One candidate.Several constraints.

A solution has to be the same integer in every description of it. One description says X is one more than a square. Another says X is a coefficient in a power of a quadratic unit, which makes it a term of a recurrence. Elementary arithmetic rules out some recurrence indices at once. For two further families of indices, an explicit algebraic construction forces a sign that a square cannot carry.

The proof works by making the descriptions meet, not by checking ever larger integers.

Three consecutive integers

An integer is powerful when every prime in its factorisation appears at least twice. 25 = 5², 27 = 3³ and 72 = 2³·3² are powerful; 26 = 2·13 is not. Whether three consecutive integers can all be powerful is an open question.

Paper I works inside one corner of it, the square-middle case (X² − 1, X², X² + 1), where the middle is automatically powerful, and adds three requirements:

The last one is the first description of the candidate: X = y² + 1. The paper obtains its restriction without assuming that the lower neighbour is powerful; a genuine triple would have to satisfy that too.

Article: §1.1 and Corollary 1.5.

A second description of the same integer

Every powerful number is uniquely K³V² with K squarefree. If the upper neighbour is powerful,

a negative Pell equation. In the quadratic arithmetic ℤ[√K] the element η = X + KV√K has norm −1, so it is a unit. Under the paper's hypotheses (K > 1, K ≡ 5 mod 8) every positive unit is a power of a least unit ε = u + v√K, hence

N is odd because both units have norm −1 and an even power would have norm +1. That exponent is where the whole story ends up. The order ℤ[√K] is chosen deliberately: a generator from a larger ring could change the exponent.

Article: §6.2, equations (6.2) to (6.4).

Multiplying a unit produces a recurrence

Write εⁿ = X_n + Y_n√K. From u² + 1 = Kv² one gets ε² = 2uε + 1, so the first coefficients obey X_n+2 = 2u·X_n+1 + X_n. Paper I studies the whole family

where A selects a sequence and n a term; the Pell coefficient is X_n = F_n(u). For A = 2 the terms run 1, 2, 9, 38, 161, 682, … The two descriptions now meet in one identity:

The square witness, the Pell coefficient and the recurrence term are the same integer. So the question becomes: for which A and N can F_N(A) − 1 be a square at all?

Article: Corollary 1.4 and §§6.1 to 6.2.

What the recurrence theorem says

These are restrictions, not a classification of what remains. F₁(2) − 1 = 1 is a square, so the index conditions cannot simply be dropped. The primes 17 and 41 play two linked roles: as fixed recurrence indices and as the degrees of two number fields used in the proof. They are not claimed to be the primes that locally detect every non-square.

Article: Theorems 1.1 to 1.3.

Count the factors of two

A square contains every prime an even number of times. Write v₂(m) for the number of factors of 2 in m; for instance v₂(40) = 3. For A ≡ 2 (mod 4) and even N > 0 the paper proves an exact count:

The right side is odd, so F_N(A) − 1 cannot be a square. The mechanism is the doubling identity F_2n(A) = 2F_n(A)² − (−1)ⁿ: the first doubling from an odd index contributes valuation 3, and every further doubling adds exactly 2. At A = 2 the shifted terms at indices 2, 4, 8 are 8, 160, 51 840, with valuations 3, 5, 7, and the proof explains that pattern at every height rather than extrapolating from it.

Article: §2.2 to §2.3 and Theorem 1.3.

What survives modulo 8

For odd N the shifted term is odd, and an odd square is always 1 modulo 8. Running the recurrence modulo 8 leaves two surviving phases, (A mod 8, N mod 4) = (2, 1) and (6, 3); the other two land on 5 and are excluded.

Surviving is weaker than it sounds and stronger than it looks. Stronger: an odd number that is 1 modulo 8 is a square in the 2-adic integers, so no higher power of two will ever rule it out. Weaker: it says nothing about odd primes: F₄₁(2) − 1 ≡ 20 (mod 23), and 20 is not a square modulo 23.

Both fixed indices are 1 modulo 4. So for N = 17 or 41 the phase A ≡ 6 (mod 8) is settled right here, and A ≡ 2 (mod 8) is the hard phase the rest of the proof must handle.

Article: Proposition 2.1 and §2.4; the full 2-adic criterion is strong Hensel (Conrad, Theorem 9.1).

