Erdős problem #365 — wave 8w

Live-page access and research date: 2026-07-28 UTC.

Outcome: partial progress, not a solution. The mandatory live-page gate passed. The first question on the page is already known to have a negative answer; the remaining polylogarithmic counting question stays open. The concrete contribution here is an exhaustive, independently guarded computation through \(10^{16}\), together with an exact reduction that identifies the dyadic integer-point estimate missing from current methods.

Claim labels:

• [a] elementary-rigorous — proved directly below;
• [b] rigorous-modulo-named-theorem — relies on the theorem explicitly

named and linked;

• [c] plausible/structural-unverified — heuristic or a non-exhaustive

literature-search conclusion;

• [d] computational-only — finite deterministic computation, with no

asymptotic inference.

No computational list or heuristic is treated as a theorem about all \(x\).

0. Mandatory live-page gate

I used the Bright Data cloud-browser route to read:

• the live problem page;
• its live LaTeX-source view;
• the live discussion thread.

The page showed:

• status: OPEN;
• claimed proofs: 0;
• “Currently working on this problem”: None;
• “Interested in collaborating”: None;
• all other interest/formalisation markers: None;
• formalised statement: No;
• related OEIS sequences: A060355, A060859, A175155;
• last problem-text edit: 31 October 2025;
• three comments.

Thus none of the required stop conditions was present.

Verbatim current statement

Copied from the live LaTeX-source route:

Do all pairs of consecutive powerful numbers \(n\) and \(n+1\) come from solutions to Pell equations? In other words, must either \(n\) or \(n+1\) be a square?

Is the number of such \(n\leq x\) bounded by \((\log x)^{O(1)}\)?

Known results stated on the page

The following are transcriptions of what the live page says, not assertions that its bibliography is complete.

1. Erdős asked Mahler whether infinitely many consecutive powerful pairs

exist. Mahler used the infinitely many solutions of \(x^2=2^3y^2+1\). [b: Pell equation]

1. The starting members \(n\) are listed in OEIS A060355.
2. The first question has a negative answer:

\[ 12167=23^3,\qquad 12168=2^3\,3^2\,13^2. \] [a] for the displayed arithmetic.

1. Walker proved that

\[ 7^3x^2=3^3y^2+1 \] has infinitely many solutions, giving infinitely many pairs in which neither endpoint is square. [b: Walker 1976]

1. The page points to problem #364 and problem B16 of Guy's collection.

All live comments

The site warns that comments are user-supplied and unverified.

1. TerenceTao, 28 March 2026. The conjecture also appears on p. 234 of

Ribenboim's My Numbers, My Friends. The comment says the 2017 Aktaş–Murty \(O(x^{2/5})\) bound is the best known.

1. Alfaiz, 16 October 2025. This points to Sentance's work on pairs of

odd powerful numbers at distance \(2\), and to McDaniel's paper, which the comment says disproves Golomb's conjecture that infinitely many integers cannot be written as a difference of two powerful numbers.

1. StijnC, 16 October 2025. This explains that Sentance's distance-\(2\)

result says every powerful pair \(n-1,n+1\) is associated with a Pell equation, sketches why \(n^2-1\) has the required powerful form, and stresses that this does not address the distance-\(1\) problem here.

None is a claimed proof or a current-worker marker. The best-bound assertion in the first comment is stale; primary sources below show why.

1. Primary-source literature audit

Original and classical sources

Erdős's original source is “Problems and results on number theoretic properties of consecutive integers and related questions”, Proceedings of the Fifth Manitoba Conference on Numerical Mathematics (1976), pp. 25–44. On p. 31, Erdős asks for a \((\log x)^c\) upper bound and asks whether infinitely many examples fail to come from Pellian equations. [b: direct inspection of the primary source]

Golomb's paper is “Powerful Numbers”, American Mathematical Monthly 77 (1970), 848–852. The live page's numerical counterexample was also multiplied and factored afresh by the verifier. [a]

Walker, “Consecutive integer pairs of powerful numbers and related Diophantine equations”, Fibonacci Quarterly 14 (1976), 111–116, develops the generalized equation \(mX^2-nY^2=\pm1\). Its final example raises \(2\sqrt7+3\sqrt3\) to the seventh power and obtains

Because \(7\mid2637362\) and \(3\mid4028637\), this is the page's \(7^3x^2=3^3y^2+1\) family after rescaling. Walker proves that the relevant odd powers give infinitely many such solutions. [b: Walker's theorem]

Counting bounds

Write

The primary-source progression is:

• Chan,

“Twin squareful numbers”, J. Aust. Math. Soc. 93 (2012), Theorems 3, 4, and 6, proves \[ S(x)\ll x^{2/5}(\log x)^2,\qquad S(x)\ll x^{7/19}\log x, \] and, assuming \(abc\), \(S(x)\ll_\epsilon x^\epsilon\). [b: Chan's theorems]

• Blomer and Schöbel,

“Twins of powerful numbers”, improve the unconditional exponent to every \(\gamma>61/180\). [b: their Theorem 3]

