Erdős problem 887 — wave 7r
Access/research date: 2026-07-27 UTC.
Claim labels used throughout:
- (a) elementary-rigorous: proved here from explicit identities and inequalities.
- (b) rigorous-modulo-named-theorem: quoted from a verified primary source.
- (c) plausible/structural-unverified: heuristic or a non-uniform structural diagnosis.
- (d) computational-only: exact browser or computer output, reproducible by the supplied checker.
0. Mandatory live-page gate
Authoritative statement
(d: direct Bright Data browser observation.) I fetched the rendered live page, its
LaTeX view, and the full discussion through Bright Data. The page was last edited
10 April 2026. Its verbatim LaTeX statement is:
> Is there an absolute constant $K$ such that, for every $C>0$, if $n$ is sufficiently large then $n$ has at most $K$ divisors in $(n^{1/2},n^{1/2}+C n^{1/4})$.
Live sources:
- https://www.erdosproblems.com/887
- https://www.erdosproblems.com/latex/887
- https://www.erdosproblems.com/forum/discuss/887
Stop-rule fields
(d: direct live-page observation.)
- Status: OPEN.
- Claimed proofs: 0.
- Currently working on this problem: None.
- Interested in collaborating: None.
- Likes / difficult / tractable / formalisable / working on formalisation: all None.
Thus neither mandatory stop condition fired.
Results actually listed on the live page
**(d: this is a transcription of what the page says, not an endorsement of every
summary sentence.)**
1. Erdős and Rosenfeld [ErRo97] are credited with infinitely many \(n\) having four
divisors in
\((n^{1/2},n^{1/2}+n^{1/4})\), and with the upper bound \(1+C^2\) for sufficiently
large \(n\) depending on \(C\).
2. Chan [Ch14] is credited with at most five divisors for square \(n\) in
\[ [n^{1/2}-n^{1/4}(\log n)^{1/7}, n^{1/2}+n^{1/4}(\log n)^{1/7}]. \]
3. Chan [Ch15] is credited with at most eighteen divisors in the analogous
\((\log n)^{1/14}\) window for an almost-square hypothesis.
4. The page points to problem 886.
There are two source-level corrections worth recording.
- (a) The live page's literal \(C=1\) four-divisor sentence is incompatible with
its own \(1+C^2\) eventual upper bound: at \(C=1\), that upper bound is two, while
an infinite set of witnesses is unbounded. In the actual [ErRo97] Proposition 4.2,
the four factor differences are bounded by \(16n^{1/4}\); their largest
one-sided divisor displacement is \((8+o(1))n^{1/4}\), not \(n^{1/4}\).
- (b) The live page renders the [Ch15] almost-square hypothesis with two minus
signs. Chan's primary paper states \(n=(N-a)(N+b)\), \(0\leq a\leq b\).
All four live comments
**(d: direct transcription/summary of user comments, which the site itself labels
unverified.)**
1. m-czech, 6 July 2026. Claims an infinite family with five one-sided divisors
for \(C=31\):
\[ n=(x^2-1)(x^2-4)(x^2-16)(x^2-49),\qquad z^2-56x^2=65. \]
Four pairs are balanced splits and the fifth is Pell-tuned. The comment explicitly
says problem 887 remains open and links
https://czechu.blog/erdos/problems/887/note887.pdf.
2. Alfaiz, 25 June 2026. Points to Patrick Letendre's 2025 preprint
arXiv:2503.12146 as a possible partial result.
3. old-bielefelder, 25 June 2026. Says the same Letendre paper helped on problem
873.
4. KoishiChan, 9 December 2025. Points to Chan's square/almost-square work and
describes a reduction to simultaneous square values of linear polynomials; the
commenter suggests Bombieri--Lang should settle the relevant conditional
finiteness.
The m-czech comment also points to formal-conjectures PR 1942. Direct inspection
shows that PR 1942 is titled “ErdosProblems: 886,” not 887. The current 887 Lean file
has corrected the lower-bound variant to “there exists \(C>0\),” rather than \(C=1\):
https://github.com/google-deepmind/formal-conjectures/blob/main/FormalConjectures/ErdosProblems/887.lean.
1. Primary-source audit
(b) The following sources exist and say the following.
