 20 is not a square residue modulo 23.

Both fixed indices 17 and 41 are 1 modulo 4. Thus A\equiv6\pmod8 is elementary, while A\equiv2\pmod8 is the hard fixed-index phase. The next construction handles that entire parameter class uniformly.

Article: Proposition 2.1 and §2.4. The complete 2-adic square criterion uses strong Hensel; see Conrad, Theorem 9.1, also cited in the article.

7. Turn a sequence term into a polynomial norm

The recurrence can be expressed through the Dickson polynomials

Here T is a polynomial variable. Direct comparison gives D_n(2A)=2F_n(A).

Fix p\in\{17,41\} and define

Then the square equation becomes

In the hard phase, x\ge4 and x\equiv4\pmod{16}.

The polynomial g_p is monic, meaning that its leading coefficient is 1, and is irreducible over the rationals. It has degree p and constant term -2. Adjoin a root \theta_p to the rational numbers to form

This is a number field: arithmetic in which expressions involving the root are legitimate numbers. Every element can be expressed using rational coefficients and the basis 1,\theta_p,\ldots,\theta_p^{p-1}.

The payoff is a norm identity. The field norm multiplies the images of an element under all embeddings of the field. If \theta_1,\ldots,\theta_p are the roots of g_p, then

The constant term gives \operatorname{Norm}(\theta_p)=2. Therefore

Under a hypothetical solution, this norm is y^2. Dividing by \theta_p removes exactly the unwanted factor of two.

The order of subtraction is deliberate. In odd degree, replacing x-\theta_p by \theta_p-x reverses the norm’s sign. The proof keeps the orientation that produces the positive square.

Article: §2.1, §3.3 and §4.1.

8. Why a square norm is not enough

A product can be a square even when its individual factors are not: 3\cdot12=36. Likewise, a square field norm does not automatically imply even valuation at every prime of the number field. Contributions from several primes above one rational prime could add up to an even number.

The proof prevents that ambiguity before transferring parity.

The exact field calculations establish

where \mathcal O_{L_p} is the ring of algebraic integers in the field. In words, every algebraic integer is an integer polynomial in the specified root. The specified root generates the entire integer ring; no additional algebraic integers are missing from its power order.

Temporarily write g=g_p and \theta=\theta_p. Imposing the relation \theta=x gives

The quotient notation means arithmetic after setting the element in parentheses equal to zero. Once \theta=x, its defining equation g(\theta)=0 becomes g(x)=0.

This identifies exactly one prime ideal \mathfrak P above each rational prime \ell that supports x-\theta. Its residue field is \mathbb F_\ell, so its residue degree is one. Other primes above \ell may exist; they simply do not divide this particular element.

Since \theta has norm 2, it is a unit at all odd primes. The norm valuation is therefore concentrated at that single supporting prime:

Here v_{\mathfrak P} counts prime-ideal divisibility, while v_\ell counts rational-prime divisibility. If the norm is y^2, the right side is even. At primes outside the support the valuation is zero.

The result is exactly what is needed: every odd-prime valuation is even. It is not the stronger, unwarranted assertion that \delta_x is a square in every local field.

Article: §3.3, equations (3.5)–(3.6).

9. The detector changes a sign without changing the support

A unit can be multiplied into an algebraic integer without introducing a new prime-ideal factor. The proof uses that freedom to change a quadratic character.

For each fixed field it supplies an explicit polynomial U_p and forms

The subscripted u_p is an algebraic unit in the degree-p field. It is distinct from the integer Pell coefficient u introduced earlier. Also, U_p here names a detector polynomial, not a Lucas U-sequence.

Exact computation gives

The resultant is an exact polynomial calculation; for monic g_p it is the product of U_p evaluated at the roots of g_p. Norm -1 makes u_p a unit. Consequently b_p has the same principal ideal as \theta_p, and remains a unit at every odd prime.

For degree 17, the choice has the compact form

The degree-41 choice has a longer exact coefficient list, printed in article equations (4.7)–(4.8). Its constant coefficient multiplies 1, not D_0=2.

