Erdős problem 978 — live-page audit, exact finite data, and the remaining tail

Access date: 2026-07-28 UTC

Live page: <https://www.erdosproblems.com/978>

Discussion: <https://www.erdosproblems.com/forum/discuss/978>

Verifier: runs/erdos978_wave7y_reverify.py

Outcome

The live page is OPEN, lists 0 claimed proofs, and lists **Currently

working: None. I therefore proceeded. This report does not** solve the

uniform problem.

The verifiable progress is:

1. [d, deterministic computation] Exactly

\[ 75\,670 \]

of the integers \(1\leq n\leq100\,000\) have \(n^4+2\) squarefree. Every

one of the 100,000 complete factorizations is checked by a second layer:

PARI only proposes factors, while the standalone Python verifier proves

primality in the applicable deterministic Miller--Rabin range and checks

exponents and multiplication back to \(n^4+2\).

2. [a, elementary-rigorous] The exact local factor is

\[ \rho(p^2)=\#\{x\bmod p^2:x^4+2\equiv0\pmod {p^2}\} = \begin{cases} 0,&p=2,\\ 2,&p\equiv3\pmod8,\\ 0,&p\equiv5,7\pmod8,\\ 4,&p\equiv1\pmod8\text{ and }(-2)^{(p-1)/4}\equiv1\pmod p,\\ 0,&p\equiv1\pmod8\text{ otherwise}. \end{cases} \]

3. [d, deterministic computation resting on the elementary formula] If

\[ c=\prod_p\left(1-\frac{\rho(p^2)}{p^2}\right), \]

directed rounding through \(p\leq10^7\), followed by a rigorous elementary

bound for the omitted tail, gives

\[ 0.75668370260818737477048117621050903107604397865209 \leq c\leq 0.75668400528178948748627617072097731946697177451523. \]

4. [a, elementary-rigorous] The problem reduces cleanly to controlling

primes \(p>\sqrt X\) for which \(p^2\mid n^4+2\). The exact missing

estimate is displayed below.

5. [a, elementary-rigorous] The only search hit I found that explicitly

claims an unconditional solution for \(n^4+2\), arXiv:2310.16952, has a

concrete sign error in its decisive fractional-part estimate. For one of

its own lifted roots, its asserted identity reads

\(\{-29/121\}=-29/121\), whereas the left side is \(92/121\).

Here and below:

• [a] means elementary-rigorous;
• [b] means rigorous modulo the explicitly named published theorem;
• [c] means plausible/structural but unverified;
• [d] means computational-only, even when the computation is exhaustive.

Step 0: mandatory live-page check

I fetched both the problem and its discussion with a Bright Data browser

session, not datacenter curl. The page title was 978 | Erdős Problems;

the page says it was last edited 31 March 2026.

The live markers were:

• status: OPEN (the status tooltip also says that it cannot be resolved by

a finite computation);

• comments: 16;
• claimed proofs: 0;
• interested in collaborating: None;
• currently working: None;
• working on formalising: None.

Verbatim current statement

The following is copied verbatim from the live page's LaTeX view:

> Let \(f\in \mathbb{Z}[x]\) be an irreducible polynomial of degree \(k>2\)

> (and suppose that \(k\neq 2^l\) for any \(l\geq 1\)) such that the leading

> coefficient of \(f\) is positive.

>

> Does the set of integers \(n\geq 1\) for which \(f(n)\) is

> \((k-1)\)-power-free have positive density?

>

> If \(k>3\), and for all primes \(p\) there exists \(n\) such that

> \(p^{k-2}\nmid f(n)\), then are there infinitely many \(n\) for which

> \(f(n)\) is \((k-2)\)-power-free?

>

> In particular, does

> \[ > n^4+2 > \]

> represent infinitely many squarefree numbers?

The final sentence makes clear that the power-of-two degree exclusion is meant

for the first question and its exceptional fixed-divisor phenomenon, not to

exclude the stated quartic special case.

