23-летний программист разгадал задачу Эрёша за 80 минут с помощью ChatGPT 5.4 Pro

23-летний разработчик из США стал первым, кто решил одну из нерешённых задач легендарного математика Пола Эрёша, используя ChatGPT 5.4 Pro. Задача, которая оставалась открытой с 1960-х годов, была решена всего за 80 минут — время, которое модель потратила на анализ и генерацию решения. Уникальность подхода заключалась в том, что ChatGPT применил хорошо известную в математике формулу, но ранее никто не использовал её для решения именно этой задачи.

По словам разработчика, он не ожидал, что модель справится с задачей с первой попытки. «Я просто задал вопрос и получил ответ, который полностью соответствовал условиям задачи. Это было невероятно», — отметил он. Диалог с ChatGPT включал подробное объяснение решения, а также подтверждение корректности подхода.

Задача Эрёша, сформулированная в середине XX века, долгое время считалась одной из самых сложных в теории графов. Её решение могло бы иметь значительные последствия для смежных областей математики и компьютерных наук. Теперь, благодаря ChatGPT, задача официально закрыта, что вызвало бурную реакцию в научном сообществе.

Эксперты отмечают, что этот случай демонстрирует растущий потенциал ИИ в решении сложных математических задач. Однако они подчёркивают, что для полноценного понимания решения всё же требуется участие человека, который сможет интерпретировать и верифицировать результаты, полученные от модели.

Разработчик уже подготовил публикацию с подробным описанием решения и отправил её в рецензируемый математический журнал. Если работа будет принята, это станет первым случаем, когда ИИ сыграл ключевую роль в доказательстве математической гипотезы.

Вот диалог с ChatGPT:

don’t search the internet. This is a test to see how well you can craft non-trivial, novel and creative proofs given a «number theory and primitive sets» math problem. Provide a full unconditional proof or disproof of the problem. Problem: «Is it true that, for any $x$, if $A\subset [x,\infty)$ is a primitive set of integers (so that no distinct elements of $A$ divide each other) then\[\sum_{a\in A}\frac{1}{a\log a}< 1+o(1),\]where the $o(1)$ term $\to 0$ as $x\to \infty$?» information you may or may not need to help with the above problem «It is proved that\[\sum_{a\in A}\frac{1}{a\log a}< e^{\gamma}\frac{\pi}{4}+o(1)\approx 1.399+o(1).\]» «It is proved that if $A$ is the set of all integers with exactly $k$ prime factors (so that $A\subset [2^k,\infty)$ and $A$ is a primitive set) then\[\sum_{a\in A}\frac{1}{a\log a}\geq 1+O(k^{-1/2+o(1)}),\]» «It is proved that\[\sum_{a\in A}\frac{1}{a\log a}= 1-(c+o(1))k^22^{-k}\]where $c\approx 0.0656$ is an explicit constant.» REMEMBER — this unconditional argument may require non-trivial, creative and novel elements.

I’m treating this as a genuine extremal proof search, with two live paths: either compress primitive tails by prime-factor layers to force a universal $1 + o (1)$ bound, or construct a counterexample beating 1 by a fixed amount.

I found a promising probabilistic route: use a “first-entry into $[x,∞)[x,\infty)$ ” random divisibility chain built from prime powers, weighted by $Λ(q)\Lambda(q)$ . Then $∑q∣nΛ(q)=log⁡n\sum_{q\mid n}\Lambda(q)=\log n$ converts the target sum into hitting probabilities, with only an $O(1/log⁡x)O(1/\log x)$ boundary loss.

A competing construction now looks much less viable: because unions of disjoint prime-support blocks stay primitive, any counterexample would need to add a second large-prime block to a near-extremal $Ω(n)=k\Omega(n)=k$ layer, but the scale mismatch makes that extra contribution too small to overcome the known deficit.

Yes.

In fact one can prove something slightly stronger.

