gpt pro 5.4 attempt on Math-5499

Erdős Problem #208

1) ROUND-2 OBJECTIVE

Path pursued: (A) full proof of the Round 1 target statement.

Round 1 treated the statement \[ s_{n+1}-s_n=o(s_n^{1/5}), \] where $s_1<s_2<\cdots$ is the increasing sequence of squarefree integers, as unresolved and only obtained the much weaker bound $s_{n+1}-s_n\ll \sqrt{s_n}$.

The decisive new fact is that the Round 1 target statement is now a theorem: Pandey proved a fixed exponent improvement beyond $1/5$. Therefore the correct Round 2 objective is to convert that theorem into a short, gap-free proof of the original statement.

2) Round-1 FOUNDATION USED

I use only the following Round 1 items.

• The definition of the sequence $s_1<s_2<\cdots$ of squarefree integers.
• The precise target statement from Round 1: \[ s_{n+1}-s_n=o(s_n^{1/5}). \]

I do not reuse the Round 1 proof of the weaker bound $s_{n+1}-s_n\ll \sqrt{s_n}$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is Pandey's 2024 theorem:

There exists $\eta>0$ such that, for all sufficiently large $X$, the interval \[ [X,\ X+X^{1/5-\eta}] \] contains a squarefree integer.

This is strictly stronger than the original Round 1 target. Indeed, it immediately implies \[ s_{n+1}-s_n\ll s_n^{1/5-\eta}, \] and therefore \[ s_{n+1}-s_n=o(s_n^{1/5}). \]

4) ATTACK PLAN (ROUND-2)

The Round 1 gap was simple: no theorem beyond the exponent $1/5$ was available inside the write-up.

To close that gap, I will:

1. state Pandey's theorem precisely;
2. substitute $X=s_n+1$ (not $X=s_n$);
3. deduce a direct upper bound for $s_{n+1}-s_n$;
4. divide by $s_n^{1/5}$ and let $n\to \infty$.

The choice $X=s_n+1$ is important: if one took $X=s_n$, the guaranteed squarefree integer in $[X,X+X^{1/5-\eta}]$ could be the left endpoint $s_n$ itself, which would not control the next gap. Shifting by $1$ avoids that issue completely.

5) WORK (ROUND-2)

External theorem used (Pandey).

There exists a constant $\eta>0$ such that for all sufficiently large real numbers $X$, the interval \[ [X,\ X+X^{1/5-\eta}] \] contains at least one squarefree integer.

Theorem 5.1.

For the increasing sequence $s_1<s_2<\cdots$ of squarefree integers, one has \[ s_{n+1}-s_n=o(s_n^{1/5}). \]

Proof.

Let $\eta>0$ be as in Pandey's theorem. Fix $n$ large enough that $X:=s_n+1$ lies in the range where Pandey's theorem applies. Then the interval \[ [s_n+1,\ s_n+1+(s_n+1)^{1/5-\eta}] \] contains a squarefree integer.

Let $t$ be such a squarefree integer. Since $t\ge s_n+1$, we have $t>s_n$. By definition, $s_{n+1}$ is the least squarefree integer strictly larger than $s_n$, so \[ s_{n+1}\le t\le s_n+1+(s_n+1)^{1/5-\eta}. \] Therefore \[ s_{n+1}-s_n\le 1+(s_n+1)^{1/5-\eta}. \] Now divide by $s_n^{1/5}$: \[ \frac{s_{n+1}-s_n}{s_n^{1/5}} \le \frac{1}{s_n^{1/5}} + \frac{(s_n+1)^{1/5-\eta}}{s_n^{1/5}}. \] As $n\to\infty$, we have $s_n\to\infty$, so \[ \frac{1}{s_n^{1/5}}\to 0, \] and also \[ \frac{(s_n+1)^{1/5-\eta}}{s_n^{1/5}} = \left(1+\frac{1}{s_n}\right)^{1/5-\eta}s_n^{-\eta}\to 0. \] Hence \[ \frac{s_{n+1}-s_n}{s_n^{1/5}}\to 0, \] which is exactly \[ s_{n+1}-s_n=o(s_n^{1/5}). \] $\square$

What this does and does not settle.

The Round 1 target statement is therefore solved affirmatively. However, the current stronger formulation recorded on the Erdős problem page asks whether \[ s_{n+1}-s_n\ll_\varepsilon s_n^\varepsilon \qquad\text{for every }\varepsilon>0, \] which is much stronger. Pandey's theorem gives a single fixed exponent $1/5-\eta$, not arbitrary $\varepsilon$. So the original Round 1 target is solved, while the stronger modern formulation remains open.

6) ADVERSARIAL VERIFICATION

(i) Why was $X=s_n+1$ chosen instead of $X=s_n$?

Because if $X=s_n$, Pandey's theorem only guarantees a squarefree number somewhere in \[ [s_n,\ s_n+s_n^{1/5-\eta}], \] and that number could be $s_n$ itself. Then no bound on $s_{n+1}-s_n$ would follow. The shift by $1$ removes this possibility.

(ii) Does Pandey's theorem apply to integer $X$ only?

The statement is formulated for sufficiently large $X$. Taking $X=s_n+1$ is legitimate since $s_n+1\to\infty$.

(iii) Is the deduction from the existence of $t$ to the bound on $s_{n+1}$ valid?

Yes. By definition, $s_{n+1}$ is the smallest squarefree integer greater than $s_n$, and $t$ is one such squarefree integer. Therefore $s_{n+1}\le t$.

(iv) Do the limits really give little-$o$?

Yes. Both terms on the right-hand side after division by $s_n^{1/5}$ tend to $0$, so the whole quotient tends to $0$.

(v) Any hidden issue with endpoint conventions?

No. Since $t\ge s_n+1$, we have $t>s_n$ regardless of whether the interval is interpreted with closed or half-open endpoints at the left.

The proof is therefore gap-free.

7) FINAL (EXACTLY ONE)

FULL SOLUTION --- FULL PROOF.

Pandey's theorem implies \[ s_{n+1}-s_n\le 1+(s_n+1)^{1/5-\eta} \] for some fixed $\eta>0$ and all sufficiently large $n$, and hence \[ s_{n+1}-s_n=o(s_n^{1/5}). \]

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 100%

9) REFERENCES

1. M. Pandey, Squarefree numbers in short intervals, arXiv:2401.13981, 2024.
2. T. F. Bloom, Erdős Problem #208, erdosproblems.com, accessed 2026-03-07.