Signs that cannot all agree

A term becomes a norm. Fix p ∈ {17, 41} and put x = 2A. Through the Dickson polynomials, F_p(A) − 1 = y² becomes g_p(x) = 2y² for a monic irreducible polynomial g_p of degree p with constant term −2. Adjoining a root θ_p gives a number field L_p in which g_p(x)/2 is the norm of δ_x = (x − θ_p)/θ_p. A hypothetical solution makes that norm a perfect square, y².

A square norm is not a square element. 3·12 = 36 is a square although neither factor is. So the paper first proves that ℤ[θ_p] is the whole ring of integers and that above each odd prime exactly one prime ideal, of residue degree one, divides x − θ_p. The norm's valuation is then concentrated in one place, and every odd-prime valuation of δ_x is even.

A detector and a product law. The paper supplies an explicit unit u_p = U_p(θ_p) of norm −1 and pairs δ_x with b_p = θ_pu_p through quadratic Hilbert symbols, one sign (a, b)_w = ±1 for each place w of the field. Hilbert reciprocity says all those signs multiply to +1. Under the square hypothesis every odd-prime sign is +1 (even valuations, unit partner), the real sign is +1 (0 < θ_p,ℝ < 2 ≤ x), and complex signs are trivially +1. Everything hinges on the dyadic places, above 2.

One reference controls every parameter. Take x = 4 as a reference. Then δ_x/δ₄ = 1 + (x − 4)/(4 − θ_p), and because 16 | x − 4 this quotient is a square at every dyadic place by Hensel's lemma; the exact field data give the margins 3 > 2 and 8 > 4, strict as the lemma requires. So δ_x and δ₄ have the same dyadic signs for every admissible x. For the reference, the dyadic product is computed to be −1. A hypothetical solution inherits that −1, has +1 everywhere else, and so has total −1. Reciprocity forbids it.

Article: §§3 to 4, Proposition 3.1, Proposition 4.1, Theorem 4.2, Appendix A. Background: Milne, Class Field Theory; Conrad, Hensel's Lemma.

The same obstruction as a Jacobi symbol

The argument can be written without number fields. For odd Q the Jacobi symbol J(c, Q) multiplies Legendre symbols over the prime factors of Q; if Q is a square coprime to c, every factor appears an even number of times and J = +1. The paper proves

for every A ≡ 2 (mod 8) and p ∈ {17, 41}. A negative value forces some odd prime to divide F_p(A) − 1 an odd number of times, so it is not a square. Which prime does the work can change with A; the certificate does not need to find it.

This is the Hilbert argument reduced to integers, not a second, independent piece of evidence: at the odd supporting primes the algebraic detector reduces to xU_p(x), and the factor x contributes +1. It is also the route the Lean development formalises. For degree 17 the detector is compact: U₁₇(T) = −1 − D₂ + D₃ + D₄ − D₅ − 2D₁₂ + D₁₄, in Dickson polynomials of T.

Article: Proposition 7.1 and §§7.1 to 7.2; the degree-41 coefficients are equations (4.7) and (4.8).

Two fixed indices reach infinitely far

The family satisfies a composition identity,

The odd inner index keeps the conjugate product at −1; for even inner indices it fails: F₄(2) = 161 but F₂(F₂(2)) = 163. Now take odd N with p | N, write N = pm and set B = F_m(A). Then B ≡ 2 (mod 4) and F_N(A) = F_p(B), which the fixed-index theorem already excludes.

The mod-8 phase must be read off B, not A: F₅₁(6) = F₁₇(882), and 882 ≡ 2 (mod 8) sends an elementary-looking parameter into the hard phase. Even N are handled by the valuation count, so the two branches together prove the full theorem.

Article: Lemma 5.1 and §5.3.

What this says about the triple

For the selected powerful-number family the normal form was y² + 1 = X = F_N(u) with u ≡ 2 (mod 4) and odd N > 0. The recurrence theorem therefore forces

neither 17 nor 41 divides the Pell exponent. The upper-powerful condition adds a divisibility rank R = K/gcd(K, v) with N = R·E, so the same conclusion reads 17 ∤ E, 41 ∤ E and 41 ∤ R.

These are necessary conditions. Avoiding 17 and 41 does not build a candidate; it removes some. Other square-middle families, and the full three-consecutive problem, are not decided by this result.

Article: Corollary 1.5 and §6.3.