Fix any constant $0<δ<log⁡20<\delta<\log 2$ . Then for every primitive set $A⊂[x,∞)A\subset [x,\infty)$ ,

$⁣(1log⁡x)(x→∞).\sum_{a\in A}\frac1{a(\log a-\delta)}\le 1+O_\delta\!\left(\frac1{\log x}\right) \qquad (x\to\infty).$

Since

$1alog⁡a≤1a(log⁡a−δ),\frac1{a\log a}\le \frac1{a(\log a-\delta)},$

this immediately gives

$∑a∈A1alog⁡a≤1+o(1).\sum_{a\in A}\frac1{a\log a}\le 1+o(1).$

So the statement in the problem is true.

Proof

Write $Λ\Lambda$ for the von Mangoldt function. I will sum over all integers $q≥1q\ge1$ ; only prime powers contribute because $Λ(q)=0\Lambda(q)=0$ otherwise.

The key identity is the exact divisor sum

$∑q∣nΛ(q)=log⁡n.\sum_{q\mid n}\Lambda(q)=\log n.$

The proof is a weighted “first-entry” decomposition on the divisibility poset, but with one important twist: I only allow large prime-power jumps. The finitely many small prime powers are absorbed into the initial mass, and that costs only $O(1/log⁡x)O(1/\log x)$ .

1. A tail estimate for prime powers

Let

$M(t):=∑q≤tΛ(q)q.M(t):=\sum_{q\le t}\frac{\Lambda(q)}{q}.$

A standard unconditional estimate (Mertens’ first theorem) says

$M(t)=log⁡t+O(1).M(t)=\log t+O(1).$

For $m≥1m\ge1$ , $y≥2y\ge2$ , set

$(log⁡(mq)−δ).T_\delta(m,y):=\sum_{q\ge y}\frac{\Lambda(q)}{q\,\log(mq)\,(\log(mq)-\delta)}.$

I claim that, uniformly as $my→∞my\to\infty$ ,

$⁣(1log⁡2(my)).(1)T_\delta(m,y) = \frac1\delta\log\frac{\log(my)}{\log(my)-\delta} + O_\delta\!\left(\frac1{\log^2(my)}\right). \tag{1}$

To prove this, write $L=log⁡mL=\log m$ , $U=log⁡yU=\log y$ , and

$F(u):=1(L+u)(L+u−δ).F(u):=\frac1{(L+u)(L+u-\delta)}.$

Then

$dM(eu).T_\delta(m,y)=\int_{U}^{\infty} F(u)\,dM(e^u).$

Since $M(e^u)=u+E(u)$ with $E (u) = O (1)$ , we get

$dE(u).T_\delta(m,y)=\int_U^\infty F(u)\,du+\int_U^\infty F(u)\,dE(u).$

On any finite interval, $E$ has bounded variation, so integration by parts gives

$du.\int_U^\infty F(u)\,dE(u) = -F(U)E(U)-\int_U^\infty E(u)F'(u)\,du.$

Now

$F(U)≪δ1(L+U)2,F′(u)≪δ1(L+u)3,F(U)\ll_\delta \frac1{(L+U)^2}, \qquad F'(u)\ll_\delta \frac1{(L+u)^3},$

and $E (u) = O (1)$ , so

$⁣(1(L+U)2).\int_U^\infty F(u)\,dE(u) = O_\delta\!\left(\frac1{(L+U)^2}\right).$

Also

$∫U∞du(L+u)(L+u−δ)=1δlog⁡L+UL+U−δ.\int_U^\infty \frac{du}{(L+u)(L+u-\delta)} = \frac1\delta\log\frac{L+U}{L+U-\delta}.$

This is exactly (1).

2. Choose a cutoff $Y$ and define the jump kernel

Fix $δ∈(0,log⁡2)\delta\in(0,\log2)$ .

