Erdős problem #979 — wave 7y
Access/search date: 2026-07-28 UTC.
Claim labels used throughout:
- (a) elementary-rigorous: a complete elementary argument is given here.
- (b) rigorous-modulo-named-theorem: the claim is a consequence of the cited theorem/source.
- (c) plausible/structural-unverified: heuristic, search miss, or otherwise not established.
- (d) computational-only: established by the finite computation and checker described below.
Outcome
(d) For the first open exponent \(k=4\), I exhaustively computed the
smallest integer with at least \(r\) unordered prime-quadruple
representations for \(1\leq r\leq4\):
\[ \begin{array}{c|r|c} r&M_4(r)&\text{number of representations at }M_4(r)\\ \hline 1&64&1\\ 2&726724&2\\ 3&141339844&3\\ 4&199898912404&4 \end{array} \]Here
\[ M_4(r)=\min\{n:U_4(n)\geq r\},\qquad U_4(n)=\#\{p\leq q\leq r\leq s:\ p^4+q^4+r^4+s^4=n\}. \](d) The same exhaustive search, extended through every prime below
1511, proves the sharp concrete-range statement
\[ U_4(n)\leq4\quad\text{for every }n<5\,212\,641\,500\,689. \]Equivalently, \(M_4(5)\geq5\,212\,641\,500\,689\), with \(M_4(5)=\infty\)
allowed if no fivefold value exists.
(c) These finite results do not establish that \(U_4\), or the
original \(f_4\), is unbounded.
Step 0: mandatory live-page audit
(d) I fetched the rendered live page through the Bright Data browser,
not through datacenter curl: problem #979
and its discussion thread.
The observations below are from the live render on 2026-07-28.
Verbatim statement
(d), verbatim live-page transcription:
> Let \(k\geq 2\), and let \(f_k(n)\) count the number of solutions to
> \[ > n=p_1^k+\cdots+p_k^k, > \]
> where the \(p_i\) are prime numbers. Is it true that
> \(\limsup f_k(n)=\infty\)?
The page attributes the problem to [Er65b, p. 224].
Live status and participation markers
(d) The page says:
- status: OPEN;
- claimed proofs: 0;
- interested in collaborating: None;
- currently working on this problem: None;
- likes: None;
- “looks difficult”: None;
- “looks tractable”: None;
- “results could be formalisable”: None;
- “working on formalising”: None;
- formalised statement: Yes;
- related OEIS sequence: A385316 possible;
- last edited: 19 September 2025.
Thus the requested claimed-proof/current-worker stop condition did not
apply.
Known result printed on the page
(d), verbatim live-page transcription:
> Erdős [Er37b] proved this is true when \(k=2\), and also when \(k=3\)
> (but this proof appears to be unpublished).
(b) I treat the published \(k=2\) result as a theorem of Erdős.
(c) I do not treat the \(k=3\) assertion as audited theorem-level
ground truth because the page and Erdős both say that its proof was
unpublished.
Discussion/comments read before doing mathematics
(c) The page declares 11 comments; the rendered thread also contains
one deleted-post placeholder. The substantive content, which the page
itself warns is unverified, is:
1. StijnC reported two numbers having at least five representations by
three prime cubes and suggested an OEIS sequence.
2. Thomas Bloom asked whether only positive primes and genuinely
distinct representations were intended. StijnC confirmed positive
prime cubes and different prime multisets.
3. The thread gives the claimed minimal fivefold example
\(10588881419\), with triples
\[ \{59,1669,1811\},\{83,1567,1889\},\{139,1427,1973\}, \{349,1091,2099\},\{479,929,2131\}. \]
A later comment says OEIS A385316 was
added.
4. StijnC asked how \(k=3\) was proved. Bloom quoted Erdős's statement
that the proof was unpublished and used special properties of
primes. Bloom suggested trying to combine Mahler's parametric
three-cube identity with sieve theory, while noting that simultaneous
prime values of the resulting higher-degree polynomials are out of
reach.