Results listed on the live page

These are treated as ground truth for the problem's stated background:

• Erdős [Er53] proved infinitely many \((k-1)\)-power-free values, except

when \(k=2^l\) and \(2^{k-1}\mid f(n)\) for every \(n\). The page gives

\(f(x)=k!(\binom{x}{k}+1)\) as a possible exceptional example.

• Hooley [Ho67] proved the asymptotic formula, settling the first question.
• Heath-Brown [He06] proved the second assertion for \(k\geq10\), and

Browning [Br11] extended it to \(k\geq9\), under the necessary local

condition now written into the statement.

• The page also records Erdős's comments about \(2^n\pm1\) and \(n!\pm1\).

All 16 discussion comments checked

The comments matter because the statement changed in March 2026. In

newest-first substance:

1. Moritz Firsching reported a DeepMind/Lean counterexample to the older

second question:

\[ X^6+33X^5+21X^4+63X^3+18X^2+24X+48. \]

It is Eisenstein at 3 and is always divisible by \(16\).

2. old-bielefelder congratulated the construction and asked how it was

found.

3. Dogmachine observed that one should exclude a fixed divisor.

4. Thomas Bloom explained that the standard conjecture assumes no fixed

\(r\)-th-power prime divisor.

5. Moritz Firsching proposed adding the precise \(p^{k-2}\) local condition.

6. Thomas Bloom confirmed that the page had been changed and knew of no

further implicit hypothesis.

7. old-bielefelder relayed ChatGPT/Gemini claims conditional on \(abc\).

8. qrdl linked Browning--Shparlinski's random-polynomial paper and

Granville's \(abc\) paper.

9. Moritz Firsching noted that the second part needs \(k>3\); the page was

updated.

10. Adenwalla corrected the fixed divisor from a mistyped exponent to

\(2^{k-1}\); the page was updated.

11. Yongxi Lin requested the positive-leading-coefficient assumption and

the natural-number interpretation; the page was updated.

12. Dogmachine asked why powers of two are excluded when the special case

is quartic.

13. StijnC explained the Erdős fixed-divisor example and why the final

quartic lacks that obvious obstruction.

14. StijnC noted that Bunyakovsky's conjecture would make \(n^4+2\) prime,

hence squarefree, infinitely often.

15. Dogmachine asked about cyclotomic examples.

16. StijnC answered that both examples are natural and pointed to related

literature.

The first comment is not a claimed resolution of the live statement. Its

polynomial has the fixed fourth-power divisor \(2^4\), so it violates the

condition now imposed when \(k=6\). The verifier independently checks the

comment's arithmetic: the polynomial is Eisenstein at 3, agrees modulo 16

with \(X(X-1)\cdots(X-5)\), and hence is always divisible by 16; its value

at zero is 48, so it is not always divisible by \(2^5\).

Primary-source literature audit

I searched for the exact fixed-polynomial question, not merely the tags.

1. Erdős and Hooley. The cited papers exist with the stated metadata:

P. Erdős, Arithmetical properties of polynomials, JLMS 28 (1953),

416--425, DOI 10.1112/jlms/s1-28.4.416;

C. Hooley, On the power free values of polynomials, Mathematika 14

(1967), 21--26,

DOI 10.1112/S002557930000797X.

2. General determinant-method range. Heath-Brown's

Theorem 16

proves the expected asymptotic for an irreducible degree-\(d\) polynomial

when the power-free exponent \(r\) satisfies

\(r\geq(3d+2)/4\). Thus \(r=d-2\) is covered for \(d\geq10\).

Browning's paper is real and has the cited bibliographic data:

T. D. Browning, Power-free values of polynomials, Arch. Math. 96

(2011), 139--150,

DOI 10.1007/s00013-011-0224-7.