• Reuss,

“Pairs of \(k\)-free Numbers, consecutive square-full Numbers”, arXiv:1212.3150v2, Theorem 4, proves \[ \boxed{S(x)\ll_\epsilon x^{29/100+\epsilon}}. \] [b: Reuss's approximate-determinant theorem]

• Aktaş and Murty,

“Fundamental units and consecutive squarefull numbers”, Int. J. Number Theory 13 (2017), prove the elementary \(O(x^{2/5})\) bound and an \(abc\)-conditional \(O_\epsilon(x^\epsilon)\) bound. Crucially, their own introduction explicitly records Reuss's stronger \(29/100+\epsilon\) exponent. [b]

• Wongcharoenbhorn and Meemark,

“Square-full values of quadratic polynomials”, arXiv:2405.06968v2 (2025), now Bull. Aust. Math. Soc. 113 (2026), pp. 40–54, again state that their \(29/100+\epsilon\) result in the relevant special setting is the same as Reuss's bound for consecutive square-full pairs. [b]

Therefore the March 2026 comment's claim that \(2/5\) is best is false: \(29/100=0.29<0.4\). The recent 2025/2026 primary paper still recognizes the Reuss exponent. [b]

I also searched exact title/statement phrases, arXiv-restricted variants, post-2014 citations of Reuss, and 2025–2026 papers using “consecutive powerful/square-full”. The 2025 Chan paper concerns three consecutive powerful numbers, and the 2026 van Doorn paper concerns arithmetic progressions of powerful numbers; neither improves \(S(x)\). I found no primary source with an exponent below \(29/100\), a polylogarithmic bound, or a claimed proof of this problem. This is an honest search result, not proof that no uncatalogued result exists. [c]

Finally, the \(abc\)-conditional statement \(S(x)\ll_\epsilon x^\epsilon\) does not imply \(S(x)\ll(\log x)^A\): its implicit constant depends on each fixed \(\epsilon\), with no uniform control as \(\epsilon\to0\). [a]

2. Elementary exact reductions

Unique square–cube representation

Lemma. Every positive powerful integer has a unique representation

Proof. If \(p^e\Vert n\), then \(e\ge2\). If \(e\) is even, put all of \(p^e\) into \(a^2\). If \(e\) is odd, then \(e\ge3\), put \(p^3\) into \(b^3\), and put \(p^{e-3}\) into \(a^2\). Thus

The parity of every prime exponent uniquely determines whether that prime occurs in \(b\), proving uniqueness. \(\square\) [a]

Exact generalized-Pell bijection

Apply (1) canonically to both endpoints:

where \(b,d\) are squarefree. Then

Setting \(X=cd\) and \(Y=ab\) gives

Conversely, any positive solution of (3), with squarefree \(b,d\), gives

The divisibilities in (3) make both integers powerful. Canonical uniqueness in (1) shows that this is a bijection, not merely a necessary condition. Also \(\gcd(b,d)=1\), since any common divisor would divide the left side of (3). [a]

Equivalently, because \(\gcd(n,n+1)=1\),

Thus the problem is exactly the square-full-value problem for the quadratic polynomial \(f(t)=t(t+1)\). [a]

For fixed \(b,d\), Walker's generalized-Pell theorem implies only \(O(\log x)\) positive solutions with \(bY^2\le x\). Indeed

and Walker expresses successive positive solutions as odd powers of the least one. The divisibility conditions merely select a subsequence. [b: Walker]

A rigorous logarithmic lower bound

Let

Then \(U_k^2-2V_k^2=1\). The recurrence

and \(V_1=2\) show that every \(V_k\) is even. Hence

are consecutive powerful numbers. Since \(2V_k^2<(3+2\sqrt2)^{2k}\),

whenever the right side is positive. [a] This confirms that logarithmic growth is genuinely present; it supplies no upper bound.

3. Exact computation through \(10^{16}\)

The standalone verifier is erdos365_wave8w_reverify.py.

Exhaustive algorithm and correctness

For \(L=X+1\), (1) implies that every powerful number at most \(L\) occurs exactly once among

The program marks squarefree \(b\) by multiples of integer squares, emits all values in (5) as exact unsigned 64-bit integers, sorts them, asserts strict uniqueness, and takes adjacent differences equal to one. Generating through \(X+1\) handles a possible pair whose larger endpoint is exactly \(X+1\). Equations (1) and (5) prove finite exhaustiveness. [a]

The following checks are separate from the main generation path:

1. An SPF table factors every integer through \(10^6+1\), tests the

powerful definition literally, and reproduces both the powerful-number count and all eight pairs through \(10^6\).

1. Every endpoint in the \(10^{16}\) output is factored afresh by a complete

trial-division prime table.

1. Those factors reconstruct the canonical \(a,b,c,d\), verify (2) and (3),

and determine square/non-square status using isqrt.

1. A hard-coded expected list is used only as a regression assertion after

recomputation; it is not an input to enumeration.

