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I would like to propose the following conjecture and seek insights or references related to it:

Conjecture:

Every non-zero integer $n$ can be expressed as the difference of a semiprime $q$ and a prime $p$, that is, $$n=q−p,$$

where:

q is a semiprime (the product of two primes, not necessarily distinct), and
p is a prime number.

Numerical Verification:

To gather empirical support for this conjecture, I conducted extensive numerical experiments within the range $−10,000,000≤n≤10,000,000$ (excluding $n=0$). In every case within this range, I successfully found pairs of $p$ and $q$ satisfying the equation $n=q−p$. No counterexamples were observed.

Example: For instance, when $n=15$:

Choose p=7 (a prime),
Then q=7+15=22, which is a semiprime (22=2×11).

Thus, $15=22−7$.

Questions:

In seeking a proof or counterexample for this conjecture, what approaches or existing theories would be effective? Especially, I would appreciate advice from the perspectives of additive number theory or multiplicative number theory.

Additional Information:

Implementation Details:
The numerical experiments were conducted using a Python program optimized with the Sieve of Eratosthenes for prime generation and parallel processing to handle large ranges efficiently. While I can provide the code upon request, the focus here is on the mathematical aspect of the conjecture.

Exclusion of Zero: The case n=0 is excluded since it would require q=p, and while p is prime, q would need to be semiprime, which only occurs if p itself is a semiprime. Given that primes greater than 2 are not semiprimes, n=0 serves as an inherent exception to the conjecture.

Graphical Representation:

Upon observation, there is a tendency for $n$ to stick to the upper range when it is a multiple of $6$ (Especially multiples of $30$), and to the lower range when $n$ is a prime number. enter image description here enter image description here enter image description here Motivation

The motivation for this conjecture stems from the robustness of semiprimes used in RSA encryption. Specifically, semiprimes are known for their resistance to division by prime numbers, as they rarely divide evenly. This led me to consider a different approach: what if, instead of division, we use subtraction? Given that semiprimes do not divide cleanly, I hypothesized that their differences with primes might display some kind of pattern or periodicity, resulting in jumps to other numbers.

From this idea, I further speculated that these "jumps" could cover all integers in a somewhat uniform manner. However, recognizing that semiprimes and primes belong to different classes, it became clear that expressing zero was not feasible. Thus, zero was excluded from the conjecture's conditions.

If proven, this conjecture could offer new perspectives in number theory, with possible connections to cryptographic systems like RSA encryption.

Additionally, this conjecture may inherently include basic memory management functionality.

I appreciate any feedback, references, or discussions that can shed light on the validity and potential avenues for this conjecture. Thank you!

  • What sense does it make to write, "$20$, which is a semiprime ... hence not a semiprime"? – Gerry Myerson Sep 28 '24 at 12:30
  • For a numerical verification, I’d like to know how the number of solutions changes. Or how the smallest prime changes. – gnasher729 Sep 28 '24 at 12:31
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    Somewhat related: Chen Jingrun proved "every sufficiently large even number can be written as the sum of a prime and a semiprime." Also, https://math.stackexchange.com/questions/4427229/is-every-odd-number-of-the-form-2pq – Gerry Myerson Sep 28 '24 at 12:33
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    For the sum of a prime and a semi prime you only have a finite number of choices, for the difference there is an infinite number of choices. So whatever large n you start with, heuristically there would be a small chance that n’ >= n is not such a sum. – gnasher729 Sep 28 '24 at 12:43
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    @GerryMyerson Thank you for your reply. The translation wasn't working properly. I'll fix it. – Akira Sukigi Sep 28 '24 at 13:28
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    The only differences that are odd are odd semi prime - 2, and 2q - p for odd primes p, q. Differences that are even are all odd semiprimes qr minus odd prime p, and 2q -2. There are much fewer cases 2q - p. – gnasher729 Sep 28 '24 at 17:22
  • Thank you for thinking about dividing numbers into patterns. I think this problem can be divided into 4 patterns: even numbers, odd numbers, positive and negative numbers. – Akira Sukigi Sep 28 '24 at 22:30
  • I have started to consider whether this conjecture might also hold not only for the difference between a semiprime and a prime, but also for differences between other classes, such as a composite number made of three primes, four primes, and so on. – Akira Sukigi Oct 03 '24 at 07:26
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    In terms of $n$, try plotting the proportion of primes $p < N$ such that $n+p$ is semiprime (for a large $N$). There are interesting patterns. – healynr Oct 03 '24 at 17:59
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    @healynr Thank you for the suggestion! I gave it a try and plotted the data—it really does look like it could be from an audio file! It makes me wonder if we could actually listen to it as sound. I might experiment with that and see what we can discover. – Akira Sukigi Oct 05 '24 at 16:16
  • If you try plotting without the bars (just the dots), you will more clearly see the tiers (relating to prime factors). – healynr Oct 05 '24 at 20:15
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    Instead of plotting the proportion $r(n)$ itself, which falls as $N$ increases, I plotted $r(n)/r(1)$, the ratio of the proportion at $n$ to the proportion at $1$, for $N = 10000000$, $0<=n<=1000$. – healynr Oct 05 '24 at 20:47
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    This should follow pretty easily from Chen's techniques (I'm not sufficiently well-versed in the details to be sure). – pzq_alex Oct 06 '24 at 13:54
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    This conjecture can be compared to Goldbach's conjecture. For Goldbach's conjecture, proving it is true is very very complex. But, playing with Goldbach, trying to estimates how many couples $(p_1,p_2)$ primes are such that $p_1+p_2=n$, for a given $n$, it is an interesting game. And the answer involves prime-factors of $n$. Playing with your conjecture, each integer $n$ can be written as $n=q-p$ ; this conjecture is certainly true. We have even certainly an infinity of couples (q,p). Estimating how many couples $(q,p)$, with p<n is an interesting game... and prime factors of $n$ will impact – Lourrran Oct 06 '24 at 15:12
  • A simple solution for prime n: n = 2n - n – Akira Sukigi Oct 17 '24 at 10:16

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