Using (1), choose once and for all an integer $Y≥2Y\ge 2$ , large enough in terms of $δ\delta$ , so that for all sufficiently large $m$ ,

$Tδ(m,Y)≤1.(2)(\log m-\delta)\,T_\delta(m,Y)\le 1. \tag{2}$

Why is this possible? By (1),

$⁣(1log⁡2m)).(\log m-\delta)T_\delta(m,Y) = (\log m-\delta)\left( \frac1\delta\log\frac{\log(mY)}{\log(mY)-\delta} +O_\delta\!\left(\frac1{\log^2 m}\right) \right).$

Using $log⁡(1+t)≤t\log(1+t)\le t$ ,

$1δlog⁡log⁡(mY)log⁡(mY)−δ≤1log⁡(mY)−δ,\frac1\delta\log\frac{\log(mY)}{\log(mY)-\delta} \le \frac1{\log(mY)-\delta},$

$⁣(1log⁡m).(\log m-\delta)T_\delta(m,Y) \le \frac{\log m-\delta}{\log(mY)-\delta} + O_\delta\!\left(\frac1{\log m}\right) = 1-\frac{\log Y}{\log(mY)-\delta}+O_\delta\!\left(\frac1{\log m}\right).$

Since $log⁡Y\log Y$ is a fixed positive constant which we may choose as large as we like, the deficit from the main term dominates the error for large $m$ , giving (2).

Now fix $x$ so large that (2) holds for all $m≥xm\ge x$ .

Define

$W(n):=1n(log⁡n−δ).W(n):=\frac1{n(\log n-\delta)}.$

For $m≥xm\ge x$ and $q≥Yq\ge Y$ , define

$\frac{(\log m-\delta)\Lambda(q)} {q\,\log(mq)\,(\log(mq)-\delta)}.$

Because of (2), for every $m≥xm\ge x$ ,

$∑q≥YP(m,mq)≤1.\sum_{q\ge Y}P(m,mq)\le 1.$

So this is a genuine sub-Markov jump kernel on $[x,∞)[x,\infty)$ : from $m$ , the chain either jumps to $m q$ with the above probability, or stops.

Every jump multiplies by an integer $q≥Y>1q\ge Y>1$ , so every path is strictly increasing and each visited state divides every later one.

3. The entrance mass

Define the “source” mass on integers $n≥xn\ge x$ by

$bx(n):=1nlog⁡n(log⁡n−δ)(∑q∣nq<YΛ(q)+∑q∣nq≥Yn/q<xΛ(q)).(3)b_x(n):= \frac1{n\log n(\log n-\delta)} \left( \sum_{\substack{q\mid n\\ q<Y}}\Lambda(q) + \sum_{\substack{q\mid n\\ q\ge Y\\ n/q<x}}\Lambda(q) \right). \tag{3}$

This means:

all small prime-power divisors $q < Y$ are inserted directly at $n$ , rather than represented by transitions;
among the large prime powers $q≥Yq\ge Y$ , those whose parent $n / q$ lies below $x$ are also inserted directly at $n$ .

Let

$Bx:=∑n≥xbx(n).B_x:=\sum_{n\ge x} b_x(n).$

We will show

$⁣(1log⁡x).(4)B_x=1+O_\delta\!\left(\frac1{\log x}\right). \tag{4}$

First assume this and finish the main argument.

Start the sub-Markov chain with initial distribution

$P(N0=n)=bx(n)Bx.\mathbb P(N_0=n)=\frac{b_x(n)}{B_x}.$

Let $v (n)$ be the probability that the chain ever visits $n$ .

Because the chain is increasing, the only ways to hit $n$ are:

start at $n$ , or
come from $n / q$ via one last jump with $q≥Yq\ge Y$ .

$P(n/q,n).(5)v(n)=\frac{b_x(n)}{B_x} + \sum_{\substack{q\mid n\\ q\ge Y\\ n/q\ge x}} v(n/q)\,P(n/q,n). \tag{5}$

Now I claim that

$v(n)=W(n)Bx(n≥x).(6)v(n)=\frac{W(n)}{B_x}\qquad(n\ge x). \tag{6}$

This is proved by induction on $n$ . If it holds for all smaller integers, then from (5),

$P(n/q,n)).v(n)=\frac1{B_x}\left( b_x(n)+ \sum_{\substack{q\mid n\\ q\ge Y\\ n/q\ge x}} W(n/q)\,P(n/q,n) \right).$

But

$P(n/q,n)=1(n/q)(log⁡(n/q)−δ)⋅(log⁡(n/q)−δ)Λ(q)qlog⁡n(log⁡n−δ)=Λ(q)nlog⁡n(log⁡n−δ).W(n/q)\,P(n/q,n) = \frac{1}{(n/q)(\log(n/q)-\delta)} \cdot \frac{(\log(n/q)-\delta)\Lambda(q)} {q\log n(\log n-\delta)} = \frac{\Lambda(q)}{n\log n(\log n-\delta)}.$

Hence

$P(n/q,n)=1nlog⁡n(log⁡n−δ)∑q∣nΛ(q)=1n(log⁡n−δ)=W(n),b_x(n)+ \sum_{\substack{q\mid n\\ q\ge Y\\ n/q\ge x}} W(n/q)\,P(n/q,n) = \frac1{n\log n(\log n-\delta)} \sum_{q\mid n}\Lambda(q) = \frac1{n(\log n-\delta)} = W(n),$

because the two sums in $b_x(n)$ together with the transition sum partition all divisors $q∣nq\mid n$ . Thus (6) holds.

Now let $A⊂[x,∞)A\subset[x,\infty)$ be primitive. A sample path cannot hit two distinct elements of $A$ , because along the path each earlier state divides each later state. Therefore

$∑a∈Av(a)≤1.\sum_{a\in A} v(a)\le 1.$

Using (6),

$∑a∈A1a(log⁡a−δ)=∑a∈AW(a)=Bx∑a∈Av(a)≤Bx.\sum_{a\in A}\frac1{a(\log a-\delta)} = \sum_{a\in A}W(a) = B_x\sum_{a\in A}v(a) \le B_x.$

So by (4),

$⁣(1log⁡x).\sum_{a\in A}\frac1{a(\log a-\delta)} \le 1+O_\delta\!\left(\frac1{\log x}\right).$

Finally,

$⁣(1log⁡x)=1+o(1).\sum_{a\in A}\frac1{a\log a} \le \sum_{a\in A}\frac1{a(\log a-\delta)} \le 1+O_\delta\!\left(\frac1{\log x}\right) = 1+o(1).$

This proves the theorem, assuming (4).

4. Estimating the total entrance mass $B_x$

Split $B_x$ according to the two terms in (3):

$Bx=Bxsmall+Bxent.B_x=B_x^{\mathrm{small}}+B_x^{\mathrm{ent}}.$

The small- $q$ part

We have

$Bxsmall=∑q<YΛ(q)q∑m≥x/q1mlog⁡(mq)(log⁡(mq)−δ).B_x^{\mathrm{small}} = \sum_{q<Y}\frac{\Lambda(q)}{q} \sum_{m\ge x/q} \frac1{m\log(mq)(\log(mq)-\delta)}.$

For each fixed $q < Y$ , the inner summand is decreasing in $m$ , so by comparison with the integral,

$∑m≥x/q1mlog⁡(mq)(log⁡(mq)−δ)≪δ∫x/q∞dttlog⁡(tq)(log⁡(tq)−δ).\sum_{m\ge x/q} \frac1{m\log(mq)(\log(mq)-\delta)} \ll_\delta \int_{x/q}^{\infty} \frac{dt}{t\log(tq)(\log(tq)-\delta)}.$

Substitute $u=log⁡(tq)u=\log(tq)$ , $d t / t = d u$ . Since $t = x / q$ gives $u=log⁡xu=\log x$ ,

$⁣(1log⁡x).\int_{x/q}^{\infty} \frac{dt}{t\log(tq)(\log(tq)-\delta)} = \frac1\delta\log\frac{\log x}{\log x-\delta} = O_\delta\!\left(\frac1{\log x}\right).$

Because $Y$ is fixed,

$⁣(1log⁡x).(7)B_x^{\mathrm{small}} = O_\delta\!\left(\frac1{\log x}\right). \tag{7}$