Thus the large-range completeness uses the elementary lemma, while an implementation-independent literal scan guards the code on a smaller full interval. [a,d]

Reproduction:

python runs/erdos365_wave8w_reverify.py

The clean run printed VERIFIED and used 28.80 seconds wall time (22.84 user, 7.43 system) with peak RSS 3,431,996 KB on this VM. [d]

Verified table

[d]

The exact 26 starting values \(n\le10^{16}\) are:

8
288
675
9800
12167
235224
332928
465124
1825200
11309768
384199200
592192224
4931691075
5425069447
13051463048
221322261600
443365544448
865363202000
8192480787000
11968683934831
13325427460800
15061377048200
28821995554247
48689748233307
511643454094368
1558709801289000

[d] No claim of novelty is made for terms already present in OEIS; the point is that this table is recomputed exhaustively rather than trusted.

There are exactly five pairs in this range for which neither endpoint is a square. Their independently reconstructed canonical certificates \(n=a^2b^3,\ n+1=c^2d^3\) are:

[d] The last row is Walker's displayed family member. The other 21 pairs have exactly one square endpoint.

Additional reproducibility certificates:

powerful integers <= 10^16+1: 217004842
SHA256(sorted native uint64 array):
910455aec8191294aad15822b3ef45776d52b19a9e831fce5ae9e709ebe6341a
SHA256(newline-joined pair list):
d2ace3a2a76e7485c64aab255774e915eaec83b4f565cb8febd4956b9e048072
independent SPF check <= 10^6+1: 2027 powerful integers, 8 pairs

[d]

4. Exact analytic bottleneck

Partition the squarefree kernels in (2) into dyadic intervals. For powers of two \(B,D\), define

There are \(O((\log x)^2)\) possible dyadic boxes because \(b,d\le(x+1)^{1/3}\). Therefore the Erdős conjecture is equivalent, up to changing the fixed logarithmic exponent by \(2\), to a uniform polylogarithmic bound for every \(M_x(B,D)\). One direction follows because each box is at most \(S(x)\); the other follows by summing all boxes. [a]

Inside a box,

and the equation is the four-variable integer-point problem

Fixed-\((b,d)\) Pell theory gives \(O(\log x)\), but it does not bound how many coefficient pairs \((b,d)\) make (7) soluble. [a,b: Walker]

Reuss applies the approximate determinant method to precisely

in dyadic boxes. In his proof, writing \(DE=x^\psi\), the optimized exponent is maximized at

Thus the current method loses its positive power in the balanced coefficient regime

This is a statement about the proved Reuss optimization, not a claim that the actual solutions concentrate there. [b: Reuss, proof of Theorem 4]

The precise missing lemma is therefore:

Prove \(M_x(B,D)\ll(\log x)^A\), uniformly in all dyadic \(B,D\le x^{1/3}\), in particular when \(BD\asymp x^{2/5}\); or supply a comparably strong global argument showing that the sum of these critical boxes is polylogarithmic.

Neither the individual Pell bound, Chan's Thue-equation estimates, nor the current determinant method supplies this uniform coefficient-average saving. Reuss reaches \(x^{29/100+\epsilon}\); \(abc\) reaches \(x^\epsilon\) for each fixed \(\epsilon\), still without logarithmic uniformity. [b]

This also explains why extending the finite table is not a route to closure. The Erdős–Szekeres asymptotic, in the sharpened Bateman–Grosswald form, gives

[b: Erdős–Szekeres/Bateman–Grosswald] A direct representation-list sweep at \(X=10^{22}\) would therefore contain about \(2.17\times10^{11}\) 64-bit integers, need about 1.74 TB merely for the raw array, and perform roughly \(8\times10^{12}\) comparison-scale sorting operations. Extrapolating the verified \(10^{16}\) run gives an optimistic lower estimate around 10–20 core-hours and, realistically with multi-terabyte/external sorting, tens of core-hours. It would only extend a finite table and could not prove the required uniform bound, so it was not run. [b,d]

5. Verified state

• The live gate permits work: OPEN, no claimed proof, no worker. [d]
• The page's first question is already answered negatively and even has

infinitely many counterexamples by Walker. [b]

• The current unconditional literature bound located is

\(S(x)\ll_\epsilon x^{29/100+\epsilon}\), not \(x^{2/5}\). [b]

• The exhaustive computation gives \(S(10^{16})=26\), with exactly five

non-square/non-square pairs. [d]

• The remaining conjecture is reduced exactly to a uniform dyadic

integer-point estimate for (7), with the known determinant method stalling at the critical kernel product \(BD\asymp x^{2/5}\). [a,b]

PARTIAL: \(S(10^{16})=26\) is exhaustively certified (five pairs have neither endpoint square), the best located theorem is \(S(x)\ll_\epsilon x^{29/100+\epsilon}\), and the unresolved step is a uniform polylogarithmic bound for the critical dyadic boxes \(d^3c^2-b^3a^2=1\) with \(BD\asymp x^{2/5}\).