Mathematical proof..

[Alex K]

Concerning your problem with the plus one.
I think you can argue that the product of all primes smaller than x, plus one, is divided by all those primes with remainder one. Is that sufficient? (long ago)

It's more important to say x+1 is not divisible by ANY of the prime numbers in x's set. I believe this is sufficient.

edit:

Assume that there are finitely many primes.
If p_1...p_N are all existing prime numbers and x = p_1 * ... * p_N

x /p_i = p_1 * .... * p_{i-1} * p_{i+1} * ... * p_N

and hence

(x+1) /p_i = p_1 * .... * p_{i-1} * p_{i+1} * ... * p_N with Remainder 1

so we have constructed a number which is not divisible by any of the primes, and there's your contradiction, and there can't be finitely many.

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So paint isn't my discovery, but I challenge you to tell me how to describe my new result..

I can't really do that without knowing your argument. But usually, one would introduce an abstract set {1,2,3,4,...} standing for the colors, and maybe a set of areas {A1,A2....}, a function which sets the color f(A_i)=1...4 and a relation which defines adjacency, such that A1°A2 =0 if it is not adjacent and A1°A2=1 if they are. That's just an example of what such things typically look like in principle, I'm not saying that that's how it actually works.

A often used approach towards such a type of proof is to assume that there is one scenario where 4 colours are NOT sufficient, and then bring this to a contradiction. In other words: If you can start from the assumption that there is one set of areas where there is no four color covering scheme, and you can conclude from this a statement like 1=0 using only valid logical steps, then you've shown that the assumption is false and the four color theorem is proven.

Alternatively, one might not use the areas as the fundamental object, but the boundaries and their vertices.

Maybe a proof by induction is possible by starting with an arbitrary number of areas which can be covered by four colors and then adding areas in succession and proving that one never needs to introduce a fifth color in each single step. The problem here might be that the way to color the given areas is not unique, and adding one might not be possible for an arbitrary coloring scheme of the existing areas.