Question and Summary

I recently made the following heuristic observations:

Let, $$ xy = p_1^{a_1} p_2^{a_2} \dots p_n^{a_n} $$ where $a_i\geq1$

Conjecture $1$: then there must exist $x-y=p_{n+1}$ where $x$ and $y$ co-prime to each other.

Conjecture $2$: for all solutions of different $(a_1,\dots,a_n)$ of $x-y= p_{n+1}$ then $x+y$ must also be a prime, $x+y = p_k$.

Can anyone provide any counter examples or reference anything relevant?

$1$st Conjecture

Let the product of $x$ and $y$ be:

$$ xy = p_1^{a_1} p_2^{a_2} \dots p_n^{a_n} $$

Where $p_i$ is the $i$'th the prime and $a_i$ is the power of the prime such that $a_i \geq1$. We choose $x$ and $y$ co-prime to each other I conjecture there must exist $x$ and $y$ such that:

$$ x-y = p_{n+1}$$

For example given the first prime is $2$:

Where then we get the next prime: $$ 2^2 - 1 = 3 \implies xy = 2^2$$ For the next prime: $$ 2 . 3 -1 = 5 \implies xy = 2.3$$ OR $$ 3^2 - 2^2 = 5 \implies xy = 3^2 2^2$$ OR $$ 2^5 - 3^3 = 5 \implies xy = 3^3 2^5 $$ And so on ... I believe any prime can be constructed by this.

$2$nd Conjecture

However, I also observed if $x-y=p_{n+1}$ then $x+y= p_k$ usually I would think this to only be true if $p_k < p_{n+1}^2$ but I think this might be true even beyond that (?).

For example:

$$ 2^5 + 3^3 = 59 > 25 $$


If both conjectures are true one can easily say Goldbach's conjecture and twin prime conjecture must be true!

  • $\begingroup$ Reason for down vote? Was there a simple counter-example I missed? $\endgroup$ – More Anonymous Jan 20 at 19:10
  • $\begingroup$ Why should $x-y$ be coprime to $x$ and $y$? I see no assertion that $x$ and $y$ are coprime. $\endgroup$ – Matt Samuel Jan 20 at 19:11
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    $\begingroup$ Ah sorry what I meant to say was we choose $x$ and $y$ coprime to each other... and thus $x-y$ is co-prime both to $x$ and $y$ ... I'll edit that $\endgroup$ – More Anonymous Jan 20 at 19:13
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    $\begingroup$ The first counterexample I can see for conjecture 2 is when $p = 7$, $x = 2\cdot 5^3=250$, and $y=3^5=243$. $xy=493=17\cdot 29$. I would bet conjecture 1 is true if I could, but the proof will not be too easy if there is one. $\endgroup$ – Hw Chu Jan 20 at 20:04
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    $\begingroup$ I think your Conjecture 1 is equivalent to an idea I had long ago. I asked one of our colloquium speakers about it, and he was unfamiliar with it, but later IIRC I posted it to sci.math (or similar newsgroup) and got a solution that was surprisingly strong. Again if I remember correctly there are primes $p$ that cannot be represented as the difference of two products of powers of distinct primes smaller than $p$. I'll see if Google can help me find that newsgroup thread. $\endgroup$ – hardmath Jan 20 at 20:20

I strongly suspect that you have tried only very small $n$. The minimal product $xy$ grows very rapidly. For example, for $n=10$ the minimum product $xy$ is $6469693230$, the product of the primes up to $29$. That would mean $x$ and $y$ are about $80434$. The first case where there are no $x,y$ with all exponents $1$ is $n=7$, where we would want $y(y+19)=510510$. It does not have a solution in the integers. There might be a solution with larger exponents, but I am skeptical. I was surprised to find $n=6$ works with $x=182, y=165, xy=30030$

Your second conjecture will be true when the exponents are small enough. You know that $x+y$ has no prime factors $p_{n+1}$ or smaller, so you are guaranteed that it is prime when $x+y \lt p_{n+1}^2$. However, as you get more primes, $x$ and $y$ will increase because their product is at least the primorial. Say we let $n=10$, so the primes are $2,3,5,7,11,13,17,19,23,29$. We have $xy$ is at least $6469693230$, so if we can use all first powers we have $x,y$ are about $80434$ and $x+y$ will be about $160868$, so there are lots of primes that could be factors.

I don't see how these two would solve Goldbach's conjecture or the twin primes problem, but if they would I would be even more surprised if they were true.


This post suggests an approach to the 1st Conjecture, too long for a comment. It is not intended to be a full answer.

OP requires $\alpha_i \ge 1$. Let $s:=\prod{q_j};\ t:= \prod {r_k};\ (q_j,r_k)\in \mathbb P; q_j \ne r_k;\ st=p_n\#$

Plainly $\gcd(s,t)=1$, so there exist $a,b$ such that $(as-bt)=|1|$

Accordingly, there will be $c,d$ such that $(cs-dt)=|p_{n+1}|$. We can let $x,y$ correspond to appropriate terms of $(cs-dt)$ or $(dt-cs)$ so as to conform to the nomenclature of the 1st Conjecture and result in a positive value for $p_{n+1}$.

The question devolves to whether $c$ can be found which contains only prime factors among $q_i$ and simultaneously $d$ can be found which contains only prime factors among $r_j$

If I make progress on this, I will update this post; I invite others to present solutions if they are aware of any.


I am providing this as an answer rather than as a comment as it's too long for that. Note there's a fairly similar question from Sep. $10, 2015$ on MathOverflow of Constructing prime numbers. The differences are that this other question supports $x \pm y = p_{n+1}$, plus for $xy = p_1^{a_1}p_2^{a_2}\ldots p_n^{a_n}$, it requires the $a_i = 1$ rather than allowing $a_i \ge 1$.

However, one of the answers, the $2$nd one down as I'm writing this, by Asterios Gkantzounis, says

maybe you could take the primes on the products in any powers you want , it is a thought that i had before some years and it is also a natural question that one can make reading Euclid's proof since still the difference of the products is not divisible by any of these primes do you take in this way always the next prime?

I believe this is basically what you are asking. A comment by François Brunault says

If you allow arbitrary exponents (and if I understand well your suggestion) then it is indeed always possible to get the next prime, just because of Fermat's little theorem. More complicated explanation : $\left(\text{Z}/p\text{Z}\right)^{*}$ is generated by the classes of the primes $< p$.

As my knowledge of number theory is quite limited, I can't really comment on how potentially valid his comment is. However, his profile indicates he's an experienced number theorist, so I don't have any reason to doubt his statement, although hardmath's comment to the question indicates it might not be true (of course, if I'm understanding everything correctly).

Note I am leaving a comment to Asterios that you're also asking a very similar question here.


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