Heath-Brown's later introduction records the Salberger/Browning

improvement to \((3d+1)/4\), hence \(d\geq9\). [b]

3. A relevant improvement for binomials. D. R. Heath-Brown,

Powerfree Values of Polynomials,

arXiv:1103.2028,

later Q. J. Math. 64 (2013), 177--188,

DOI 10.1093/qmath/har030,

proves the asymptotic for \(X^d+c\) when

\[ r\geq\frac{5d+3}{9}. \]

Taking \(r=d-2\) covers these special binomials for every \(d\geq6\).

The paper explicitly says that the quartic squarefree case is the desired

next case. At \(d=4,r=2\), its required inequality is

\(2\geq23/9\), which fails. This is useful partial literature not

mentioned on the live page. [b]

4. Conditional result. A. Granville,

ABC allows us to count squarefrees, IMRN 1998, 991--1009,

DOI 10.1155/S1073792898000592,

proves the expected fixed-polynomial squarefree asymptotic under the

\(abc\)-conjecture. Thus \(abc\) gives a positive answer for \(n^4+2\),

but this is conditional. [b]

5. Average results are not a fixed-polynomial solution.

Browning--Shparlinski,

Square-free values of random polynomials,

arXiv:2305.15493, J. Number Theory

261 (2024), 220--240,

DOI 10.1016/j.jnt.2024.02.013,

says in its introduction that fixed polynomials of degree at least four

have only the conditional \(abc\) treatment and proves an average-over-

polynomials theorem. Jelinek,

arXiv:2308.15146, is likewise an

average theorem. Neither specializes to prove the assertion for the

single polynomial \(X^4+2\).

6. Pandey is conditional/roadmap, not a closure. P. P. Pandey,

arXiv:2303.06610, explicitly labels

its key estimate a conjecture and gives conditional consequences and a

quadratic demonstration. It does not unconditionally settle \(X^4+2\).

7. An explicit but invalid solution claim. N. A. Carella,

Squarefree Values Of Polynomials,

arXiv:2310.16952v1, claims

\[ \sum_{x\leq n\leq2x}\mu^2(n^4+2) =c_fx+O(x^{1/2+\varepsilon}). \]

It is a single 2023 submission in math.GM; I found no journal version.

More importantly, its proof has the exact fatal gap in the next section.

The 2024 peer-reviewed primary literature and the 2026 live page still

describe the fixed quartic problem as open.

The searches found no verified paper proving or disproving the amended

question. The inaccessible full text of Browning's Springer article was

cross-checked through its DOI metadata, the live citation, and Heath-Brown's

primary-source summary; I do not claim to have independently reconstructed

that paper's proof.

Exact local analysis

Let

Proposition

The displayed formula for \(\rho(p^2)\) in the Outcome section holds.

Proof [a].

For \(p=2\), an even \(a\) gives \(a^4+2\equiv2\pmod4\), while an odd \(a\)

gives \(a^4+2\equiv3\pmod4\). Thus \(\rho(4)=0\).

Let \(p\) be odd. A root \(a\) of \(a^4+2\equiv0\pmod p\) is nonzero and

The elementary one-step Hensel calculation therefore gives one and only one

lift \(a+tp\bmod p^2\). Consequently \(\rho(p^2)=\rho(p)\).

The group \(\mathbb F_p^\times\) is cyclic of order \(m=p-1\). In a cyclic

group, \(x^4=b\) has \(g=\gcd(4,m)\) solutions exactly when

\(b^{m/g}=1\), and otherwise has none. Taking \(b=-2\) gives

The elementary quadratic-character identity

\((\frac{-2}{p})=1\) exactly for \(p\equiv1,3\pmod8\) now yields the four

cases. \(\square\)

For squarefree \(d\), the Chinese remainder theorem gives

In particular \(\rho(d^2)\leq4^{\omega(d)}\ll_\varepsilon d^\varepsilon\).

Also \(\sum_p\rho(p^2)/p^2<\infty\), no factor vanishes, and therefore

\(c>0\). These are all [a].

