Erdoes #887 - working report - Erdős Problem a Day

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.)

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 $C$.

Chan [Ch14] is credited with at most five divisors for square $n$ in

\[ [n^{1/2}-n^{1/4}(\log n)^{1/7}, n^{1/2}+n^{1/4}(\log n)^{1/7}]. \]

Chan [Ch15] is credited with at most eighteen divisors in the analogous

$(\log n)^{1/14}$ window for an almost-square hypothesis.

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.)

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.

Alfaiz, 25 June 2026. Points to Patrick Letendre's 2025 preprint

arXiv:2503.12146 as a possible partial result.

old-bielefelder, 25 June 2026. Says the same Letendre paper helped on problem

873.

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

0<y_i\le24T.

For $y_5$, this follows from

z^2=15T+154<16T,\qquad y_5=3Xz<12T.

Equations (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<Cn^{1/4} \iff (d^2+n)^2<(2d+C^2)^2n \qquad(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 $d-\sqrt n<Cn^{1/4}$, then $y=O_C(n^{1/4})$ and

0<u-2\sqrt n=\frac{y^2}{u+2\sqrt n}=O(C^2).

Thus the problem asks for an absolute bound, independent of the length parameter $C^2$, 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.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger