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Motivated by questions such as this one, I have developed a proof of the following theorem, which I need to be subject to peer review:

Let $n,k$ be positive integers, and consider the interval $$ I = [kn, (k+1)n] = \{kn, kn+1, kn+2, \dots, (k+1)n\}. $$ Assume $(k+1)n < p_{\pi(n)+1}^2$, where $p_{\pi(n)+1}$ is the first prime greater than $n$. Then, the number of composite numbers in $I$ cannot exceed $n$. Therefore, at least one number in $I$ must be prime.

Proof

For each prime $p \leq n$, let: $$ A_p = \{x \in I \mid p \text{ divides } x\} $$ $$ B_p = \{x \in [0,n] \mid p \text{ divides } x\} $$ Lemma: $|A_p| \leq |B_p|$

Lemma proof

  • Any $x \in A_p$ has the form $x = kn + r$ with $kn \leq x \leq (k+1)n$ and $p \mid x$.

  • Define a function: $$ f \colon A_p \longrightarrow B_p, \quad f(kn + r) = r. $$

The function

(i) is well-defined: if $kn + r \in A_p$, then $r$ satisfies $r \equiv -kn \pmod{p}$ and $0 \leq r \leq n$, so $r \in B_p$.

(ii) It is injective: suppose $f(kn + r_1) = f(kn + r_2)$, meaning $r_1 = r_2$. Then, since $kn + r_1 = kn + r_2$, we have $x_1 = x_2$, so the function is one-to-one.

Hence $|A_p| \leq |B_p|$, as desired. $\square$

Using the Lemma, together with the fact that $$B_{p_1} \cap B_{p_2} \leftrightarrow A_{p_1} \cap A_{p_2}$$ we have that $$ \left|\bigcup_{p\leq n} A_p\right| \leq \left|\bigcup_{p \leq n} B_p\right| $$

Key fact: Notice that 1 is not divisible by any prime. The set $\bigcup_{p \leq n} B_p$ is exactly $[0,n]$ with $1$ removed, because $1$ has no prime divisors. Thus, $$ \bigcup_{p \leq n} \{y \in [0,n] : p \mid y \} = [0,n] \setminus \{1\}. $$

Thus, $\left|\bigcup_{p \leq n} B_p\right| = [0,n] \setminus \{1\}$, which has only $n$ elements. Therefore, by the Lemma and its corollary, we deduce that there can be at most $n$ composite numbers in the interval $I$. Since $I$ has $n+1$ distinct numbers, there must be at least one integer in $I$ which is not composite; such an integer must then be prime.

Hence we conclude that there is at least one prime in the interval $[kn,(k+1)n]$, completing the proof of the theorem.

$\square$

I have sent it to a top tier journal -as (if correct) the theorem directly implies several classical conjectures in number theory, including Oppermann's, Legendre's, Brocard's, and Andrica's conjectures, as well as new bounds on prime gaps- and I was told literally:

Because so many authors have submitted false solutions to the problem addressed in your manuscript, we can only consider such solutions if the exposition is exceptionally clear. If you are convinced that your solution is correct, and wish to continue to pursue publication, then you should have someone else (for instance a mathematically literate friend or colleague, or perhaps a mathematician at a local university) read your manuscript and give you suggestions for improving the readability. You should submit your manuscript again to a journal only if that person is able to understand your manuscript well enough to certify its correctness.

Therefore, I would appreciate any kind (or unkind) peer-review, suggestion, etc., that you could provide, as unfortunately I do not have any close colleague that can review it and certify / discard the validity of the theorem and its proof.

Thanks in advance for your help!

Juan Moreno
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    This is not a great place for peer review - it is designed for specific questions. – Thomas Andrews Feb 21 '25 at 23:38
  • Looks like the corollary is false, or at least not transparently true, because the sets are not disjoint. The size of $\bigcup A_p$ is an inclusion-exclusion value, $$\sum_{d\mid p_1\cdots p_n}\mu(d)|A_d|,$$ where we extend $A_d$ to non-primes, and you get $|A_d|\leq |B_d|,$ for all $d,$ and half of the values are $\mu(d)$ are $-1.$ So you are going to need a more detailed argument for your corollary. – Thomas Andrews Feb 21 '25 at 23:53
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    If this proof worked, you'd have a remarkably simple proof of Bertrand's postulate, which makes me think it will be hard to prove your corollary. – Thomas Andrews Feb 22 '25 at 00:02
  • In the inclusion-exclusion from my previous comment, it should have been $d\mid p_1\cdots p_{\pi(n)}.$ – Thomas Andrews Feb 22 '25 at 00:04
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    Proving four outstanding conjectures using high school arithmetic should set off multiple alarms. The "proof" is wrong, and it's your job to find the mistake(s), not ours. – Gerry Myerson Feb 22 '25 at 00:26

1 Answers1

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It is not true that if $S_1,S_2,T_1,T_2$ are sets with $|S_i|\leq |T_i|$ for $i=1,2$ that $|S_1\cup S_2|\leq |T_1\cup T_2|.$

For example, $S_1=\{1,2\}, S_2=\{3,4\}$ and $T_1=\{1,2\},T_2=\{1,2,3\}.$

The problem is that the size of the union is not the sum of the sizes. It would be true always if $T_1,T_2$ were disjoint.

To count a union, in general, you need an inclusion-exclusion argument. In your case this can be written explicitly if we expand $A_d$ and $B_d$ to non-primes $d.$

Then you get:

$$|\bigcup A_p|=\sum_{1<d\mid p_1\cdots p_{\pi(n)}}(-1)^{\omega(d)-1}|A_d|$$ where $\omega(d)$ is the number of distinct prime factors of $d.$ A similar formula for the union of $B_p.$

You get, as with primes, you can easily show $|A_d|\leq |B_d|,$ which is a problem because this means the cases when $\omega(d)$ is even, we might be subtracting more in this sum.

It is not hard to show the $|A_d|\leq |B_d|\leq |A_d|+1.$ You have to ensure that there are at least as many cases where $|B_d|-|A_d|$ is $+1$ when $\omega(d)$ is odd as when $\omega(d)$ is even. That is a tough sell.

Thomas Andrews
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  • thanks for your useful answer. I have added a short (but fundamental) inclusion-exclusion argument. Do you think it works? – Juan Moreno Feb 22 '25 at 01:38
  • Your "along with the fact that" is not a complete statement. $X\leftrightarrow Y$ doesn't mean anything when $X,Y$ are sets. – Thomas Andrews Feb 22 '25 at 01:52
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    It is highly unlikely this approach will work, because when $k=1,$ it would give a very easy proof of Bertrand's postulate, and the current simplest proof, while technically elementary, takes a long time to read through. There isn't even a simple elementary proof that there are at least $n$ primes less than $2^n,$ which follows from Bertrand, but is much weaker. – Thomas Andrews Feb 22 '25 at 01:59