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]
2. The starting members \(n\) are listed in OEIS A060355.
3. The first question has a negative answer:
\[ 12167=23^3,\qquad 12168=2^3\,3^2\,13^2. \]
[a] for the displayed arithmetic.
4. 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]
5. 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.
2. 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.
3. 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”](https://www.renyi.hu/~p_erdos/1976-39.pdf),
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
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”](https://www.fq.math.ca/Scanned/14-2/walker.pdf),
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
\[ 7(2637362)^2=3(4028637)^2+1. \]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
\[ S(x)=\#\{n\le x:n\ \text{and}\ n+1\ \text{are powerful}\}. \]The primary-source progression is:
- Chan,
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,
improve the unconditional exponent to every
\(\gamma>61/180\). [b: their Theorem 3]
- Reuss,
[“Pairs of \(k\)-free Numbers, consecutive square-full
Numbers”](https://arxiv.org/abs/1212.3150), 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”](https://mast.queensu.ca/~murty/fundamental-units.pdf),
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”](https://arxiv.org/abs/2405.06968),
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
\[ \boxed{n=a^2b^3,\qquad b\ \text{squarefree}.} \tag{1} \]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
\[ b=\prod_{e\ {\rm odd}}p,\qquad a=\prod_{e\ {\rm even}}p^{e/2} \prod_{e\ {\rm odd}}p^{(e-3)/2}. \]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:
\[ n=a^2b^3,\qquad n+1=c^2d^3, \]where \(b,d\) are squarefree. Then
\[ c^2d^3-a^2b^3=1. \tag{2} \]Setting \(X=cd\) and \(Y=ab\) gives
\[ \boxed{dX^2-bY^2=1,\qquad d\mid X,\quad b\mid Y.} \tag{3} \]Conversely, any positive solution of (3), with squarefree \(b,d\), gives
\[ n=bY^2,\qquad n+1=dX^2. \]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 \(\gcd(n,n+1)=1\),
\[ \boxed{n,n+1\text{ powerful}\iff n(n+1)\text{ powerful}.} \tag{4} \]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
\[ X\sqrt d+Y\sqrt b\le2\sqrt{x+1}, \]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
\[ (3+2\sqrt2)^k=U_k+V_k\sqrt2. \]Then \(U_k^2-2V_k^2=1\). The recurrence
\[ U_{k+1}=3U_k+4V_k,\qquad V_{k+1}=2U_k+3V_k \]and \(V_1=2\) show that every \(V_k\) is even. Hence
\[ 2V_k^2,\quad U_k^2 \]are consecutive powerful numbers. Since
\(2V_k^2<(3+2\sqrt2)^{2k}\),
\[ \boxed{ S(x)\ge \left\lfloor\frac{\log x}{2\log(3+2\sqrt2)}\right\rfloor } \]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
Exhaustive algorithm and correctness
For \(L=X+1\), (1) implies that every powerful number at most \(L\) occurs
exactly once among
\[ b^3a^2,\qquad 1\le b\le\lfloor L^{1/3}\rfloor,\quad b\text{ squarefree},\quad 1\le a\le\left\lfloor\sqrt{L/b^3}\right\rfloor. \tag{5} \]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 \(10^6\).
2. Every endpoint in the \(10^{16}\) output is factored afresh by a complete
trial-division prime table.
3. Those factors reconstruct the canonical \(a,b,c,d\), verify (2) and (3),
and determine square/non-square status using isqrt.
4. 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
\[ \begin{array}{c|rrrrrrrrrrrrrrrr} k&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16\\ \hline S(10^k)&1&1&3&4&5&8&9&10&12&14&15&18&19&24&25&26 \end{array} \][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:
| \(n\) | \(a\) | \(b\) | \(c\) | \(d\) |
|---:|---:|---:|---:|---:|
| 12167 | 1 | 23 | 39 | 2 |
| 5425069447 | 3977 | 7 | 26041 | 2 |
| 11968683934831 | 4927 | 79 | 31364 | 23 |
| 28821995554247 | 48671 | 23 | 1898091 | 2 |
| 48689748233307 | 1342879 | 3 | 376766 | 7 |
[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
\[ \begin{aligned} M_x(B,D)=\#\{(a,b,c,d)\in\mathbb N^4:\;& B\le b<2B,\ D\le d<2D,\\ &b,d\text{ squarefree},\ a^2b^3\le x,\ c^2d^3-a^2b^3=1\}. \end{aligned} \tag{6} \]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,
\[ a\asymp x^{1/2}B^{-3/2},\qquad c\asymp x^{1/2}D^{-3/2}, \]and the equation is the four-variable integer-point problem
\[ d^3c^2-b^3a^2=1. \tag{7} \]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
\[ e^3v^2-d^3u^2=1 \]in dyadic boxes. In his proof, writing \(DE=x^\psi\), the optimized exponent
is maximized at
\[ \boxed{\psi=2/5,\qquad f(2/5)=29/100.} \]Thus the current method loses its positive power in the balanced
coefficient regime
\[ \boxed{BD\asymp x^{2/5}}. \tag{8} \]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
\[ \#\{m\le X:m\text{ powerful}\} \sim\frac{\zeta(3/2)}{\zeta(3)}\sqrt X \approx2.173\sqrt X. \][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 \(BD\asymp x^{2/5}\).