Erdős problem #364 — wave 8w

Accessed 2026-07-28. This report distinguishes:

• (a) elementary-rigorous;
• (b) rigorous modulo the named theorem/source;
• (c) plausible/structural-unverified;
• (d) computational-only.

0. Mandatory live-page gate

I fetched both the problem page and its discussion thread through the Bright Data browser path, not datacenter curl:

• <https://www.erdosproblems.com/364>
• <https://www.erdosproblems.com/forum/discuss/364>

The live page was last edited 13 April 2026 and showed:

• status/open-page warning: open (the problem card is marked

VERIFIABLE);

• 0 claimed proofs;
• Currently working on this problem: None;
• Interested in collaborating: None;
• 5 comments.

Thus none of the mandatory stop conditions fired.

Verbatim live statement

Are there any triples of consecutive positive integers all of which are powerful (i.e. if \(p\mid n\) then \(p^2\mid n\))?

Results listed on the live page

The following is a faithful inventory, not an extrapolation.

1. Erdős asked Mahler whether infinitely many consecutive powerful pairs exist;

Mahler used the infinitely many solutions of \(x^2=8y^2+1\).

1. Mollin and Walsh also made the no-triple conjecture. If \(n_k\) is the

\(k\)-th powerful number, Erdős believed the stronger gap assertion \[ n_{k+2}-n_k>n_k^c \] for some \(c>0\).

1. The \(abc\) conjecture implies only that there are finitely many such

triples, not that there are none.

1. Four consecutive powerful numbers are impossible because one is

\(2\bmod 4\).

1. Chan proved nonexistence when the middle term is a cube and the two outer

terms have forms \(p^3y^2,q^3z^2\), with \(p,q\) prime.

1. She proved nonexistence when the middle term is a cube and the two outer

terms have forms \(p^2a^3,q^2b^3\), with \(p,q\) prime.

1. The page says, citing OEIS A076445, that no center occurs below

\(7.38\times 10^{28}\).

1. It cross-references problems 137, 365, and 938.

All five live comments

The newest comment's construction is correct (a):

and \(4n(n+1)\) is powerful when the coprime integers \(n,n+1\) are powerful.

1. Primary-source and literature audit

I searched the exact problem, “three consecutive powerful/squarefull numbers,” the square-centered subcase, and the cited paper titles. I then opened the primary PDFs rather than relying on search snippets.

1. Erdős (1976). The cited paper exists:

Problems and results on number theoretic properties of consecutive integers and related questions. Its powerful-number discussion records Mahler's Pell observation and Erdős's stronger two-gap belief. Local PDF SHA-256: 2c098f2c3e5079ef09a44f693c7db43a29428bb1c8bf7df7ab57b020306a9ad0.

1. Mollin–Walsh (1986). The cited

On powerful numbers, Internat. J. Math. Math. Sci. 9 (1986), 801–806, exists. Section 2 states that the three-consecutive problem remained open and exhibits, among other outer pairs, \(130576327,130576329\) and \(13837575261123,13837575261125\). Those pairs are independently rediscovered below.

1. Chan (2025). arXiv:2503.21485

exists and matches the published INTEGERS 25, A7. Theorem 1 is exactly the \(p^3y^2,q^3z^2\) cube-centered result stated above (b). Downloaded journal PDF SHA-256: ea09dbd38f23e373b8c0b40343cca68a7173cc99bef8b061ceebf840c9d1ba2f.

1. She (2025).

arXiv:2507.16828 exists and matches INTEGERS 25, A103. Theorem 1 is exactly the \(p^2a^3,q^2b^3\) cube-centered result (b). Downloaded journal PDF SHA-256: 2b94a588e870e52680b9f10fd0be93083c211188092e29e31822de50fbf022f9.

1. Beckon (2019).

On Consecutive Triples of Powerful Numbers proves that a possible first term is \(7,27,\) or \(35\bmod 36\) (b). Section 2 below rederives this result elementarily and the verifier recomputes the residue list.

1. Aktaş–Murty (2017).

Fundamental units and consecutive squarefull numbers gives the unique \(a^2b^3\) representation, generalized Pell reductions, and discusses the conditional \(abc\) finiteness result (b). It does not prove zero triples.