The first-entry part

Now

$Bxent=∑m<x1m∑q≥max⁡(Y,x/m)Λ(q)qlog⁡(mq)(log⁡(mq)−δ).B_x^{\mathrm{ent}} = \sum_{m<x}\frac1m \sum_{q\ge \max(Y,x/m)} \frac{\Lambda(q)}{q\log(mq)(\log(mq)-\delta)}.$

Apply the tail estimate (1) with $y=max⁡(Y,x/m)y=\max(Y,x/m)$ . Then

$y=max⁡(mY,x),m\,y=\max(mY,x),$

so if we put

$M_m:=\log(m\,y)=\max(\log(mY),\log x),$

we have

$log⁡x+log⁡Y].M_m\in[\log x,\ \log x+\log Y].$

Therefore, uniformly in $m < x$ ,

$⁣(1log⁡2x).\sum_{q\ge \max(Y,x/m)} \frac{\Lambda(q)}{q\log(mq)(\log(mq)-\delta)} = \frac1\delta\log\frac{M_m}{M_m-\delta} + O_\delta\!\left(\frac1{M_m^2}\right) = \frac1{\log x}+O_\delta\!\left(\frac1{\log^2 x}\right).$

Hence

$⁣(1log⁡2x))∑m<x1m.B_x^{\mathrm{ent}} = \left(\frac1{\log x}+O_\delta\!\left(\frac1{\log^2 x}\right)\right) \sum_{m<x}\frac1m.$

Since $∑m<x1/m=log⁡x+O(1)\sum_{m<x}1/m=\log x+O(1)$ ,

$⁣(1log⁡x).(8)B_x^{\mathrm{ent}} = 1+O_\delta\!\left(\frac1{\log x}\right). \tag{8}$

From (7) and (8),

$⁣(1log⁡x),B_x=1+O_\delta\!\left(\frac1{\log x}\right),$

which is exactly (4).

The proof is complete. ∎

Conclusion

So the answer to your problem is yes:

$primitive∑a∈A1alog⁡a≤1+o(1).\sup_{\substack{A\subset [x,\infty)\\ A\ \text{primitive}}} \sum_{a\in A}\frac1{a\log a} \le 1+o(1).$

The argument actually gives the stronger weighted form

$⁣(1log⁡x)\sum_{a\in A}\frac1{a(\log a-\delta)}\le 1+O_\delta\!\left(\frac1{\log x}\right)$

for every fixed $0<δ<log⁡20<\delta<\log2$ .

This is also sharp at the constant level, since the usual $k$ -almost-prime examples already show that the extremal value approaches $1$ .

no internet: Please write a full correct resolution to the problem formatted as a publishable maths research paper using amsart using a4paper, margin=1in. Keep the title brief and to the point. The abstract should only be at most 6 sentences. Use section headings sparingly. Do not add an author entry. Be rigorous and self-contained. Ensure to address any issues you raised. Give the LaTeX in a code markdown block.

I’m checking the argument carefully before I typeset it, especially the normalization and the tail estimate, so the paper version is fully watertight.

I’m also stress-testing the cutoff and sub-Markov construction, including boundary terms, to ensure the first-entry decomposition gives an exact $1+O(1/log⁡x)1+O(1/\log x)$ majorant and the weight recursion closes correctly.

I’ve now shaped the core argument into a paper draft, and I’m polishing the prime-power Mertens lemma so that the analytic input is stated cleanly and proved in a publishable form.

Автор

Сергей Сергеев

Сергей Сергеев — практикующий трейдер и автор FinTerminal. С 2013 года работает с Forex, криптовалютами и акциями, анализируя рыночные риски, волатильность, исполнение сделок и поведение цены в новостные периоды. В материалах сочетает собственный торговый опыт, статистику рынков и проверку условий брокеров для частных трейдеров и инвесторов. При подготовке публикаций опирается на данные регуляторов, рыночную статистику, биржевые котировки и публичные условия брокеров, отдельно отмечая риски, ограничения и спорные моменты, которые важно проверить перед принятием решения.

23-летний программист разгадал задачу Эрёша за 80 минут с помощью ChatGPT 5.4 Pro