Erdős problem #939 — live audit, exact \(r=4\) search, and residual cases

Access date: 2026-07-28 UTC. The live page and its discussion thread were loaded through a rendered Bright Data browser session because the live page, not the stale tracker YAML, is authoritative.

0. Mandatory live-page gate

Live URL: <https://www.erdosproblems.com/939>

The page was marked OPEN, was last edited 28 May 2026, listed 0 claimed proofs, and displayed:

• Interested in collaborating: None
• Currently working on this problem: None
• I am working on formalising the results on this problem: None

It listed nine comments. Therefore neither mandatory stop condition applied.

Verbatim live statement

The following is copied verbatim from the page's “View the LaTeX source” display:

Let $r\geq 2$. An $r$-powerful number $n$ is one such that if $p\mid n$ then $p^r\mid n$.

If $r\geq 4$ then can the sum of $r-2$ coprime $r$-powerful numbers ever be itself $r$-powerful? Are there at most finitely many such solutions?

Are there infinitely many triples of coprime $3$-powerful numbers $a,b,c$ such that $a+b=c$?

Results incorporated on the live page

The page says:

1. Nitaj proved that the \(r=3\) question has infinitely many solutions, with

the displayed example \[ 2^3 3^5 73^3+271^3=919^3. \]

1. Cohn produced infinitely many such triples in which none of the three

entries is a cube; Walsh later gave another construction.

1. Lander and Parkin disproved Euler's sum-of-powers conjecture at exponent

five with \[ 27^5+84^5+110^5+133^5=144^5. \]

1. Cambie found \(r=5,7,8\) examples. The displayed \(r=5\) example is

\[ 2^8 3^{10}5^7+2^{12}23^6+11^5 13^5=3^7 61^5. \] The page does not display the \(r=7,8\) data.

1. A construction in the comments gives infinitely many representations for

every \(r\geq6\), using exactly \(r-2\) summands (and initially only \(\lceil r/2\rceil+1\) before splitting a term).

The \(r=5\) examples and the \(r\geq6\) construction use joint coprimality, not pairwise coprimality. This is the interpretation incorporated by the page. For \(r=4\) there are only two summands, so joint coprimality is just \(\gcd(a,b)=1\), and then \(a,b,a+b\) are automatically pairwise coprime.

All nine live comments/markers

The thread is nested, so its rendered order is not chronological. These are all nine mathematical/comment items; claims made only in comments remain unverified unless explicitly audited below.

| Author/time shown | Content | Audit here | |---|---|---| | StijnC, 11:36 26 May 2026 | Corrects Walsh's arXiv identifier to 2404.03970. | Verified. The page bibliography currently has the extra digit 2404.039701. | | KentaKitamura, 01:50 25 May 2026 | Gives the smaller \(r=5\) identity \(2^5 17^6+7^{11}+2^5 3^6 7^8=3^7 13^7\), joint gcd \(1\), with a GitHub checker. | Every integer, factorisation, sum, and gcd was independently recomputed. | | StijnC, 09:40 26 May 2026 | Confirms the example; reports an exhaustive \(r=4\) bound \(\max(a,b)>10^{14}\); reports a meet-in-the-middle \(r=5\) classification for target \(d<10^{16}\). | The \(r=4\) bound is independently reproduced and extended here to \(10^{16}\). The \(r=5\) classification was not rerun and is not promoted to a theorem here. | | Liam Price, 23:17 24 May 2026 | Posts the GPT-5.5 Pro binomial construction for all \(r\geq6\), plus Overleaf details and a Lean formalisation. | Reproved from scratch below and checked numerically for \(6\leq r\leq10\). | | StijnC, 19:59 25 May 2026 | “Nice argument :) !” | No mathematical claim. | | Alfaiz, 05:32 13 Oct 2025 | Points to Walsh's preprint for the third question. | Paper and theorem verified. | | StijnC, 11:54 13 Oct 2025 | Notes that Walsh cites the earlier Nitaj and Cohn solutions. | Verified in Walsh's paper. | | Dogmachine, 12:13 13 Oct 2025 | Asks whether an \(n\)-conjecture could imply finiteness. | A question, not a claim. | | StijnC, 13:00 13 Oct 2025 | Explains that Ramaekers' variant would have sufficed, but Hölzl–Kleine–Stephan disprove it and current bounds are too weak. | The cited paper and its explicit disproof of Ramaekers' conjecture were verified; the application is discussed under “wall”. |

The page also records Likes this problem: KStar; every difficulty, tractability, and formalisation-worker marker is None.

1. Claim labels

I use the requested labels:

• (a) elementary-rigorous: proved directly in this report.
• (b) rigorous-modulo-named-theorem: a claim taken from a verified

primary theorem.

• (c) plausible/structural-unverified: conditional or not independently

proved.

• (d) computational-only: exhaustive finite computation with a

reproducible checker, not a uniform theorem.

2. Primary-source literature audit

1. Original problem — verified. Erdős, “Problems and results on number

theoretic properties of consecutive integers and related questions,” original PDF, p. 8, asks whether \(u_i^{(4)}+u_j^{(4)}=u_k^{(4)}\) has no or only finitely many solutions and then states the \(r-2\) generalisation.

1. Nitaj 1995 — verified. The primary journal abstract explicitly says it

proves infinitely many coprime \(3\)-powerful solutions to \(x+y=z\): A. Nitaj, “On a Conjecture of Erdős on 3-Powerful Numbers,” Bull. London Math. Soc. 27 (1995), 317–318, doi:10.1112/blms/27.4.317. Thus the third live-page question is settled (b).

1. Cohn 1998 — verified from the full two-page paper. Cohn states and

proves that one non-cube seed gives infinitely many pairwise-coprime \(3\)-powerful triples, and supplies a seed: J. H. E. Cohn, “A conjecture of Erdős on 3-powerful numbers,” Math. Comp. 67 (1998), 439–440, doi:10.1090/S0025-5718-98-00881-3.

1. Walsh — verified. The correct identifier is

arXiv:2404.03970, not the 2404.039701 currently printed in the page bibliography. Walsh's Theorem 1.1 gives infinitely many pairwise-coprime solutions \(x^3+y^3=p^4z^3\) from positive-rank curves, hence \(3\)-powerful triples. A journal version is doi:10.21857/y7v64t4jky. Its introduction also says that \(abc\) implies only finitely many pairwise-coprime \(4\)-powerful triples and that no example is known.

1. Lander–Parkin identity — verified. The arithmetic was recomputed.

The readily verifiable primary AMS publication is L. J. Lander and T. R. Parkin, “Counterexample to Euler's conjecture on sums of like powers,” Bull. Amer. Math. Soc. 72 (1966), 1079, doi:10.1090/S0002-9904-1966-11654-3. This primary bibliographic record differs from the page's [LaPa67] metadata, but it contains exactly the displayed identity.

1. Hölzl–Kleine–Stephan 2025 — verified. Their abstract states that

Ramaekers' conjecture is false for every \(n\geq5\): R. Hölzl, S. Kleine, F. Stephan, “Improved lower bounds for strong \(n\)-conjectures,” J. Aust. Math. Soc. 119 (2025), 61–81, doi:10.1017/S1446788725000084.

Exact-title and exact-equation searches of arXiv, AMS, journal pages, and the web found no primary paper giving an unconditional \(r=4\) example, unconditional \(r=4\) finiteness, or an infinite \(r=5\) family. The only current bounded \(r=4,5\) claims found were the May 2026 page comments and KentaKitamura's exploratory repository. This is a search miss, not a proof that no other literature exists.

3. What is now settled

3.1 The \(r\geq6\) construction is elementary-rigorous

This independently verifies the live comment.

For \(X>Y>0\), binomial expansion gives

There are \(h=\lceil r/2\rceil\) terms in the sum. Put

Replace the \(j=3\) term by \(t\) terms with coefficients

They are positive and distinct, and \(\sum_i v_i=2\binom r3\). The number of right-hand summands is now

Let \(B\) be divisible by every prime occurring in any coefficient \(v_i\) or \(2\binom rj\) for odd \(j\ne3\). Choose any prime \(q>B\) and set

The first summand and the left side of (1) are literal \(r\)th powers. Every other summand has the form \(cX^aY^b\) with \(b\geq1\); every prime of \(c\) already divides \(B\). Hence each prime exponent in that summand is at least \(r\), so it is \(r\)-powerful.

Finally,

because \(\gcd(X,Y)=1\). All non-first summands have prime support contained in that of \(XY\), while the first is \((X-Y)^r\). Thus the joint gcd of all \(r-2\) summands is \(1\). Infinitely many choices of \(q\) give distinct identities.

This proves (a) that for every \(r\geq6\), the answer to existence is yes and the answer to “at most finitely many?” is no.

The checker instantiated the construction at \(r=6,\ldots,10\), obtaining:

| \(r\) | number of terms | radical base \(B\) | prime \(q\) | joint gcd | |---:|---:|---:|---:|---:| | 6 | 4 | 30 | 31 | 1 | | 7 | 5 | 210 | 211 | 1 | | 8 | 6 | 1554 | 1559 | 1 | | 9 | 7 | 7014 | 7019 | 1 | | 10 | 8 | 16590 | 16603 | 1 |

3.2 The exact residual problem

• \(r=3\): infinitely many, by Nitaj/Cohn/Walsh (b).
• \(r=4\): no example is known. Unconditional existence/nonexistence and

unconditional finiteness remain open. The new finite exclusion below is (d).

• \(r=5\): existence is yes from the two checked examples (a), but

finite versus infinite remains open.

• Every \(r\geq6\): infinitely many by the construction above (a).

Thus only \(r=4\) and the infinitude/finitude part of \(r=5\) remain after incorporating the live results.

4. A clean coefficient reduction

Lemma (a). Every \(r\)-powerful positive integer has a unique expression

where \(d\) is \(r\)th-power-free and \(\operatorname{rad}(d)\mid x\).

Indeed, if \(v_p(n)=rq+s\), \(0\leq s<r\), put \(p^q\) into \(x\) and \(p^s\) into \(d\). If \(s>0\), \(r\)-powerfulness forces \(rq+s\geq r\), hence \(q\geq1\), so \(p\mid x\). Uniqueness follows prime by prime.

Consequently, the \(r=4\) case is exactly the union of equations

where each \(d_i\) is fourth-power-free, \(\operatorname{rad}(d_i)\mid x_i\), and the three terms are pairwise coprime. This explains why Fermat's theorem for pure fourth powers does not settle the problem: the coefficients \(d_i\) are unbounded, so (2) is not a finite list of generalized Fermat curves.

Similarly, \(r=5\) asks for

with fifth-power-free, unbounded coefficients and only joint coprimality of the first three terms. A uniform theorem over these coefficients is the missing step; treating any fixed finite coefficient list cannot close the problem.

5. New exact computation for \(r=4\)

Result

Computational certificate (d). There is no solution in positive \(4\)-powerful integers

with

Equivalently, any \(r=4\) solution must have \(\max(a,b)>10^{16}\). This extends the live comment's \(10^{14}\) bound by a factor of \(100\). It does not imply that no solution exists.

Complete enumeration

Let \(B=10^{16}\). Every prime divisor of a \(4\)-powerful \(n\leq2B\) obeys \(p^4\leq2B\). The checker sieves every prime through \(\lfloor(2B)^{1/4}\rfloor\), recursively chooses distinct primes in increasing order, and for each chosen prime tries every exponent \(e=4,5,\ldots\) allowed by the bound. Unique factorisation proves this enumerates every \(4\)-powerful \(n\leq2B\), exactly once.

The resulting exact counting table is:

| \(x\) | number of \(4\)-powerful \(n\leq x\), including \(1\) | |---:|---:| | \(1\) | 1 | | \(10\) | 1 | | \(10^2\) | 5 | | \(10^3\) | 11 | | \(10^4\) | 25 | | \(10^5\) | 57 | | \(10^6\) | 117 | | \(10^7\) | 235 | | \(10^8\) | 464 | | \(10^9\) | 906 | | \(10^{10}\) | 1,741 | | \(10^{11}\) | 3,312 | | \(10^{12}\) | 6,236 | | \(10^{13}\) | 11,654 | | \(10^{14}\) | 21,661 | | \(10^{15}\) | 40,049 | | \(10^{16}\) | 73,699 | | \(2\cdot10^{16}\) | 88,472 |

Now assume a solution has \(\max(a,b)\leq B\), and order \(a\leq b\). Then \(a,b\) lie in the 73,699-element summand list and \(c=a+b\leq2B\) lies in the 88,472-element target list.

The only pruning is a proved necessary condition modulo

For \(p=2,3\):

1. \(a,b\) cannot both be divisible by \(p\), since \(\gcd(a,b)=1\).
2. If \(p\mid a+b\), then \(p^4\mid a+b\), since the sum must be

\(4\)-powerful.

Residue buckets discard only pairs violating one of these statements. Every remaining unordered pair is tested by exact integer addition, hash-set membership for \(a+b\), and math.gcd(a,b).