1. Stephens–Williams (1988). Their

Some Computational Results on a Problem Concerning Powerful Numbers, Math. Comp. 50, 619–632, verifies a Pell/fundamental-unit computation for squarefree \(D<10^8\). Its introduction explicitly presents the triple problem as open and records the necessary Pell-unit condition coming from Mollin–Walsh (b). It is not an exhaustive search by size of the potential triple.

1. Current-status cross-check. Wouter van Doorn's May 2026 paper

arXiv:2605.06697, page 1, explicitly says the consecutive-integer problem is “still open in general”; its subject is instead three-term arithmetic progressions of powerful numbers (b). Downloaded PDF SHA-256: 90ea2c2e06467d36de53592c234edbbade11fd206c21d756f0648dfba595cfd2.

1. OEIS audit. A076445 lists the smaller

members of powerful pairs differing by 2. Its official data currently contain 13 terms through \(73840550964522899559001927225\). The linked 33-term extension is explicitly labelled conjectural: those extra values belong but are not known to be consecutive sequence terms. I found no proof certificate on OEIS for the ordering of the official 13. Therefore the live page's \(7.38\times10^{28}\) statement is recorded as its cited computational ground truth (d), but it is not used in my result. My code independently proves completeness of the first eight pairs through \(10^{17}\).

I found no primary source claiming a complete solution or a global square-centered theorem. This is a search miss, not a proof that no such source exists.

2. Elementary reductions

2.1 Local restrictions

Lemma 1 (a). If \(N-1,N,N+1\) are powerful, then \(N\equiv0\bmod4\).

Proof. A powerful even integer is divisible by \(4\), so it cannot be \(2\bmod4\). Unless \(N\equiv0\bmod4\), one of \(N-1,N,N+1\) is \(2\bmod4\). \(\square\)

Among three consecutive integers exactly one is divisible by 3, and if it is powerful it is divisible by 9. Combining this with Lemma 1 gives, for the first term, exactly

This is the Beckon restriction, now obtained directly (a). The verifier enumerates the 36 residue classes and asserts this exact list.

2.2 Canonical square-cube representation

Lemma 2 (a). Every powerful positive integer has a unique representation

Proof. Write \(m=\prod p^{e_p}\), where every \(e_p\ge2\). Put \(p\) in \(b\) exactly when \(e_p\) is odd. For even \(e_p\), put \(p^{e_p/2}\) in \(a\); for odd \(e_p\ge3\), put \(p^{(e_p-3)/2}\) in \(a\). Parity of each exponent also proves uniqueness. \(\square\)

Thus a general candidate can be written uniquely as

where the three squarefree kernels are pairwise coprime. They satisfy

Pairwise coprimality follows because \(N\) is even, \(N\pm1\) are odd, and the three integers are pairwise coprime (a).

For a fixed kernel triple, homogenizing (1) gives

The gradients of the two diagonal quadrics cannot be dependent at a projective point: dependence first forces \(X=W=0\), and the two equations then force \(Y=Z=0\). Hence (2) is a smooth intersection of two quadrics, so it is a genus-one curve (a). Siegel's theorem gives only finitely many integral points for each fixed kernel triple (b). It supplies no uniform exclusion as the three squarefree kernels vary.

2.3 A useful reduction when the center is a square

Lemma 3 (a). If

are all powerful, then \(x-1\) and \(x+1\) are powerful.

Proof. Lemma 1 makes \(x\) even. Consequently \(\gcd(x-1,x+1)=1\). Their product \(x^2-1\) is powerful, so every prime occurring in either coprime factor occurs there with the same exponent that it has in the product, at least 2. \(\square\)

This lowers a search for square centers as large as \(x^2\) to a complete search for powerful pairs differing by 2 only as large as \(x\).

3. Exact finite result

Computer-assisted theorem (d). There is no triple of powerful integers

with \(1\le x\le10^{17}\). Equivalently, Erdős #364 has no square middle term at most \(10^{34}\).

This is a restricted theorem; it does not settle nonsquare middle terms. For comparison, \(10^{34}\) is about \(1.35427\times10^5\) times the live page's general \(7.38\times10^{28}\) center bound.

