I may delete post if equations do not render in LateX, standby, preview doesn't show anything.
Greetings Science Forums and Mathematics Community!
I am Edward Solomon, and in my research into prime numbers and the Andrica Conjecture I geometrically derived a formula concerning the interior sum of partitions inside a rectangular rank-n tenor, which coincidentally made a striking statement equivalent to Fermat's Last Theorem.
Here is the formula:
Let c be an integer.
Let a be an integer, such that a < c.
Let k = c - a.
\[ 0 = \sum^{j=k-2}_{j = 0} (-1)^j\binom{k-2}{j}(c-2-j)^n \]
From the above equation it follows that:
\[(c-k)^n = |\sum^{j=k-3}_{j = 0} (-1)^j\binom{k-2}{j}(c-2-j)^n| \]
Geometrically this states that a n-cube of side-length (c-k) , is the alternating (k-2) binomial sum of first (k-3) n-cubes (of the same power) greater than itself.
From the above formula, we can make two statements (since Andrew Wiles already proved Fermat's Last Theorem):
1: Given c, and let n = 2, then there exists only one of pair of integers, (k1, k2), such that their sum (when inputted into the formula) = c^2, if there exists any solution at all.
2: Given c, and let n > 2, then there exists no solutions, (k1, k2), such that their sum (when inputted into the formula) = c^n.
How we would even proceed to prove these statements (without Wiles, 1995)...I have no idea! (well I do have ideas, since I know how to disprove a more generalized version involving rank-n tensors when all the dimensions of a n-rectangular prism are distinct prime numbers, but I do not see how to bridge the gap from a rectangular tensor to a cubic tensor, and furthermore why the case of n = 2 permits solutions for cubic tensors when the general case involving rectangular tensors does not).
Personally, what I find most interesting is that there even exists solutions when n = 2, since the method I used to derive that formula would suggest otherwise. For instance, test the Pythagorean Triple {39, 80, 89}, (k1 = 50, k2 = 9). It's hard to believe that some expression containing 47 = (k1-3) binomial expansions, when added to another expression containing only 6 = (k2-3) binomial expansions, can somehow add up to square, especially if you saw the geometry of the implied processes in live action.
I derived this result from my work on the Andrica Conjecture, Totatives and the counting of prime gaps by manipulating rank-n tensors (n be the size of the sieving set of primes) with prime number dimensions (n-dimensional rectangular prisms). If wish to see this work let me know, thankfully all of the information pertaining to Fermat's Last Theorem occurs in the first chapter of my book.
To make a long story short, I was confronted with having to count the number of viable systems of congruences that permitted coprime gaps (gaps between consecutive totatives) of certain lengths to occur, and the geometry involved in the manipulation of the tensors (the rank of the tensor is determined by the number of primes in the sieving set) containing the solutions which yielded a rather large collection of simple formulas, since Chinese Remainder Theorem demands that every viable system of congruences exists within the tensor, and exists uniquely (making the geometry and counting rather trivial).
I myself do not have time to currently research this topic any further (at least alone), as I am devoted to finishing my work on the Andrica and Cramer Conjectures. So, citizens of the world, unite!
If you notice something interesting I'll be glad to hear and forward you Chapter 1, Appendix A, Appendix B and Appendix F from my book on the Andrica Conjecture, as Chapter 1 contains all things related to Fermat's Last Theorem.
Sincerely,
Edward Solomon
P.S.
I apologize if these formulas were already known. I however cannot see how one could have derived them, other than by accident while working on something else (as I did haha).
P.S.2 Also be warned that I have submitted a formal paper on prime numbers and the Andrica Conjecture containing some of these observations, and have sent the same to several of my friend's and colleagues, so do not attempt to use this work without giving me credit. You can use it without my permission...it's public, but give me credit for deriving this monstrous formula. Thank you!