Why not use b_p=\theta_p alone? Proposition 4.1 proves that its dyadic reference product is +1. The chosen unit changes that product to -1 while preserving odd-prime unitness. This is the specific job of the detector.

Norm -1 alone is not the detecting condition. The unit -1 also has norm -1 in an odd-degree field, but has trivial reference character here. The successful units supply a particular sign, not just a particular norm.

Article: Proposition 4.1 and §§4.2–4.4.

10. Hilbert reciprocity links the local signs

Ordinary real arithmetic is one way to examine a number. Arithmetic organised around divisibility by a prime gives another. A completion makes one such notion of closeness into a complete number system. In a number field, the different real, complex and prime-adic viewpoints are called its places. Places above 2 are called dyadic.

A quadratic Hilbert symbol attaches a sign to a pair of nonzero numbers at one place:

We use w for a place here, to distinguish it from the Pell coefficient v. The symbol is a local quadratic compatibility test. In its norm interpretation, +1 means that a is a norm from the local quadratic extension obtained by adjoining \sqrt b; when b is already a square, the symbol is automatically +1. It is not simply a test of whether a itself is a square.

Hilbert reciprocity says that when a,b belong to the original number field, all their local signs satisfy

The product covers every place. Only finitely many factors differ from +1, so it is well-defined. This is a compatibility law for the same pair of global numbers, not a rule saying that locally soluble equations always have a global solution.

Apply the pairing to (\delta_x,b_p). Under the square hypothesis, the odd-prime valuations of \delta_x are even, and b_p is a unit there. The standard odd-prime Hilbert formula then gives +1 at every odd finite place.

There is exactly one real embedding in each fixed field. Its distinguished root satisfies 0<\theta_{p,\mathbb R}<2, while x\ge4. Hence

which makes the real Hilbert symbol +1. Complex Hilbert symbols are automatically +1.

Thus a hypothetical square has already forced every non-dyadic sign to be positive. The remaining question is the product of the dyadic signs.

Article: Proposition 3.1 and §§3.1–3.4. Classical background: Milne, Class Field Theory, Chapter III §4 and Chapter VIII §5.10.

11. Why one reference calculation controls every parameter

A finite calculation at one integer would normally say little about infinitely many others. Here there is a theorem explaining exactly what stays unchanged.

Use the reference x=4 and put

For any admissible x, direct division gives

At a dyadic prime \mathfrak P, define e=v_{\mathfrak P}(2) and d=v_{\mathfrak P}(4-\theta_p). The valuation is normalised to count one copy of the prime ideal as one. Since 16\mid x-4,

The exact field data give (e,d)=(1,1) at one dyadic prime and (e,d)=(2,0) at the other two. Therefore the respective lower bounds are

or, uniformly, 4e-d>2e.

Strong Hensel’s lemma implies that 1+t is a local square when v_{\mathfrak P}(t)>2v_{\mathfrak P}(2). To see the threshold, apply the lemma to f(Z)=Z^2-(1+t) at Z=1: its value is -t, and its derivative is 2.

The strict inequality matters. In \mathbb Q_2, 5=1+4 lies exactly on the excluded boundary and is not a square. The argument requires divisibility beyond that boundary, not equality with it.

Consequently \delta_x/\delta_4 is a square at every dyadic place. Two numbers whose quotient is a square have the same squareclass, and Hilbert symbols depend only on squareclasses. Thus

The reference is allowed to be a nonsolution. Its role is to determine the dyadic character shared by the entire progression. When x=4 itself, the quotient is exactly 1, so that endpoint is included too.

Article: §3.5 and Table 4. Classical background: Conrad, Hensel’s Lemma, Theorem 9.1.

12. The finite arithmetic and the final contradiction

The fixed fields have the following certified structure. A signature (r,s) means r real embeddings and s conjugate pairs of complex embeddings, accounting for degree r+2s.