Exact run ledger

Command:

/usr/bin/time -f 'wall=%e sec cpu_user=%U sec maxrss=%M KB' \
  python3 runs/erdos939_wavew023_reverify.py --progress

Output summary:

independent enumerator audit through 200,000: PASS
summand_count: 73699
target_count: 88472
all_unordered_pairs: 2715808150
active_residues: 495
locally_surviving_pairs: 631038029
powerful_sum_candidates_after_local_filters: 6692
coprime_hits: []
summand_sha256:
  c0fba29de66348f3ea560c6536434205c5b942d0cd7105b1ae9718bcb895d958
target_sha256:
  597520077e30c93d7ad4e8b2ddd9985bd728cb08130d6b993ffd668f37a5f441
wall: 92.72 sec
cpu_user: 92.64 sec
maxrss: 29032 KB

The digests encode each sorted integer as an unsigned 64-bit little-endian word. “Powerful-sum candidates” is deliberately the post-local-filter count, not the count before the safe congruence and local-coprimality filters.

Independent checks

The standalone checker contains checks separate from the main large scan:

1. It trial-factors every integer through \(200{,}000\) and compares the

resulting definition-based list with the recursive enumerator. Both give 72 terms and SHA-256 4950fe69b39b205ecba68ec0d10e9644346903608cc7551404c39a253e90cbea.

1. At \(B=10^8\), a completely unpruned scan checks all 107,880 unordered

pairs. It sees 959 sums that are \(4\)-powerful before local restrictions; the pruned scan sees 20 after them; both see zero coprime hits.

1. During development, an independent unpruned scan at the live comment's

\(B=10^{14}\) checked all 234,610,291 unordered pairs, found 97,425 \(4\)-powerful sums before the gcd test, and found zero coprime hits in 64.13 seconds. This independently reproduces the comment before the faster residue search extends it.

1. The checker recomputes from integer powers and trial factorisation the

Nitaj, Lander–Parkin, Cambie, and Kitamura identities, their required powerfulness, and their gcd claims.

The complete standard-library verifier is erdos939_wavew023_reverify.py.

Its exhaustive core is, schematically:

values = enumerate_r_powerful(2 * B, 4)
A = [n for n in values if n <= B]
S = set(values)
for a in A:
    for residue_b in locally_allowed_residue_buckets[a % 1296]:
        for b in that_bucket_with_b_at_least_a:
            if a + b in S and gcd(a, b) == 1:
                hits.append((a, b, a + b))

The proof above, rather than an assumption about density, is what makes the finite scan exhaustive.

6. Conditional finiteness and the exact wall

\(abc\) implication for \(r=4\)

For a coprime \(4\)-powerful solution \(a+b=c\),

and \(a,b<c\), so

The \(abc\) conjecture with, for example, \(\varepsilon=1/4\) would give

which bounds \(c\) and hence gives only finitely many solutions. This is a rigorous implication but has a conjectural premise, so it is labelled (c), not an unconditional solution.

Why the multi-term machinery does not finish \(r=5\)

For \(s\leq r-2\), the hoped-for strong \(n\)-conjecture estimate would need an exponent strong enough that a height \(H\) obeys roughly

whose exponent is \(<1\). The live comment correctly points out that the Ramaekers variant which would supply this route is false: Hölzl–Kleine– Stephan prove its failure for every \(n\geq5\). Moreover, the problem assumes only joint coprimality for three or more summands; large pairwise common factors are possible, so standard pairwise-coprime formulations do not apply for free.

The exact missing lemma is therefore a uniform height/radical bound for the unbounded coefficient family (3), under joint rather than pairwise coprimality, with exponent strictly below one after using \(5\)-powerfulness. No verified theorem found in the literature supplies that statement.

Computational wall and cost

The number of \(4\)-powerful integers through \(B\) grows on the order of \(B^{1/4}\), so a direct pair scan grows approximately as \(B^{1/2}\). Using the measured single-core 92.7 seconds at \(10^{16}\):

• \(B=10^{18}\) is estimated at about 15.5 CPU-minutes

(\(0.26\) core-hours) in this Python implementation.

• \(B=10^{20}\) is estimated at about 2.6 core-hours.

Those computations would only move a lower bound and cannot establish uniform nonexistence or finiteness. Closing \(r=4\) needs either an explicit identity or an unconditional uniform theorem for (2); closing \(r=5\) needs an infinite construction or the missing uniform height theorem for (3).

PARTIAL: (d) no coprime 4-powerful solution \(a+b=c\) has \(\max(a,b)\leq10^{16}\), a 100-fold extension of the live bound; (a) the \(r\geq6\) infinite construction is independently proved, leaving global \(r=4\) and finite-versus-infinite \(r=5\) open.