- P. Erdős and M. Rosenfeld, The factor-difference set of integers, Acta Arith.
79 (1997), 353–359, DOI 10.4064/aa-79-4-353-359:
https://matwbn.icm.edu.pl/ksiazki/aa/aa79/aa7944.pdf.
Proposition 4.1 proves the \(1+C^2\) type bound. Proposition 4.2 gives four small
factor differences for \(a(a+1)\cdots(a+7)\). Proposition 4.3 proves that eight
distinct weights have at most four equal-sum \(4+4\) splits.
- T. H. Chan, Factors of a perfect square, Acta Arith. 163 (2014), 141–143,
DOI 10.4064/aa163-2-4:
https://www.impan.pl/en/publishing-house/journals-and-series/acta-arithmetica/all/163/2/82006/factors-of-a-perfect-square.
It proves the live page's five-divisor bound for sufficiently large squares.
The preprint is arXiv:1303.2069.
- T. H. Chan, Factors of almost squares and lattice points on circles,
Int. J. Number Theory 11 (2015), 1701–1708; preprint
https://arxiv.org/abs/1406.2230. It proves the stated eighteen-divisor
almost-square result and uses Turk's simultaneous-Pell bound.
- P. Letendre, Divisors of an Integer in a Short Interval,
https://arxiv.org/abs/2503.12146 (2025). Its general theorem gives a
\(\tau(n)^{1-\xi}\)-type estimate (with additional \(V(n)\log\tau(n)\) factors);
it does not give an absolute constant at the \(n^{1/4}\) endpoint for arbitrary
\(n\).
Exact-title, phrase, arXiv, and citation searches found no further primary paper
specifically resolving problem 887. This is an honest search miss, not evidence that
no such paper exists.
2. New result
Theorem
(a: elementary-rigorous.) Define
\[ (X_0,z_0)=(135,523),\qquad X_{j+1}=4X_j+z_j,\qquad z_{j+1}=15X_j+4z_j, \]and
\[ n_j=\frac{(X_j^2-1)(X_j^2-9)(X_j^2-25)(X_j^2-81)}{16}. \]Then every \(n_j\) has at least five distinct divisors in
\[ (\sqrt{n_j},\sqrt{n_j}+25n_j^{1/4}). \]More sharply, for every fixed real \(C>24\), all sufficiently large \(j\) have at
least five divisors in
\[ (\sqrt{n_j},\sqrt{n_j}+Cn_j^{1/4}). \]Consequently this gives an explicit infinite five-divisor family with asymptotic
window threshold \(24\), strengthening the live comment's explicit \(C=31\) family.
It does not improve the lower bound \(K\geq5\), and it does not answer whether
a uniform \(K\) exists.
An equivalent integral product form makes the affine-cube origin visible. With
\[ t_j=\frac{X_j-9}{2},\qquad A=\{0,2,3,4,5,6,7,9\}, \]we have
\[ n_j=16\prod_{a\in A}(t_j+a). \]3. Proof of the theorem
Everything in this section is (a) elementary-rigorous.
3.1 Pell orbit
The seed satisfies
\[ 523^2-15\cdot135^2=154. \]The recurrence preserves the equation because
\[ (15X+4z)^2-15(4X+z)^2=z^2-15X^2. \]It preserves oddness, positivity, and strictly increases \(X\). Hence it supplies
infinitely many odd solutions of
\[ z^2-15X^2=154 \]with \(X\ge135\), and the resulting \(n_j\) are strictly increasing.
Because \(X\) and \(1,3,5,9\) are odd, every \(X^2-r^2\) is divisible by
\(8\), so \(n_j\) is an integer.