The residue degree f describes the size of the local residue field; over 2 it has 2^f elements. The number e is the ramification index, equal here to the valuation of 2. The sum of the products ef accounts for the field degree.

The discriminant is an exact integer associated with the polynomial or field. The square of the power-order index divides the polynomial discriminant, so only 2 and p could obstruct the assertion that \mathbb Z[\theta_p] is the full ring of integers. Appendix A.2 checks both primes using Dedekind’s index criterion. Its polynomial congruences prove index one, and then determine the dyadic prime types and valuations used above. No unsupported maximal-order assumption is hidden inside the argument.

The negative reference product can also be recovered without a direct dyadic-symbol routine. Set

For degree 17,

The two odd-prime reference signs are -1 and +1, giving product -1. For degree 41,

and the five signs are +1,-1,-1,-1,+1, again with product -1. They are ordinary quadratic-residue calculations at the listed rational primes, using the residue 4U_p(4).

The reference has positive infinite signs. Reciprocity therefore requires its dyadic product to be -1 as well: the two negative products multiply to +1.

Now suppose an admissible square solution existed. Its odd product would have to be +1, by the square-norm argument. Its dyadic product would remain -1, by Hensel stability. Its infinite product would be +1.

This explains why reciprocity can be used in both the reference calculation and the contradiction without circularity. The reference is not assumed to satisfy the square condition. A hypothetical solution would impose an additional condition that is incompatible with its inherited dyadic sign.

That proves the hard phase at both fixed indices. Combining it with the elementary 6\bmod8 phase proves the fixed-17 and fixed-41 exclusions for all A\equiv2\pmod4.

Article: Theorem 4.2, Appendix A and §5.1. The displayed field and character values are finite certificate inputs, not numerical estimates.

13. The Jacobi identity is the arithmetic shadow

The same obstruction can be written using ordinary integers. For a positive odd denominator

the Jacobi symbol is

Here \ell ranges over the prime factors of Q, a_\ell is its exponent, and the Legendre symbol (c/\ell) is +1 or -1 according to whether a nonzero residue c is a square or nonsquare modulo \ell. If \ell\mid c, the symbol is zero; the detector construction proves coprimality and prevents that case.

If Q is a square and \gcd(c,Q)=1, every a_\ell is even. Every factor contributes +1, so J(c,Q)=+1. A positive Jacobi value is not a proof that the denominator is square; the argument uses the opposite implication, that a negative value rules it out.

The paper instead proves

for every nonnegative A\equiv2\pmod8 and p\in\{17,41\}. This identity does not assume that the denominator is a square. It is an unconditional certificate excluding that possibility.

The link to the number field is exact. Put x=2A and Q=g_p(x)/2. At the unique odd supporting prime above \ell, the reduction \theta_p\mapsto x turns the algebraic detector into xU_p(x). The local Hilbert formula gives

where the product is over odd finite places. The factor x has Jacobi value +1 here. Indeed, write x=4m with m\equiv1\pmod4; then Q\equiv-1\pmod m, and quadratic reciprocity gives

The case m=1 uses J(c,1)=1. Thus the odd Hilbert product is exactly J(U_p(x),Q). Its value is -1 because the dyadic product is -1 and the infinite product is +1.

This is why “arithmetic shadow” is useful language. It is the same obstruction expressed in integers, not a further independent vote against the candidate. Lean’s integer-Jacobi argument provides a formal arithmetic route to the exclusions; a few numerical evaluations of the identity would only be checks.

The identity also clarifies the local story. A negative product forces at least one odd prime factor of Q to occur to an odd exponent. That prime makes Q a nonsquare locally. Its identity can vary with A.

At p=17,A=2, for example, F_{17}(2)-1=13\cdot1751444197 has exactly one factor of 13. The specialised square equation already fails at 13. The theorem’s strength is a fixed certificate that works for every parameter, without first finding and factoring its individual obstruction.

Article: Proposition 7.1 and §§7.1–7.2.

14. Why two fixed indices exclude infinitely many indices

The sequence has a composition identity: