Erdős problem #936 — wave7u report
Date: 2026-07-28 (UTC)
Outcome: partial progress, not a solution. The live-page stop check passed. I
obtain an exact valuation criterion for both exponential variants, isolate a
specific open Wieferich-prime obstruction, and give replayable exact
classifications in finite ranges:
\[ \begin{array}{c|c|c} \text{family}&\text{range checked}&\text{powerful indices}\\ \hline 2^n-1&1\le n\le1000&\{1\}\\ 2^n+1&1\le n\le1000&\{3\}\\ n!+1&1\le n\le139&\{4,5,7\}\\ n!-1&2\le n\le139&\{2\}. \end{array} \]Here and below, a powerful number is a positive integer. Thus \(1\) is
powerful, while \(1!-1=0\) is outside the definition. This convention has no
effect on the eventual finiteness question.
0. Mandatory live-page check
I fetched both the main page and discussion thread through the Bright Data
browser on 2026-07-28. The direct sources were:
- <https://www.erdosproblems.com/936>
- <https://www.erdosproblems.com/latex/936>
- <https://www.erdosproblems.com/forum/discuss/936>
The following is the verbatim current statement from the live LaTeX page:
> Are
> \[2^n\pm 1\]
> and
> \[n!\pm 1\]
> powerful (i.e. if $p\mid m$ then $p^2\mid m$) for only finitely many $n$?
Live status and markers:
- Status: OPEN.
- Claimed proofs: 0.
- “Currently working on this problem”: None.
- “Interested in collaborating”: None.
- The page was last edited 31 October 2025.
- There were six live comments. None was a proof claim.
Therefore the mandatory stop condition did not fire.
The page lists two conditional results:
1. Cushing and Pascoe prove, assuming \(abc\), that for every fixed \(k\ge0\)
there are only finitely many pairs \((n,x)\) with \(x\) powerful and
\(\lvert x-n!\rvert\le k\).
2. P. A. CrowdMath proves the exponential question conditionally on \(abc\).
The comments, which the page itself warns are unverified, contain:
- an informal July 2026 summary of Wieferich/Zsigmondy and Wilson-quotient
restrictions, including an unsupported “63.3%” density computation;
- three comments only clarifying the OEIS “possible” marker;
- a comment pointing to the two \(abc\)-conditional papers;
- the examples \(2^1-1=1\), \(2^3+1=9\), \(2!-1=1\), and
\(4!+1=25,\ 5!+1=121,\ 7!+1=5041\).
I did not treat any comment as ground truth. The structural assertions used
below are rederived, and the finite claims are independently certificate
checked.
1. Claim labels
I use the requested labels throughout:
- (a) elementary-rigorous;
- (b) rigorous modulo the explicitly named theorem;
- (c) plausible/structural-unverified;
- (d) computational-only.
No computational list below is promoted to a theorem about all \(n\).
2. Primary-source and literature check
2.1 Original source
I downloaded and visually inspected page 32 of Erdős’s original paper:
P. Erdős, *Problems and results on number theoretic properties of consecutive
integers and related questions*, Proceedings of the Fifth Manitoba Conference
on Numerical Mathematics (1975), pp. 25–44,
<https://www.renyi.hu/~p_erdos/1976-39.pdf>.
The displayed source really has both signs:
\(Q_r(2^n\mathbin{\pm}1)\) and \(Q_r(n!\mathbin{\pm}1)\), followed by the
expectation that these can be powerful for only finitely many \(n\). This
also checks that the OCR loss of the minus bar in some search results is not
a change of statement.
2.2 The two cited conditional papers
I downloaded and checked the actual papers, not just search snippets.
- D. Cushing and J. E. Pascoe, Powerful numbers and the ABC-conjecture,
arXiv:1611.01192, <https://arxiv.org/abs/1611.01192>. Theorem 4.1 is exactly
the fixed-\(k\) statement
\[ x\text{ powerful},\qquad |x-n!|\le k \]
having only finitely many solutions, assuming \(abc\). This covers both
\(n!+1\) and \(n!-1\). (b: \(abc\))
- P. A. CrowdMath, Applications of the abc conjecture to powerful numbers,
arXiv:2005.07321, <https://arxiv.org/abs/2005.07321>. Theorem 2.3 states
that, for fixed coprime positive \(k,r\), only finitely many \(k^n+r\) are
powerful, assuming \(abc\). It directly covers \(2^n+1\). The minus case
follows by the same one-line \(abc\) radical estimate applied to
\((2^n-1)+1=2^n\): if \(z=2^n-1\) is powerful, then
\(\operatorname{rad}(2z)\le2\sqrt z\), which violates \(abc\) with any
exponent \(1+\epsilon<2\) for all sufficiently large \(z\). (b: \(abc\))
2.3 Additional directly relevant material found
A title/arXiv/full-text search and a citation-graph query found one additional
paper directly about the factorization of \(n!+1\):
W. Gerst, A conjecture on the prime factorization of \(n!+1\),
arXiv:1809.07360, <https://arxiv.org/abs/1809.07360>. Its Conjecture 2.1 says
that \(n!+1\) is squarefree outside an explicit finite set. If true, it would
be much stronger than the required finiteness for \(n!+1\), but it is only a
conjecture and supplies no unconditional progress that can be used here.