5. StijnC gave a birthday-paradox/extreme-collision heuristic suggesting
unbounded multiplicity despite vanishing average density.
6. Bloom linked problem #322, the analogous question for all integers,
and noted the special positive result for three cubes.
7. After a deleted post, Woett objected that an argument implying many
two-term fifth-power collisions would run into the much harder
equation \(a^5+b^5=c^5+d^5\), referring to problem #324.
8. StijnC emphasized that the average representation count tends to
zero: the number of prime \(k\)-tuples up to scale \(N^k\) is
\(o(N^k)\).
No comment was a claimed proof or a current-worker declaration.
Primary-source literature audit
Erdős's sources
(d) The original survey was located and read:
P. Erdős, [*Some Recent Advances and Current Problems in Number
Theory](https://users.renyi.hu/~p_erdos/1965-17.pdf), in Lectures on
Modern Mathematics*, vol. III (1965), p. 224.
(d) On p. 224 Erdős defines the same \(f_k\), cites his \(k=2\)
paper, says he can also prove \(k=3\) unpublished, and then says
“I can prove nothing for \(k>3\).” This matches the live-page summary.
(b) The published \(k=2\) source is P. Erdős,
[*On the Sum and Difference of Squares of
Primes](https://doi.org/10.1112/jlms/s1-12.1.133), J. London Math.
Soc.* 12 (1937), 133–136. Its sequel,
[*On the Sum and Difference of Squares of Primes
(II)*](https://users.renyi.hu/~p_erdos/1937-08.pdf), states in its
introduction that Part I already gave unboundedly many representations
for infinitely many \(n\), and Part II strengthens the quantitative
lower bound. This verifies that the cited paper exists and supports the
page's \(k=2\) claim.
A withdrawn apparent solution
(d) Exact-title searching found Anay Aggarwal,
[*On integers with many representations as the sum of \(k\)th powers of
primes*](https://arxiv.org/abs/2509.11558), arXiv:2509.11558. Its
abstract claimed the stronger result of unbounded representations by
only two prime \(k\)th powers.
(d) The official current arXiv record says the paper was withdrawn
on 2025-09-16 and gives the comment “Error in pigeonhole argument.”
No PDF is currently available. Therefore it is not evidence for a
solution.
Other nearby current literature
(d) A search result for arXiv:2512.14386 still exposed stale
“four fourth powers” metadata. The official current record, Yang Qu and
Rong Ma, [*Sums of eight fourth power of
primes*](https://arxiv.org/abs/2512.14386), version 5, actually concerns
eight fourth powers. It therefore does not answer the \(k=4\),
four-summand extremal-multiplicity question.
(b) Alessandra Migliaccio and Alessandro Zaccagnini,
[*The average number of representations of an integer as a sum of two
prime powers over multiples of a fixed
integer*](https://arxiv.org/abs/2603.24120), arXiv:2603.24120v2
(2026-07-21), prove under GRH a weighted first-moment asymptotic for
two prime \(k\)th powers over multiples of a fixed modulus.
(a) A first-moment asymptotic of total order \(N^{2/k}\) does not
give a lower bound for the maximum multiplicity, so this result does
not settle #979.
(d) Metadata was also verified for Patrick Shields,
[*Representation of Integers as a Sum of Two \(h\)-TH Powers of
Primes](https://doi.org/10.1112/plms/s3-38.2.369), Proc. London Math.
Soc.* 38 (1979), 369–384. The full text was closed in the sources
available here, so I make no theorem-level claim about its contents.
(c) Exact-statement/title searches over arXiv, publisher metadata,
the Erdős archive, and indexed web results found no non-withdrawn paper
claiming the required unbounded maximum for every \(k\). This is an
honest search miss, not a proof that no such paper exists.
Counting convention
(a) The statement does not say whether permuting the \(p_i\) creates
a new solution. The discussion and A385316 use unordered multisets, so
the computation uses \(p_1\leq\cdots\leq p_k\).
(a) If \(f_k^{\rm ord}\) instead counts ordered tuples, then for
every \(n\)
\[ U_k(n)\leq f_k^{\rm ord}(n)\leq k!\,U_k(n). \]Thus ordered and unordered unboundedness are equivalent.
Exact \(k=4\) computation
Displayed certificates
(d) The least twofold value is
\[ 726724 =7^4+7^4+11^4+29^4 =17^4+17^4+23^4+23^4. \](d) The least threefold value is
\[ \begin{aligned} 141339844 &=11^4+17^4+17^4+109^4\\ &=11^4+61^4+79^4+97^4\\ &=41^4+59^4+61^4+103^4. \end{aligned} \](d) The least fourfold value is
\[ \begin{aligned} 199898912404 &=23^4+281^4+397^4+641^4\\ &=137^4+383^4+467^4+601^4\\ &=151^4+227^4+557^4+563^4\\ &=257^4+317^4+347^4+643^4. \end{aligned} \](d) All 16 primes in the last display are distinct within their
own quadruples. Since the exhaustive count is exactly four unordered
representations, this same integer has exactly \(4\cdot4!=96\) ordered
representations.
Why the record claims are exhaustive
Let \(N=199898912404\).
(a) Any representation \(p^4+q^4+r^4+s^4\leq N\), with
\(p\leq q\leq r\leq s\), satisfies
\[ s^4+3\cdot2^4\leq N, \]so \(s\leq\lfloor(N-48)^{1/4}\rfloor=668\).
(d) A fresh Eratosthenes sieve finds 121 primes at most 668, ending
with 661.
(a) The number of nondecreasing quadruples drawn from these 121
primes is
\[ \binom{121+4-1}{4}=\binom{124}{4}=9\,381\,251. \](d) The checker generates all 9,381,251 sums as exact unsigned
64-bit integers, sorts them, run-length encodes equal values, and
selects the first run of length at least \(r\). It returns precisely the
four-row table above.
(a) Because every representation of every \(n\leq N\) occurs in
that finite list, the first-run calculation proves minimality within
the stated unordered convention, modulo the correctness of the finite
execution.
Lower bound for the first fivefold value
(d) The extended run sieves the 239 primes at most 1500 (last prime
1499), generates
\[ \binom{242}{4}=139\,389\,580 \]nondecreasing quadruples, sorts all their sums, and finds no run of
length five.
(a) The next prime is 1511. Any quadruple using it or a larger prime
has sum at least
\[ 1511^4+3\cdot2^4=5\,212\,641\,500\,689. \]Consequently the extended finite result proves
\[ U_4(n)\leq4\quad(n<5\,212\,641\,500\,689). \]Code and reproduction
The standalone independent checker is
erdos979_wave7y_verify.py. It contains:
- a fresh deterministic Eratosthenes sieve;
- exact integer fourth roots;
- packed exhaustive tuple generation;
- an exact integer sort and run-length check;
- a second, structurally different recovery pass that loops over the
first three primes and solves exactly for the fourth;
- an optional larger exhaustive check for the fivefold lower bound.
The initial C++ search implementation is
erdos979_search.cpp. It independently uses
four nested loops and a std::sort of uint64_t sums.
The core enumeration used by the Python checker is:
for i in range(m):
for j in range(i, m):
prefix_ij = powers[i] + powers[j]
for k in range(j, m):
width = m - k
sums[cursor:cursor + width] = prefix_ij + powers[k] + powers[k:]
cursor += width
sums.sort()
Reproduction commands:
python runs/erdos979_wave7y_verify.py
python runs/erdos979_wave7y_verify.py --extended
g++ -O3 -std=c++17 -Wall -Wextra -Werror -pedantic \
runs/erdos979_search.cpp -o /tmp/erdos979_search
/tmp/erdos979_search 668
/tmp/erdos979_search 1500
Observed independent-checker output:
base_bound=668; primes=121; tuples=9381251
least_at_least_1=64; count=1; reps=(2,2,2,2)
least_at_least_2=726724; count=2; reps=(7,7,11,29); (17,17,23,23)
least_at_least_3=141339844; count=3; reps=(11,17,17,109); (11,61,79,97); (41,59,61,103)
least_at_least_4=199898912404; count=4; reps=(23,281,397,641); (137,383,467,601); (151,227,557,563); (257,317,347,643)
VERIFIED: exhaustive unordered k=4 record table through multiplicity 4
extended_base_bound=1500; primes=239; tuples=139389580
VERIFIED: every n < 5212641500689 has at most four unordered representations
(d) On this VM the default checker took about 1.4 seconds and
329 MB maximum RSS. The extended check took about 13 seconds and
1.14 GB maximum RSS.
Source hashes:
6bf792707412c21d726d59f16cb7da5d5700c87be9c0d40b1d2f2a3170ae897f erdos979_search.cpp
199e94f5d838de5278f45148e4057761426a97fbbda8d01a68b8d5b7cf443db0 erdos979_wave7y_verify.py
Exact reduction and the remaining wall
(a) For fixed \(k\) and \(R\), define the off-diagonal \(R\)-fold
collision count
\[ \mathcal E_{k,R}(X)= \sum_{n\leq X}\binom{U_k(n)}{R}. \]It counts choices of \(R\) genuinely different unordered prime
\(k\)-tuples having a common sum of \(k\)th powers.
(a) The original problem is equivalent to:
\[ \text{for every fixed }k\geq2\text{ and }R\geq1,\quad \mathcal E_{k,R}(X)>0\text{ for some }X. \]For \(k=4,R=5\), the computation proves
\(\mathcal E_{4,5}(X)=0\) for
\(X<5\,212\,641\,500\,689\).
(b) The prime number theorem gives the first-moment bound
\[ \sum_{n\leq X}U_k(n) \leq \binom{\pi(X^{1/k})+k-1}{k} =O_k\!\left(\frac{X}{(\log X)^k}\right). \]Hence the mean of \(U_k(n)\) over \(n\leq X\) tends to zero.
(a) This identifies the precise pigeonhole failure: first-moment
counting supplies fewer than one tuple per target integer on average,
and congruence restrictions improve this by only constant factors.
Vanishing average is compatible with an unbounded maximum, but it
cannot prove one.
(c) The missing analytic lemma is a positive lower bound for the
off-diagonal high collision count \(\mathcal E_{k,R}(X)\), for every
\(R\), after excluding permutations and repeated identical
representations. Current first-moment Waring–Goldbach estimates and the
average theorem above do not supply such a bound. A constructive
alternative would be a family of arbitrarily many equal sums together
with an unconditional theorem forcing every base in those identities
to be prime; the discussion's Mahler route stalls exactly at this
prime-values step for higher-degree polynomials.
(c) Straight exhaustive search scales badly. With prime cutoff
\(B\), it stores
\(\binom{\pi(B)+3}{4}\) sums. At \(B=5000\) this is 8,421,345,240
values (62.7 GiB raw); an in-memory or external sort would reasonably
cost about 2–10 core-hours and require a 64–128 GiB node, roughly
\$5–\$30 at ordinary on-demand compute/RAM pricing. At \(B=10000\) it
is 95,524,442,860 values (711.7 GiB raw), plausibly 30–100 core-hours
and tens to hundreds of dollars. Neither cutoff is guaranteed to find
a fivefold value, so those jobs were not run here.
PARTIAL: Exhaustively proved for unordered k=4 representations that M_4(4)=199898912404 (four displayed prime quadruples) and M_4(5)>=5212641500689; the uniform unboundedness question remains open.