3.2 Five exact factor pairs
Put
\[ N=(X^2-1)(X^2-9)(X^2-25)(X^2-81),\qquad n=N/16 \]and
\[ q=X^4-58X^2+135. \]For any orbit solution \((X,z)\), define:
\[ \begin{array}{ll} d_1=(X^4-10X^2+9)/4,&e_1=(X^4-106X^2+2025)/4,\\ d_2=(X^4-26X^2+25)/4,&e_2=(X^4-90X^2+729)/4,\\ d_3=(X^4-34X^2+225)/4,&e_3=(X^4-82X^2+81)/4,\\ d_4=(X^4-58X^2+192X-135)/4, &e_4=(X^4-58X^2-192X-135)/4,\\ d_5=(q+6Xz)/4,&e_5=(q-6Xz)/4. \end{array} \]Direct multiplication gives \(d_ie_i=n\) for \(1\le i\le4\). For the fifth pair,
the exact identity
\[ q^2-N=36X^2(15X^2+154) \]and \(z^2=15X^2+154\) give
\[ (q+6Xz)(q-6Xz)=N, \]hence \(d_5e_5=n\).
All ten quantities are integers. For the first four pairs this is also immediate
from their expressions as four linear factors divided by \(4\). For the fifth,
odd \(X,z\) give
\[ q\equiv2\pmod4,\qquad 6Xz\equiv2\pmod4, \]so both \(q+6Xz\) and \(q-6Xz\) are divisible by \(4\).
The positive factor differences are
\[ \begin{aligned} y_1&=d_1-e_1=24X^2-504,\\ y_2&=16X^2-176,\\ y_3&=12X^2+36,\\ y_4&=96X,\\ y_5&=3Xz. \end{aligned} \]They are positive for \(X\ge135\), so \(d_i>\sqrt n>e_i\).
For completeness, the first four lower factors are visibly positive products of
linear factors when \(X>9\). Also \(z<4X\) for \(X\ge135\), and hence
\[ q-6Xz>X^4-82X^2+135>0, \]so the fifth lower factor is positive as well.
3.3 Distinctness
The five factor sums are
\[ d_i+e_i=\frac{X^4-58X^2+\kappa_i}{2}, \qquad (\kappa_1,\ldots,\kappa_5)=(1017,377,153,-135,135). \]The constants are pairwise distinct. Since a factor pair of fixed product \(n\) is
determined by its sum, the five pairs, and therefore the five upper divisors, are
distinct.
3.4 A uniform \(C=25\) certificate
For any \(de=n\) with \(d>e\), put \(u=d+e\) and \(y=d-e\). Then
\[ u^2-y^2=4n \]and
\[ d-\sqrt n =\frac y2+\frac{u-2\sqrt n}{2} =\frac y2+\frac{y^2}{2(u+2\sqrt n)} <\frac y2+\frac{y^2}{8\sqrt n}. \tag{1} \]Let \(T=X^2\). Since the four factors defining \(N\) are at least \(T-81\),
with three of them strictly larger,
\[ n^{1/4}>\frac{T-81}{2},\qquad \sqrt n>\frac{(T-81)^2}{4}. \tag{2} \]For \(X\ge100\), every displayed \(y_i\) satisfies
\[ 0Equations (1) and (2), together with \(T-81>T/2\), now give
\[ d_i-\sqrt n <12T+\frac{(24T)^2}{2(T-81)^2} <12T+1152. \]Finally,
\[ 12T+1152<\frac{25(T-81)}2<25n^{1/4} \]whenever \(T>4329\). Every orbit member has \(X\ge135\), so this proves the
\(C=25\) assertion for the entire stated orbit.
3.5 The sharper \(C>24\) assertion
As \(j\to\infty\),
\[ \frac{n_j^{1/4}}{X_j^2}\longrightarrow\frac12,\qquad \frac{z_j}{X_j}\longrightarrow\sqrt{15}. \]Also, the second term in (1) is \(O(1)\), whereas \(n_j^{1/4}\asymp X_j^2\).
Therefore the five normalized upper displacements tend respectively to
\[ 24,\quad16,\quad12,\quad0,\quad3\sqrt{15}. \]Since \(3\sqrt{15}<12<24\), every fixed \(C>24\) contains all five divisors
for all sufficiently large members of the orbit.
4. Exact examples and complete finite censuses
These census claims are (d) computational-only, although the five advertised
divisors also follow from the elementary theorem.