The first relevant primes are

The verifier brute-forces the formula for every prime \(p\leq10^4\) and

explicitly constructs every unique lift modulo \(p^2\) for \(p\leq1000\).

Rigorous numerical enclosure for the local constant

At \(B=10^7\), the exact root-type counts among all

\(\pi(B)=664\,579\) primes are

The zero count includes \(p=2\). Thus 249,055 primes contribute a nontrivial

factor to the truncated product.

The checker multiplies every rational factor with 50-digit directed lower

and upper rounding. For the omitted primes,

Since \(\prod_i(1-a_i)\geq1-\sum_i a_i\) for \(0\leq a_i\leq1\), the omitted

tail is greater than \(1-4/B\). This proves the numerical enclosure quoted

above, modulo only the exhaustively checked prime sieve and integer modular

arithmetic. It is still labelled [d], not a theorem about the limiting

frequency of squarefree values.

The predicted statement

is [c] unconditionally and [b] under Granville's use of \(abc\).

Exact finite computation

Define

The deterministic table is:

| \(N\) | \(S(N)\) | \(S(N)/N\) |

|---:|---:|---:|

| 10 | 8 | 0.800000 |

| 100 | 75 | 0.750000 |

| 1,000 | 759 | 0.759000 |

| 10,000 | 7,564 | 0.756400 |

| 100,000 | 75,670 | 0.756700 |

This table is [d]. It is consistent with the local constant but proves

nothing uniform as \(N\to\infty\).

Why the checker is a certificate check

For each \(n\), PARI/GP emits a complete proposed factorization

The Python program then independently checks:

1. \(q_10\);

2. \(\prod_iq_i^{e_i}=n^4+2\) exactly;

3. every \(q_i\) is prime by strong tests to the first twelve prime bases;

4. squarefreeness is equivalent to all \(e_i=1\).

Sorenson and Webster,

Strong Pseudoprimes to Twelve Prime Bases,

Math. Comp. 86 (2017), 985--1003,

DOI 10.1090/mcom/3134, proved that the

least composite passing these twelve tests is

Here the largest value is

strictly below that bound. Thus the primality layer is [b], and the

resulting exhaustive finite count is [d] with no probabilistic

primality assumption.

The full standalone code is in runs/erdos978_wave7y_reverify.py. The

central local routine is:

def local_root_count(p: int) -> int:
    if p == 2:
        return 0
    g = gcd(4, p - 1)
    return g if pow(p - 2, (p - 1) // g, p) == 1 else 0

Reproduction:

python runs/erdos978_wave7y_reverify.py

Observed output:

squarefree_counts={10: 8, 100: 75, 1000: 759, 10000: 7564, 100000: 75670}
root_type_counts_through_10000000={0: 415524, 2: 166161, 4: 82894}; relevant=249055
local_density_interval=(0.75668370260818737477048117621050903107604397865209, 0.75668400528178948748627617072097731946697177451523); prime_cutoff=10000000
preprint_fractional_part_counterexample=argument:-29/121, actual_fractional_part:92/121
VERIFIED: complete factorizations through n=100000, local product enclosure, page counterexample, and claimed-proof gap

Runtime on this VM was 35.53 seconds, about one core, with peak RSS 44,720

KB. A separate direct PARI rerun of \(S(100000)\) took 15.85 seconds and

again returned 75,670.

Exact reduction: what remains

For \(Y\geq2\), put

For fixed \(Y\), the Chinese remainder theorem and periodicity give

The two counts differ by at most \(T(X,Y)\). Consequently the single

uniform estimate

would prove the full asymptotic \(S(X)\sim cX\), and hence answer the

quartic question. This reduction is [a].

The part of (*) with \(Y \[ \begin{aligned} \#\{n\leq X:\exists\,p\in(Y,\sqrt X],\ p^2\mid n^4+2\} &\leq \sum_{YAfter division by \(X\), this is \(O(1/Y)+o(1)\).

Thus the exact unresolved range is

Necessarily \(p\ll X^2\), and every such event gives an integer point