Erdős problem #913 — wave 7t

Date of live check and computation: 2026-07-28 (UTC).

Claim labels used below:

• [a] elementary-rigorous;
• [b] rigorous modulo the explicitly named published theorem;
• [c] plausible, conditional, or structural but unverified;
• [d] computational-only.

0. Mandatory live-page gate

I fetched the rendered page with the Bright Data browser path, not datacenter curl, and then followed its comments link. [d] The live page was marked OPEN; it showed 0 claimed proofs, Interested in collaborating: None, and Currently working on this problem: None. Thus the stop condition did not fire.

The exact current statement is:

Are there infinitely many \(n\) such that if \[ > n(n+1)=\prod_i p_i^{k_i} > \] is the factorisation into distinct primes then all exponents \(k_i\) are distinct?

The page cites [Er82c, p.28]. It also remarks that the desired conclusion would follow if there were infinitely many primes \(p\) for which \(8p^2-1\) is prime: for odd \(p\), put \(n=8p^2-1\), so that the exponent set is \(\{1,2,3\}\). The page links the Lean formalisation and OEIS A359747.

I read all three comments in the discussion thread. [d] In summary:

1. Przemek Chojecki (2026-01-23) observes that exponent \(1\) occurs at most

once, hence at least one of \(n,n+1\) is powerful; gives \(n=63\) as a small example; and describes the Schinzel/Bateman--Horn conditional family \(n=2^a p^b-1\).

1. StijnC notes the conditional Mersenne-prime route and cautions that a

typical large integer has many primes to exponent \(1\).

1. Nat Sothanaphan corrects the general Schinzel family by requiring

\(\gcd(a,b)=1\), which ensures the relevant binomial is irreducible.

The page itself warns that comments are unverified. None is marked as a claimed proof or partial solution.

Live sources:

• Problem #913
• Discussion thread

1. Literature check

I verified the following primary or author-hosted sources.

• P. Erdős, Miscellaneous problems in number theory, Congr. Numer. 34

(1982), 25--45, specifically printed p. 28: author archive PDF.

• K. Aktaş and M. Ram Murty, On the number of special numbers, Proc.

Indian Acad. Sci. Math. Sci. 127 (2017), 423--430, DOI 10.1007/s12044-016-0326-z. Their “special” integers are precisely integers whose prime exponents are mutually distinct.

• C. Sanna, *On the number of distinct exponents in the prime factorization

of an integer*, arXiv: 1902.09224, later Proc. Indian Acad. Sci. Math. Sci. 130 (2020), paper 27.

• C. Sanna, *On the exponents in the factorizations of \(r\) consecutive

numbers*, Quaestiones Math. 45 (2022), 1221--1228, DOI 10.2989/16073606.2021.1938277.

• A. Schinzel and W. Sierpiński, *Sur certaines hypothèses concernant les

nombres premiers*, Acta Arith. 4 (1958), 185--208: EuDML record and full text.

• P. Mihăilescu, *Primary cyclotomic units and a proof of Catalan's

conjecture*, J. reine angew. Math. 572 (2004), 167--195, DOI 10.1515/crll.2004.048.

[b] Aktaş--Murty and Sanna prove ambient density results for special integers or for prescribed values of the number of distinct exponents. Neither theorem imposes the cross-condition that the exponent sets of \(n\) and \(n+1\) be disjoint. In fact, the elementary lemma below forces at least one of \(n,n+1\) into the exceptional, zero-density set of powerful numbers. Thus those ambient asymptotics do not give a lower bound on the quadratic subsequence \(n(n+1)\).

[d] Searches by the exact factorisation phrase, by \(n(n+1)\), through the cited papers and their bibliographies, and through the OEIS cross-links found no primary source claiming a resolution of this specific infinitude question. This is a report of the search, not a claim that no such source can exist.

2. Elementary reduction and a quantitative upper bound

For a positive integer \(m\), set

regarded initially as a multiset.

Lemma 2.1 [a]. An integer \(n\) is a solution exactly when:

1. no value repeats inside \(E(n)\);
2. no value repeats inside \(E(n+1)\); and
3. the two resulting sets are disjoint.

Proof. Since \(\gcd(n,n+1)=1\), the prime factorisation of \(n(n+1)\) is the disjoint union, on prime bases, of those of \(n\) and \(n+1\). Therefore equality of two exponents can occur only in one of the three listed ways. \(\square\)

Recall that \(m\) is powerful if every prime divisor of \(m\) has valuation at least \(2\); \(1\) is included vacuously.

Lemma 2.2 [a] (powerful-neighbour reduction). Every solution has at least one of \(n,n+1\) powerful. More exactly, every solution is in one of the following disjoint structural cases:

• both \(n\) and \(n+1\) are powerful; or
• one is powerful and the other is \(q u\), where \(q\) is a prime,

\(v_q(qu)=1\), \(u\) is powerful, and \((q,u)=1\).

The additional distinctness conditions from Lemma 2.1 still have to hold.

Proof. All primes occurring to the first power contribute the same exponent \(1\), so at most one such prime can occur in the whole product. If none occurs, both neighbours are powerful. If it occurs on one side, the other side is powerful, and deleting that one prime from its own side leaves a powerful integer. \(\square\)

This gives a useful uniform counting result. Let \(P(Y)\) count powerful integers at most \(Y\), including \(1\).

Lemma 2.3 [a]. Every powerful integer has a unique representation

Consequently,

Proof. For each prime exponent \(e\ge2\), put the prime in \(b\) exactly when \(e\) is odd. The residual exponent \(e-3\) in the odd case, or \(e\) in the even case, is nonnegative and even and so goes into \(a^2\). Parity also proves uniqueness. Summing the possible \(a\)'s gives the formula and dropping squarefreeness and the floor gives the bound. \(\square\)

Write \(S(X)\) for the number of solutions \(n\le X\).

Proposition 2.4 [a].

In particular \(S(X)=O(\sqrt X)\), so the solution set has natural density zero.

Proof. By Lemma 2.2, every candidate is either a powerful \(n\le X\), or is one less than a powerful integer at most \(X+1\). Count the union by the sum of its two parts and apply Lemma 2.3. \(\square\)

This does not prove finiteness or infinitude, but it reduces an exhaustive search from \(X\) integers to \(O(\sqrt X)\) structurally forced candidates.

3. Two exact uniform slices

3.1 All exponents at most 2

Proposition 3.1 [a]. The only positive solutions for which every exponent is at most \(2\) are

For \(n\ge2\), the only ones are \(3\) and \(4\).

Proof. The case \(n=1\) gives \(1\cdot2=2\). If \(n\ge2\), each of \(n,n+1\) contributes at least one prime. Distinct positive exponents not exceeding \(2\) must therefore be exactly \(\{1,2\}\), with one prime on each side. Thus the neighbours are a prime and the square of a prime.

If \(n=p\) and \(n+1=q^2\), then \(p=q^2-1\); this is composite for \(q>2\), while \(q=2\) gives \(n=3\). If \(n=q^2\) and \(n+1=p\), odd \(q\) makes \(p=q^2+1\) an even integer greater than \(2\), while \(q=2\) gives \(n=4\). \(\square\)

3.2 Exactly two distinct prime divisors

Let \(\omega(m)\) be the number of distinct prime divisors of \(m\).

Proposition 3.2 [b, using Mihăilescu's theorem]. The solutions \(n\ge2\) with \(\omega(n(n+1))=2\) are exactly

Moreover, in the first line \(a\) must be prime, and in the second line \(a\) must be a power of \(2\). Hence this two-prime slice is infinite if and only if there are infinitely many Mersenne primes or infinitely many Fermat primes.

Proof. Because \(n,n+1>1\), each must be a power of one prime. If both exponents exceed \(1\), Mihăilescu's theorem says the only consecutive perfect powers are \(8\) and \(9\), giving \(n=8\) and exponents \(3,2\).

Otherwise one exponent is \(1\), and distinctness makes the other at least \(2\). The even neighbour cannot be the prime \(2\), since its only positive neighbour here is \(3\), which is not a higher prime power. Thus the even neighbour is \(2^a\) and the odd neighbour is prime, giving the two displayed families.

If \(a\) is composite then \(2^a-1\) has a proper algebraic factor. If \(a=2^t u\) with odd \(u>1\), then \((2^{2^t})^u+1\) is divisible by \(2^{2^t}+1\). These give the final restrictions. \(\square\)

4. Why the standard consecutive-powerful construction does not help

There is an elementary recursive construction of infinitely many consecutive powerful pairs:

Indeed,

and if \(x_r,x_r+1\) are powerful then so are \(x_{r+1},x_{r+1}+1\).

Proposition 4.1 [a]. In this infinite family, only the seed \(x_0=8\) solves problem #913.

