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Math-5551-A3 · revision r1 · on Math-5551

gpt pro 5.2 attempt on Math-5551

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: proved. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #283
  2. 1) FORMAL RESTATEMENT
  3. 2) QUICK LITERATURE/CONTEXT CHECK
  4. 3) ATTACK PLAN
  5. 4) WORK
  6. 4.1. The divisibility hypothesis is necessary
  7. 4.2. A provable weaker statement: infinitely many $m$ are representable (for any $p$ with positive leading term)
  8. 4.3. Why this does not yet solve the conjecture
  9. 5) VERIFICATION
  10. 6) FINAL
  11. 7) COMPLETION ESTIMATE

Erdős Problem #283

1) FORMAL RESTATEMENT

Let $p\in\mathbb{Z}[X]$ be a polynomial with positive leading coefficient, and assume there is no integer $d\ge 2$ such that $d\mid p(n)$ for every $n\ge 1$ (equivalently, $\gcd\{p(1),p(2),\dots\}=1$).

The conjecture asks whether there exists $M$ (depending on $p$) such that for every integer $m\ge M$ one can find distinct integers \[ 1\le n_1<n_2<\cdots<n_k \] with \[ 1=\frac1{n_1}+\cdots+\frac1{n_k} \qquad\text{and}\qquad m=p(n_1)+\cdots+p(n_k). \]

2) QUICK LITERATURE/CONTEXT CHECK

Known special cases and related results recorded in the prompt/context include:
  • For $p(x)=x$ (so $m=\sum n_i$), Graham $1963$ proved the conjecture: all sufficiently large $m$ work.
  • Cassels $1960$ showed that the stated hypothesis on $p$ implies every sufficiently large integer is a sum of distinct values $p(n_i)$ (without the reciprocal-sum constraint).
  • Alekseyev $2019$ proved the conjecture for $p(x)=x^2$ for all $m>8542$.
  • Recent work of van Doorn $2025$ investigates quantitative bounds and proves the conjecture for many further linear/quadratic $p$.
The general case for arbitrary admissible $p$ is still listed as open.

3) ATTACK PLAN

  1. Closure operations on Egyptian fractions. Use identities like \[\frac1n=\frac1{n+1}+\frac1{n(n+1)}\] (to keep the reciprocal sum equal to $1$) to generate many different denominator sets, while tracking how $\sum p(n_i)$ changes.
  2. Additive semigroup approach. Try to show that from one base solution one can generate an additive semigroup of attainable $m$ whose gcd is $1$, hence contains all sufficiently large integers.
  3. Use Cassels to hit the $p$-sum, then tune reciprocals. First choose distinct $n_i$ so that $\sum p(n_i)=m$ (possible for large $m$ by Cassels), then adjust the set via splitting/merging operations that preserve the $p$-sum modulo small numbers while forcing the reciprocal sum to become exactly $1$.
  4. Search for counterexample polynomials. If false, the obstruction likely comes from subtle congruences tying $p(n)$ to available Egyptian-fraction denominator patterns.

4) WORK

4.1. The divisibility hypothesis is necessary

If there exists $d\ge 2$ dividing $p(n)$ for all $n\ge 1$, then any sum $\sum_i p(n_i)$ is divisible by $d$, so it cannot equal all sufficiently large integers. Thus the "no such $d$" condition is necessary.

4.2. A provable weaker statement: infinitely many $m$ are representable (for any $p$ with positive leading term)

Proposition: Let $p\in\mathbb{Z}[X]$ have positive leading coefficient. Then there exist infinitely many integers $m$ for which there are distinct $n_1<\cdots<n_k$ with \[1=\sum_{i=1}^k \frac1{n_i}\qquad\text{and}\qquad m=\sum_{i=1}^k p(n_i).\] Moreover, one can ensure all $p(n_i)>0$ and hence $m>0$.
Proof: Fix the identity \[\frac1n=\frac1{n+1}+\frac1{n(n+1)}\qquad(n\ge 2),\] which preserves the reciprocal sum.

Since $p$ has positive leading coefficient, there exists $N$ such that $p(n)>0$ for all $n\ge N$. Starting from any Egyptian fraction for $1$ (e.g. $1=\frac12+\frac13+\frac16$) and repeatedly applying the above splitting operation to the largest denominator, we can obtain an Egyptian fraction representation \[1=\sum_{i=1}^k \frac1{n_i}\] with all denominators $n_i\ge N$ (and still distinct, because splitting the current largest $n$ replaces it by $n+1$ and $n(n+1)$, both larger than all previous denominators). For this representation, set $m_0=\sum_{i=1}^k p(n_i)$, which is a positive integer.

Now iterate: at each stage, split the current largest denominator $n$ into $n+1$ and $n(n+1)$. This keeps $\sum 1/n_i=1$, preserves distinctness, and changes the $p$-sum by \[\Delta(n)=p(n+1)+p(n(n+1))-p(n).\] For $n\ge N$, all terms are positive, so $\Delta(n)>0$ and the resulting $m$ strictly increases. Hence we obtain infinitely many distinct $m$. ∎

4.3. Why this does not yet solve the conjecture

The above produces an infinite set of attainable $m$ but does not show it is cofinite. To reach "all sufficiently large $m$", one would need a mechanism to adjust $m$ by controlled small increments while keeping $\sum 1/n_i=1$ and denominators distinct.

5) VERIFICATION

  • The splitting identity is exact.
  • Distinctness under "split the current maximum denominator" is immediate because both new denominators exceed the previous maximum.
  • Positivity of $p(n)$ for large $n$ follows from positive leading coefficient.

6) FINAL

UNRESOLVED.
  1. Furthest point reached: proved a weaker theorem (infinitely many representable $m$) by a clean closure operation on Egyptian fractions; verified necessity of the "no common divisor" condition.
  2. Key gap: no argument was found that the attainable set of $m$ is eventually all integers (cofinite), for a general admissible polynomial $p$.
  3. Most promising next step: develop a finite set of "local moves" on Egyptian fraction representations that change $\sum p(n_i)$ by $\pm 1$ (or generate an additive semigroup of increments with gcd $1$), while preserving $\sum 1/n_i=1$.
  4. Explicit missing lemma that would close it: existence of a bounded "adjustment gadget"---a finite identity \[\sum_{j=1}^r \frac1{a_j}=\sum_{j=1}^s \frac1{b_j}=\text{(same rational)}\] with all $a_j,b_j$ distinct, such that $\sum p(a_j)-\sum p(b_j)=1$.

7) COMPLETION ESTIMATE

COMPLETION: 25%