gpt pro 5.4 attempt on Math-5551

Round 2 continuation for Erdős Problems #282, #283, #287

Erdős Problem #283

1) ROUND-2 OBJECTIVE

I pursue (A) proof for the constant admissible case $p(n)\equiv 1$. Round 1 produced only a general infinite-family theorem for arbitrary positive-leading-coefficient $p$, but not a cofinite result. The constant case is the cleanest place where the remaining cofinite gap can actually be closed completely by the Round-1 splitting move.

2) Round-1 FOUNDATION USED

I use two Round-1 ingredients.

1. The splitting identity \[ \frac1n=\frac1{n+1}+\frac1{n(n+1)}. \]
2. The fact that if one applies this identity to the largest denominator in a distinct Egyptian fraction expansion, then distinctness is preserved.

3) NEW INSIGHT / TOOL (ROUND-2)

For $p(n)\equiv 1$, the target quantity is simply \[ m=\sum_i p(n_i)=k, \] namely the number of terms. Thus the general polynomial problem collapses to a pure length-realisation problem for Egyptian fraction decompositions of $1$. The Round-1 splitting move increments the number of terms by exactly $1$ while preserving distinctness and the reciprocal sum, so it becomes an exact "length-raising gadget".

4) ATTACK PLAN (ROUND-2)

I prove the exact set of attainable lengths.

1. Show that $1$ term is possible: $\{1\}$.
2. Show that $2$ terms are impossible.
3. Show inductively that every length $k\ge 3$ is possible by repeatedly splitting the current largest denominator.

Since the only admissible constant polynomial is $p\equiv 1$, this completely settles the constant-polynomial case.

5) WORK (ROUND-2)

5.1. The only admissible constant polynomial.

Let $p(n)\equiv c$ be constant. The problem assumes that there is no integer $d\ge 2$ dividing all values $p(n)$. For a constant polynomial this means exactly that $c=1$. Thus the only admissible constant case is \[ p(n)\equiv 1. \]

5.2. Exact classification of attainable lengths.

Let $\mathcal{L}$ be the set of all integers $k\ge 1$ for which there exist distinct positive integers \[ 1\le n_1<\cdots<n_k, \qquad 1=\frac1{n_1}+\cdots+\frac1{n_k}. \] I claim that \[ \mathcal{L}=\{1\}\cup\{3,4,5,\dots\}. \]

First, $1\in\mathcal{L}$ because \[ 1=\frac11. \] Second, $3\in\mathcal{L}$ because \[ 1=\frac12+\frac13+\frac16. \] Now suppose $k\in\mathcal{L}$ with $k\ge 3$. Choose a corresponding decomposition and let $N$ be its largest denominator. Replace the term $1/N$ by \[ \frac1{N+1}+\frac1{N(N+1)}. \] The reciprocal sum is unchanged. Both new denominators are strictly larger than $N$, hence strictly larger than every previously occurring denominator, and they are distinct from each other. So we obtain a distinct Egyptian fraction expansion of $1$ with exactly $k+1$ terms. Therefore $k+1\in\mathcal{L}$.

By induction, every integer $k\ge 3$ lies in $\mathcal{L}$.

It remains to rule out $k=2$. If \[ 1=\frac1a+\frac1b \qquad (1\le a<b), \] then \[ ab=a+b, \] so \[ (a-1)(b-1)=1. \] This has no solution in distinct positive integers. Hence $2\notin\mathcal{L}$.

Thus the classification is proved.

5.3. Consequence for Problem #283 with $p\equiv 1$.

For $p(n)\equiv 1$, one has \[ m=\sum_i p(n_i)=k. \] Hence the attainable $m$ are exactly the attainable lengths. Therefore \[ m=1 \text{ is attainable},\qquad m=2 \text{ is not attainable},\qquad m\ge 3 \text{ are all attainable}. \] In particular the original "for all sufficiently large $m$" statement holds, with exact threshold $M=3$.

6) ADVERSARIAL VERIFICATION

1. Does the induction accidentally reuse denominators? No. The split is applied to the current largest denominator $N$, and both $N+1$ and $N(N+1)$ are $>N$, so they cannot coincide with any previous denominator.
2. Does the split work when $N=1$? I never use it starting from the one-term decomposition $\{1\}$. The induction starts from the three-term decomposition $\{2,3,6\}$.
3. Could there be a hidden two-term solution involving $1$? No. If $a=1$, then $1=1+1/b$ is impossible. The algebraic reduction to $(a-1)(b-1)=1$ also rules this out.
4. Does this solve the full polynomial problem? No. It completely settles only the unique admissible constant case $p\equiv 1$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The general polynomial conjecture remains open, but Round 2 fully closes one genuine subclass that Round 1 did not settle: the admissible constant-polynomial case. For \[ p(n)\equiv 1, \] one has an exact answer: \[ m \text{ is attainable } \Longleftrightarrow m=1 \text{ or } m\ge 3. \] So the constant case is completely solved, with exact threshold $M=3$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 40%

9) REFERENCES

Checked or used in this round:

1. T. F. Bloom, Erdős Problem #283, webpage accessed 2026-03-07.