For the first member,
\[ (X,z)=(135,523),\qquad n=6\,851\,429\,083\,699\,200. \]The five formula divisors are
\[ 82\,992\,096,\quad82\,919\,200,\quad82\,882\,800,\quad 82\,779\,840,\quad82\,879\,335. \]The verifier factors \(n\) from scratch, enumerates every divisor, and tests the
open interval using the exact integer equivalence
\[ d-\sqrt n| \(X\) | \(n\) | \(\tau(n)\) | exact number of one-sided \(C=25\) divisors |
|---:|---:|---:|---:|
| 135 | 6,851,429,083,699,200 | 24,192 | 8 |
| 1063 | 101,882,958,737,986,120,181,760 | 294,912 | 5 |
For \(X=1063\), the complete window is exactly
\[ \{319191151680,\ 319197665280,\ 319197880560,\ 319200140448,\ 319204660320\}. \]5. Reproduction and independent checker
The standalone standard-library checker is
runs/erdos887_wave7r_reverify.py. Run:
python runs/erdos887_wave7r_reverify.py
Observed output:
symbolic identities: OK
Pell recurrence and seven exact orbit members: OK
uniform C=25 certificate for every orbit member X>=100: OK
asymptotic normalized offsets: 24, 16, 12, 0, 3*sqrt(15)
X=135: tau(n)=24192, complete C=25 count=8
window divisors: 82779840 82790400 82844160 82879335 82882800 82919200 82967976 82992096
X=1063: tau(n)=294912, complete C=25 count=5
window divisors: 319191151680 319197665280 319197880560 319200140448 319204660320
ALL CHECKS PASSED
The checker implements polynomial arithmetic itself, verifies every factor identity
coefficient-by-coefficient, proves the recurrence identity at the coefficient level,
checks seven large orbit members using integers only, checks the uniform
\(X\ge100\) inequality certificate, and performs the two complete divisor censuses.
It uses no floating-point arithmetic and no third-party packages.
Core construction code (the complete source is in the standalone file):
def construction_n(X):
N = 1
for r in (1, 3, 5, 9):
N *= X*X - r*r
assert N % 16 == 0
return N // 16
def construction_pairs(X, z):
T = X*X
q = X**4 - 58*T + 135
numerators = [
(X**4 - 10*T + 9, X**4 - 106*T + 2025),
(X**4 - 26*T + 25, X**4 - 90*T + 729),
(X**4 - 34*T + 225, X**4 - 82*T + 81),
(X**4 - 58*T + 192*X - 135,
X**4 - 58*T - 192*X - 135),
(q + 6*X*z, q - 6*X*z),
]
return [(d // 4, e // 4) for d, e in numerators]
X, z = 135, 523
for _ in range(7):
assert z*z - 15*X*X == 154
n = construction_n(X)
pairs = construction_pairs(X, z)
assert len({d for d, _ in pairs}) == 5
assert all(d*e == n and d*d > n for d, e in pairs)
assert all((d*d+n)**2 < (2*d+25**2)**2*n for d, _ in pairs)
X, z = 4*X + z, 15*X + 4*z
6. Exact remaining wall
The original uniform-\(K\) question remains open.
(a) A close factor pair is equivalently an integral solution
\[ u^2-y^2=4n,\qquad u=d+n/d,\quad y=d-n/d. \]If \(d-\sqrt n
\(C^2\), on the number of integer \(u\)'s just above \(2\sqrt n\) for which
\(u^2-4n\) is a square, once \(n\) is sufficiently large depending on \(C\).
(b) Chan's special-case bounds use effective bounds for simultaneous Pell
equations (Turk) after exploiting a square/almost-square factorisation of \(n\).
Those hypotheses are absent for general \(n\).
(c) With several close pairs, subtracting one square equation from the others
produces a system of simultaneous Pell-type/square-value constraints whose
coefficients range over \(O(C^2)\). The missing unconditional ingredient is a
*uniform bound on the number of compatible square values independent of that
coefficient range*. Finiteness for each fixed higher-genus curve would not suffice:
the curve changes with \(C\), and problem 887 requires a bound uniform over all of
them. The present construction uses four polynomial balanced splits plus one Pell
condition; forcing a sixth by the same architecture introduces a second independent
square condition and supplies no such uniformity.
PARTIAL: An elementary Pell-orbit construction gives infinitely many n with five divisors in the required window for every C>24 (explicitly, the whole stated orbit works for C=25), improving the live comment's C=31 window without resolving existence of a uniform K.