(c)
The \(n!+1\) branch contains the Brocard–Ramanujan equation
\(n!+1=m^2\). Berndt and Galway report no further square solutions through
\(n=10^9\):
B. C. Berndt and W. F. Galway, *On the Brocard–Ramanujan Diophantine
Equation \(n!+1=m^2\)*, Ramanujan Journal 4 (2000), 41–42,
<https://doi.org/10.1023/A:1009873805276>.
That is a computation only for the square subfamily, not for all powerful
values, so it does not subsume the computation in this report.
For the Wieferich obstruction below, the relevant conditional benchmark is:
J. H. Silverman, Wieferich’s criterion and the abc-conjecture, Journal of
Number Theory 30 (1988), 226–237,
<https://doi.org/10.1016/0022-314X(88)90019-4>. Silverman proves under
\(abc\) that there are \(\gg\log x\) base-2 non-Wieferich primes up to \(x\).
Unconditionally, even infinitude of the non-Wieferich primes is not known; a
primary source explicitly recording that wall is H. Graves and M. R. Murty,
The abc conjecture and non-Wieferich primes in arithmetic progressions,
Journal of Number Theory 133 (2013), 1809–1813.
I found no primary source giving an unconditional resolution of any of the
four eventual-finiteness statements. This is a report of the searches I
performed, not a proof that no such source exists.
3. Exact reduction for \(2^n\pm1\)
Let \(q\) be an odd prime and write
\[ d_q=\operatorname{ord}_q(2). \]Call \(q\) base-2 Wieferich when
\[ 2^{q-1}\equiv1\pmod {q^2}. \]3.1 Minus sign
If \(q\mid2^n-1\), then \(d_q\mid n\). Since \(d_q\mid q-1\), one has
\(q\nmid d_q\). Lifting the exponent gives
\[ v_q(2^n-1) =v_q(2^{d_q}-1)+v_q(n/d_q) =v_q(2^{d_q}-1)+v_q(n). \tag{1} \]Also \(q-1=d_qt\) with \(q\nmid t\), so the same identity gives
\[ v_q(2^{q-1}-1)=v_q(2^{d_q}-1). \tag{2} \]Consequently \(v_q(2^{d_q}-1)\ge2\) exactly when \(q\) is Wieferich.
Thus:
\[ \boxed{ 2^n-1\text{ is powerful}\iff \text{every non-Wieferich prime }q\mid2^n-1\text{ divides }n. } \tag{3} \]This is (b: the elementary LTE lemma). All steps other than LTE are
included above. The checker independently tests (1)–(2) for every applicable
pair \(q\le5000,\ n\le200\).
3.2 Plus sign
If \(q\mid2^n+1\), then \(d_q\) is even. Put \(e_q=d_q/2\). Necessarily
\(n/e_q\) is odd. The plus form of LTE gives
\[ \begin{aligned} v_q(2^n+1) &=v_q(2^{e_q}+1)+v_q(n/e_q)\\ &=v_q(2^{e_q}+1)+v_q(n). \tag{4} \end{aligned} \]The two factors of \(2^{d_q}-1\) show
\[ v_q(2^{e_q}+1)=v_q(2^{d_q}-1), \]which is at least two exactly when \(q\) is Wieferich. Therefore:
\[ \boxed{ 2^n+1\text{ is powerful}\iff \text{every non-Wieferich prime }q\mid2^n+1\text{ divides }n. } \tag{5} \]This is again (b: LTE), with the same finite independent sanity check in
the verifier.
3.3 A sharp unconditional wall
Suppose, for contradiction, that there were only finitely many odd
non-Wieferich primes, and let \(K\) be their product. For every multiple
\(n\) of \(K\), every non-Wieferich prime divisor of either \(2^n-1\) or
\(2^n+1\) divides \(n\). Criteria (3) and (5) would make both terms
powerful. Hence:
> Finiteness of the powerful values in either exponential family implies
> infinitude of the base-2 non-Wieferich primes.