Proof. The first iterate is

so exponent \(2\) repeats. For \(r\ge2\), \(x_r\) is divisible by \(x_2\), and

is divisible by both \(3^2\) and \(17^2\). Thus exponent \(2\) already repeats inside \(x_r\). The seed has \(8=2^3\), \(9=3^2\), so it does qualify. \(\square\)

This is only a negative result for this standard recursive family; it does not exclude other consecutive powerful pairs.

5. Exact computation through \(10^8\)

The standalone verifier is erdos913_wave7t_reverify.py. It uses only the Python standard library.

Run:

python runs/erdos913_wave7t_reverify.py --progress \
  > /tmp/erdos913.json

Two complete methods are required to agree:

1. Signature scan. A smallest-prime-factor sieve computes a bit mask

for every \(m\le10^8+1\). Bit \(e\) records valuation \(e\); a sentinel records a repeated valuation inside \(m\). Consecutive valid masks must be disjoint.

1. Powerful-neighbour scan. Independently enumerate every

\(a^2b^3\le10^8+1\) with squarefree \(b\), take both possible neighbouring candidates, and factor every candidate from scratch by trial division. Lemma 2.2 proves this candidate list exhaustive.

The program also trial-factors every hit, brute-forces every \(n\le100000\) independently, checks the OEIS list through \(4000\), and checks the arithmetic used in Proposition 4.1.

5.1 Counts

Let \(C(X)=\#\{n\le X:n\text{ qualifies}\}\).

[d] Both full methods returned exactly 1469 hits. The second method enumerated 21044 powerful integers through \(100000001\), whose two neighbour positions formed 42077 distinct candidates, and returned the same 1469 hits.

[d] The SHA-256 digest of all lines

n|factorization(n)|factorization(n+1)\n

in increasing \(n\) order is

d400f508a1a401ab0b72feb72816191070f9889fe30039936905ba0d44f48356

The digest is regenerated only after every displayed factorisation has been multiplied back and its exponent condition rechecked.

5.2 Exact subdivisions

[d] Counts by \(\omega(n(n+1))\) through \(10^8\):

There were no hits with more than five distinct prime divisors in this range.

[d] The 12 two-prime hits were

in agreement with Proposition 3.2.

[d] Only \(n=8\) had both \(n\) and \(n+1\) powerful. This is a bounded statement only.

[d] There were 134 hits with exact exponent set \(\{1,2,3\}\). Of those, 116 came from the live page's odd-prime family \(n=8p^2-1\). The script outputs all 116 members of that subfamily.

[d] The last ten hits at or below \(10^8\) were

[d] The largest gap between consecutive recorded hits was 843912, from 53831087 to 54674999. Neither this gap nor the count data imply finiteness or infinitude.

The final clean run used Python 3.12.3, one CPU core, 73.82 seconds wall time, and a peak resident set of 798676 KiB.

6. Conditional route and exact remaining wall

For an odd prime \(p\), if \(8p^2-1\) is prime then

so the exponents are \(1,3,2\). [a] This verifies the page's sufficient construction term by term. (The exceptional \(p=2\) merges the two powers of \(2\); it still gives the solution \(n=31\), now with exponents \(\{1,5\}\).)

[c] Schinzel's Hypothesis H applied to \(x\) and \(8x^2-1\), or the Bateman--Horn conjecture, predicts infinitely many such odd primes \(p\). This is conditional, not a proof of #913. The exact missing lemma for this route is:

Even proving infinitely many prime values of one irreducible quadratic is open; the simultaneous prime restriction adds the usual sieve parity barrier. A proof of the displayed lemma would settle #913 positively, but the converse need not hold because other exponent patterns can supply solutions.

The powerful route has an equally concrete wall. [a] Lemma 2.2 reduces it to signature-separated solutions of

where \(U,V\) are powerful and \(q\) is prime, with the appropriate coprimality and disjoint-exponent constraints. Consecutive powerful numbers alone are plentiful, but Proposition 4.1 shows why the standard recursive family immediately repeats exponent \(2\). No checked theorem supplies infinitely many pairs with the required separated signatures.

[c] A linear extrapolation of the signature scan to \(10^9\) would cost about 0.21 core-hours (roughly 12--13 minutes here) and about 8 GB RAM, so I did not run it under the few-minute limit. More finite computation cannot resolve an infinitude question. What is actually needed is either the simultaneous-prime lemma above, or a new uniform theorem producing infinitely many signature-separated powerful/almost-powerful neighbours.

PARTIAL: proved an elementary O(sqrt(X)) candidate bound, classified the max-exponent-2 and two-prime slices, ruled out the standard infinite powerful recurrence beyond n=8, and independently verified exactly 1469 solutions through 10^8; infinitude remains open.