3.1 Completeness of the computation

Set \(L=10^{17}+1\). By Lemma 2, the following loops generate every powerful \(m\le L\) exactly once:

The verifier:

1. constructs the squarefree flags by an Eratosthenes square sieve;
2. generates all canonical values into an unsigned 64-bit array;
3. sorts and checks strict increase (hence no implementation duplicates);
4. extracts every adjacent difference 2;
5. independently factors every reported pair using prime/exponent

certificates;

1. for each midpoint \(x\), verifies a prime \(p\) with

\[ p\mid x^2+1,\qquad p^2\nmid x^2+1, \] which proves \(x^2+1\) is not powerful;

1. independently checks the generator against definition-level trial

factorization for every integer through 100,000.

The \(+1\) in \(L\) is important: it includes \(x+1\) when \(x=10^{17}\).

3.2 Complete pair and obstruction table

The exhaustive list below contains every powerful pair \(x-1,x+1\) with \(x\le10^{17}\). The final column is a prime occurring to exact exponent one in \(x^2+1\).

The complete factor certificates for the two pair members are embedded in the checker. For example,

and

3.3 Reproduction and independent reruns

Complete standalone source:

runs/erdos364_wave8w_reverify.py

Run:

python runs/erdos364_wave8w_reverify.py
ERDOS364_SORT_KIND=mergesort python runs/erdos364_wave8w_reverify.py

Environment and exact common output:

python=3.12.3 numpy=2.4.4
root_limit=100000000000000000
powerful_generation_limit=100000000000000001
max_squarefree_kernel=464158
canonical_powerful_count=686552743
small_oracle_limit=100000
small_oracle_powerful_count=619
locally_admissible_start_residues_mod_36=[7, 27, 35]
sorted_uint64_sha256=9851207077e836c7b47315ab1498ba0c414ee124aef281e6f3a2a87ec3ef0a67
pairs_differing_by_2=8

The quicksort run took 83.1 seconds and 6.15 GB maximum RSS. The independent stable-mergesort run took 98.3 seconds and 7.95 GB maximum RSS. Both gave the same count, byte-level sorted-array hash, pair table, and prime certificates (d).

The core enumeration in the standalone source is:

b_max = integer_cuberoot(POWERFUL_LIMIT)
squarefree = squarefree_sieve(b_max)
counts, total = canonical_counts(POWERFUL_LIMIT, squarefree)
values = generate_all_powerful(POWERFUL_LIMIT, counts, total)
values.sort(kind=SORT_KIND)
pairs, digest = scan_sorted(values)

No OEIS value is supplied to these generation or difference-scanning functions. The hard-coded pair factorizations are used only after the exhaustive scan, as independently checked certificates.

4. Exact remaining wall

The full problem remains open. Equation (2) explains the mathematical wall precisely:

• for each fixed \((b_-,b_0,b_+)\), one has a single standard

genus-one/integral-point problem (Siegel gives finiteness, but not a uniform effective calculation);

• a hypothetical triple may use any of infinitely many pairwise coprime

squarefree kernel triples;

• neither Siegel's theorem nor the Pell reductions in the cited literature

give a uniform kernel bound or a uniform no-integral-point theorem.

The missing lemma that would turn this reduction into a solution is therefore one of the following:

1. a uniform theorem ruling out affine integral points on every curve (2)

for pairwise coprime squarefree \(b_-,b_0,b_+\); or

1. an effective finite bound on those kernels, followed by certified

integral-point computations for every remaining curve.

The \(abc\) conjecture supplies finiteness, not the required zero statement. Treating a finite set of kernels or a computational list as uniform would be the invalid step.

A direct size search also stops for a quantified reason. The number of powerful integers through \(T\) is asymptotic to

At the live page's \(T\approx7.38\times10^{28}\), this predicts about \(5.9\times10^{14}\) entries, roughly 4.2 PiB for the raw 64-bit array alone (c). That is not a viable route to a from-scratch verification here.

Even extending the present square-root search to the next OEIS-scale pair near \(3.74\times10^{18}\) would require about \(4.2\times10^9\) canonical entries, 34 GB for the base array (about 68 GB for mergesort), and an extrapolated 9–10 single-core minutes, about 0.15 core-hours (c). On a 64-GB cloud node this is roughly $0.10–$0.30 of node time, but it exceeds the stipulated few-minute computation budget, so I did not run it.

PARTIAL: Exact from-scratch enumeration proves there is no square-centered powerful triple \(x^2-1,x^2,x^2+1\) for \(x\le10^{17}\) (center \(\le10^{34}\)); the unrestricted problem remains open because no uniform control over the infinitely many squarefree-kernel genus-one curves is known.