This is (b: LTE). Infinitude of non-Wieferich primes is itself open
unconditionally. It does not prove that problem #936 is equivalent to that
open problem: infinitude alone does not give the uniform divisor needed for
every sufficiently large \(n\).
The exact missing uniform statement exposed by (3) and (5) is:
\[ \exists N\ \forall n>N\ \exists q: \quad q\mid2^n\pm1,\quad q\nmid n,\quad 2^{q-1}\not\equiv1\pmod{q^2}. \tag{6} \]Statement (6), separately for either sign, would settle that exponential
branch. Current primitive-divisor machinery does not supply the final
non-Wieferich condition.
3.4 What Bang–Zsigmondy does supply
The Bang–Zsigmondy theorem says that \(2^m-1\) has a primitive prime divisor
for \(m>1\), except \(m=6\). A convenient original reference is K. Zsigmondy,
Zur Theorie der Potenzreste, Monatshefte für Mathematik und Physik 3
(1892), 265–284, <https://doi.org/10.1007/BF01692444>.
- For \(2^n-1\), a primitive divisor has order \(n\). If the term is
powerful, that primitive divisor is Wieferich by (1)–(2). Different \(n\)
give different primes because their orders differ. The exceptional
\(n=6\) is harmless: \(2^6-1=63=3^2\cdot7\) is not powerful.
- For \(2^n+1\), apply the theorem to \(2^{2n}-1\). Except when \(n=3\), a
primitive divisor has order \(2n\), lies in \(2^n+1\), and must be
Wieferich if that term is powerful. The exception is the actual example
\(2^3+1=9\).
Therefore, apart from \(2^1-1=1\) and \(2^3+1=9\), the number of powerful
terms in either family is at most the cardinality of the set of Wieferich
primes. This is (b: Bang–Zsigmondy and LTE). It is a clean injection,
not a finiteness proof, because finiteness of the Wieferich primes is also
unknown.
4. Elementary factorial filters and a second wall
Let
\[ W_p=\frac{(p-1)!+1}{p} \]be the Wilson quotient of a prime \(p\).
4.1 Two necessary congruences
If \(p=n+1\) is prime, Wilson’s theorem says
\[ p\mid n!+1. \]If \(n!+1\) is powerful, then necessarily \(p^2\mid n!+1\), or
\[ W_p\equiv0\pmod p. \]Thus \(p\) must be a Wilson prime. (a)
If \(p=n+2\) is prime, then \((p-2)!\equiv1\pmod p\), so
\[ p\mid n!-1. \]Write \((p-2)!=1+ap\). Then
\[ W_p=\frac{(p-1)(p-2)!+1}{p}\equiv1-a\pmod p. \]It follows that
\[ p^2\mid n!-1\iff W_p\equiv1\pmod p. \tag{7} \]This is (a). The checker independently tests both Wilson-quotient
equivalences for every prime \(p\le300\). These are necessary filters only;
they do not control the other prime factors.
4.2 Brocard–Ramanujan obstruction
Every square is powerful. Consequently, proving that \(n!+1\) is powerful
only finitely often would in particular prove that the Brocard–Ramanujan
equation
\[ n!+1=m^2 \]has only finitely many solutions. That finiteness question remains open,
despite the computation through \(10^9\). This implication is (a) and
explains why a purely elementary uniform argument for the \(n!+1\) branch
would already settle a famous subproblem.
For either factorial sign, every prime divisor is greater than \(n\).
The direct uniform lemma that would close the branch is:
\[ \exists N\ \forall n>N\ \exists p>n: \qquad v_p(n!\pm1)=1. \tag{8} \]No cited theorem or search result supplies (8).
5. Exact finite computation
5.1 Certificate design
The bounded result is (d). Its individual rejection certificates have
the elementary form
\[ p\mid A_n,\qquad p^2\nmid A_n, \tag{9} \]with \(p\) a proved prime. Condition (9) immediately proves that \(A_n\) is
not powerful, so no complete factorization is trusted or required.
The standalone checker is:
runs/erdos936_wave7u_verify.py
The full source, including all residual certificate primes, is in that file.
Its SHA-256 after the final run is
4ce79fec19dfc6b78e26fb7c635bab0937b0feef312bc38ff4ab49d1b700c9a3.
The discovery and verification paths are deliberately separated:
1. For \(2^n\pm1\), the checker regenerates every prime \(q\le10^6\), computes
\(d_q=\operatorname{ord}_q(2)\), and obtains exact-once witnesses from the
relevant residue classes. It then directly checks (9), so correctness
does not depend on the order-sieve derivation.
