# $n^{th}$ Dimensional Prime Nexus Conjecture (Hamilton's Path & Cycle)

The system and proposition of Prime Nexus -

We have to arrange $$n>1$$ Distinct natural numbers in a sequence such that the adjacent elements sums upto any Prime number. Convenience is when we set this problem to Graph Theoretic setting and specifically by using Hamilton's path and cycles. This system of clusters of natural numbers summing upto primes is called the Prime Nexus .

Leave the question that at which extent it's helpful , likewise if the numbers are more then it starts getting tough .

$$n^{th}$$ Dimensional Prime Nexus Conjecture

We can always generate any regular & definite n-th dimensional shape out of the nexus of Primes , specifically out of the Hamilton's path such that the nodes sums upto any Prime number.

Suppose in order to arrange Distinct natural numbers in a sequence such that their adjacent elements sums upto any prime , we can have $$1D$$ Shape , $$2D$$ , $$3D$$ and so on... generated out of the nexus , ( Preferring the extension of squares & cubes...)

I've worked out till $$3D$$ ( can also replace $$1$$ by $$9$$ In $$3D$$ ) as the number of numbers increases in the cluster , the probability of getting more higher dimensional nexus gets increased. I strongly believe that the system never saturates at any $$k^{th}$$ dimension. So If someone comes up with any counter example, actually he/she hasn't ! Because if we try generation of prime nexus for more higher number of numbers , we can always do that !

Please lemme know your opinions on it , how could anyone explain this enigmatic behavior of prime numbers making up higher dimensional shapes... And how could anyone even attack this conjecture! I'm very excited to learn new things regarding this conjecture, Everyone else here on MSE community holds more knowledge that I do , so please forgive me for any inconveniences followed up in presentation of my contents here.

Regards

• You wrote "Everyone else here on MSE community holds more knowledge that I do". From my experience so far, I've seen that the MSE community has a wide range of knowledge about basically every field in math. On almost any topic, there will be people here who know less, and others who know more, than you do. Aug 27 '19 at 18:19
• @JohnOmielan Sir thanks a lot for this beautiful comment , but I don't know anything else except the Lord's Divine Grace , Believe me , I'm so much curious to learn anything, and from everyone I can initiate the process of eternal learning, a cycle, therefore this is reason I consider everyone else here to be at higher potential of knowledge than I do... Whether learning from Junior or colleagues or seniors ^_^ Aug 27 '19 at 18:24
• Presumably all natural numbers are required to be unique? Otherwise there's a trivial solution of putting 1 on every vertex. Aug 27 '19 at 22:18
• @AlexR. Yes indeed! I forgot to mention that , but I had that thing in my mind to mention that they're all distinct of course, moreover notice that it could happen with 1 on each node only , because 2n is Prime for n = 1 only , and this again emphasizes the enigmatic behavior of number 2 Aug 28 '19 at 5:24

The vertices of a hypercube of dimension $$n$$ can be described by the set of sequences $$(\pm 1,\pm 1,\cdots,\pm 1)$$ of length $$n$$, such that two vertices are connected iff their sequences differ by a single element. Also, it is well known that hypercubes are bipartite graphs, meaning that you can color all hypercube vertices with 2 colors, such that no two same-colored vertices share an edge.
By the Green-Tao theorem, there exist arbitrarily long arithmetic progressions (APs) of primes, so that $$p_k=p_0+Ak$$ is prime for $$m=0,1,\cdots, N$$, where $$N=N(p_0,A)$$, and we can find arbitrarily large such $$N$$. So for a hypercube of size $$n$$, find an arithmetic sequence of primes of length at least $$n=N$$ (there's no actual algorithm for this, just the fact that it's possible).
Thus, split the hypercube into its bipartite vertex sets $$X\cap Y$$. For vertices $$i$$ in $$X$$, label them with $$p_0+Ai$$ and for vertices $$j$$ in $$Y$$, label them with $$Aj$$. It's clear that if you pick a large enough AP of primes, then all values on the vertices can be made distinct.