2. Residual candidate factors were located with the FactorDB API. FactorDB
is not trusted by the result. Every residual factor is independently
proved prime and both modular congruences in (9) are recomputed.
3. Residual Mersenne primes are independently certified by the
Lucas–Lehmer theorem.
4. For \(n!\pm1\), every prime \(p\le2{,}000{,}000\) is generated from
scratch and \(n!\bmod p^2\) is accumulated. Residual factors are handled
as in item 2.
5. Certificate primes below \(2^{64}\) use deterministic Miller–Rabin with
the standard complete seven-base set. The 41 distinct larger primes are
proved by PARI/GP’s isprime(p,2), which is its APRCL proof mode—not its
probable-prime mode.
The core acceptance condition in the checker is literally:
# Example: p is a certificate for 2^n-1.
assert pow(2, n, p) == 1
assert pow(2, n, p*p) != 1
# Example: p is a certificate for n!+1.
assert factorial_mod(n, p) == p - 1
assert factorial_mod(n, p*p) != p*p - 1
5.2 Verified table
The checker proves the following bounded classification (d):
\[ \begin{array}{c|c|c|c} A_n&n\text{ range}&\{n:A_n\text{ powerful}\}&\text{values}\\ \hline 2^n-1&1\le n\le1000&\{1\}&1\\ 2^n+1&1\le n\le1000&\{3\}&9=3^2\\ n!+1&1\le n\le139&\{4,5,7\}&25=5^2,\ 121=11^2,\ 5041=71^2\\ n!-1&2\le n\le139&\{2\}&1. \end{array} \]For every other index in those ranges, the checker provides and replays a
prime-exponent-one certificate.
5.3 Reproduction
Environment:
- Python 3.12.3;
- PARI/GP 2.15.4, using GMP 6.3.0;
- maximum resident memory in a timed run: about 71 MB.
Command:
python runs/erdos936_wave7u_verify.py
Final output:
VERIFIED: for 1 <= n <= 1000, powerful 2^n-1 occurs exactly at n={1},
and powerful 2^n+1 occurs exactly at n={3}.
VERIFIED: for 1 <= n <= 139, powerful n!+1 occurs exactly at n={4,5,7};
for 2 <= n <= 139, powerful n!-1 occurs exactly at n={2}.
(1!-1=0 is excluded because powerful numbers are positive.)
BARRIER: 140!+1 is composite, but has no prime p <= 2,000,000 with v_p(140!+1)=1.
TIMING: structural checks 0.007s; power certificates 1.171s; factorial certificates 6.124s; total 7.301s
The time varies slightly between runs; the mathematical output was identical
in repeated executions.
6. Precise stopping point and compute cost
The first value not classified by the factorial-plus certificate search is
\(140!+1\), a 242-digit integer.
The checker independently establishes two facts (d):
1. \(140!+1\) is composite: base \(2\) violates Fermat’s necessary
congruence for primality.
2. No prime \(p\le2{,}000{,}000\) has \(v_p(140!+1)=1\).
This does not say that \(140!+1\) is powerful. It says exactly what is
missing: a proved prime factor occurring to exponent one, or a full
powerful-factorization proof. A 55-second local PARI factor attempt found no
factor, and FactorDB returned only the unfactored composite; neither fact is
used as a mathematical certificate.
The cost of going further is factor-size dependent. A targeted ECM campaign
could find a moderate-size factor in tens to hundreds of core-hours, but it
has no guaranteed stopping bound if the least factor is large. In the
balanced worst case, a general factorization of a 242-digit number is near
RSA-240/RSA-250 scale: RSA-250 required about 2700 core-years, i.e. roughly
\(2.4\times10^7\) core-hours. This comparison uses Boudot et al.,
[*Comparing the Difficulty of Factorization and Discrete Logarithm: A
240-Digit Experiment*](https://arxiv.org/abs/2006.06197), which reports about
1000 core-years for RSA-240 and 2700 core-years for RSA-250. Only one
exponent-one prime is needed here, so that worst-case effort may be
unnecessary, but it is far beyond the few-minute budget and was not
attempted.
More importantly, extending any finite table cannot settle the original
question. The exponential branches need a uniform non-Wieferich primitive
divisor statement such as (6), and the factorial branches need a uniform
exponent-one divisor statement such as (8). Those are the exact mathematical
walls left by this run.
PARTIAL: Exact certificate checks give only n={1} for 2^n-1 and n={3} for 2^n+1 through n=1000, and only n={4,5,7} for n!+1 and n={2} for n!-1 through n=139; LTE reduces each exponential branch to a uniform non-Wieferich-divisor lemma that is currently unavailable.