Beyond Binary Claims: Mathematical Discovery Through Human-AI Collaboration Part 4

Working on the formal proof

Outline of Proof

Introduction to Part 4

In Part 3 of this series I presented work on C-Duplets and C-Triplets that was discovered on my path to understanding what limits the solutions of the equation:

\(\bf{a! \times b! = c!}\)

for integers a, b, and c:

\(1 < a < b < c - 1\)

That this equation has only the non-trivial solution:

\((a, b, c) = (6, 7, 10)\)

is known as the “ABC Factorial Conjecture.”

In this process I made observations about the characteristics of the prime factorizations of factorial numbers, and introduced the requirement that for an integer c to be equal to any n! the prime factorization of c must follow “Factorial Exponent Form” (FEF) that requires the exponents from the smallest prime factor (2) to the largest prime factor to be monotonic non-increasing, and for the exponent of the largest prime factor to have the exponent exactly 1. In this part of the series I am going to show that I need only a less restrictive property, “No Zero Exponents” (NZE) to prove my theorem.

The NZE property is also a requirement for all factorial numbers. If a number does not have NZE, it is not a factorial. This property requires that no exponent in the prime factorization of a number can be zero up to the exponent of the largest prime factor. This is another way of saying that no prime factor can be missing. This is the case for all factorials (because having a zero would break monotonic non-increasing) and also for the primorial numbers which have all exponents exactly equal to 1.

Now consider the NZE property applied to products of consecutive integers. Because of the way the notation relates to the proof of the ABC Conjecture, I will be referring to k-tuples of consecutive integers in descending order:

\(c(c-1)(c-2)...(c-k+1)\)

Novel Product Properties

What my research has shown is that, as the number c increases, the product of the k-tuple makes larger and larger leaps to higher values, and in the process the prime factorization of that product loses prime factors, so picks up zero exponents. This imposes limits on the value of c, beyond which the product cannot have NZE, and so cannot be equal to any n!.

A proof of this conjecture will be included in my formal proof of the ABC Theorem, but experimentally I have found these greatest NZE values for the k-tuple product at c to be:

• k = 2, c = 633,556

• k = 3, c = 442

• k = 4, c = 66

• k = 5, c = 15

Whereas the NZE limit for products of 2-tuples is high, the product for 3-tuples grows as a cubic function instead of a quadratic, so it makes bigger leaps and picks up zero exponents faster. This is more true for 4-tuples that leap with the 4th power of c, and worse with 5-tuples that advance by 5th power.

I refer to these as the “left side” NZE limits on k-tuple products because something odd happens when k goes past 5. On the “right side” of k = 5, the list of NZE limits on k-tuple products (starting at k = 6) is:

[16, 17, 18, 19, 26, 27, 28, 29, 30, 31, 38, 39, 46, 47, 48, 49, 50, 51, 52, 53, 62, 63, 64, 65, 66, 67, 68, 69, 74, 75, 76, 77, 78, 79, 86, 87, 94, 95, 96, 97, 98, 99, 106, 107, 108, 109, 110, 111, 112, 113, 122, 123, 124, 125, 126, 127, 128, 129, 134, 135, 136, 137, 146, 147, 148, 149, 150, 151, 152, 153, 158, 159, 166, 167, 168, 169, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 194, 195, 196, 197, 198, 199, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 254, 255, 262, 263, 264, 265, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 288, 289, 290, 291, 292, 293, 302, 303, 304, 305, 306, …]

This seems to be because the extended size of the k-tuples can hold on to prime factors better than the most limited case scenario of k = 5. Whereas the left side was dropping with a double-log function:

\( \log(c) = -11.54\log(k) + 20.40 \)

or:

\( c \approx e^{20.40}k^{-11.54} \)

the right side switches to:

\(c \approx 0.5k^{1.5}\)

which is just a bit faster than linear.

Relevance to the Proof of the ABC Conjecture

My proof of the ABC Conjecture proceeds in sections. The first section rearranges the ABC equation to work with the k-tuples:

\(a! \times b! = c!\)

becomes:

\(a! = c!/b! = c!/(c-k)! = c(c-1)(c-2)...(c-k+1)\)

for:

\(let\, k=c-b,\, then\, b=c-k\)

The proof is then a matter of showing that the k-tuple cannot be equal to any factorial for c greater than the NZE limit for a given k, and then using computer search to exclude all values of c less than those limits. This is easy on the left side, because for cases k = 2 through 5, I can show no solutions for c greater than 10 (I have now checked that to 50 million). However, on the right side it is not so easy because the NZE limit on the product of the k-tuple keeps growing (slowly) with k. That requires looking at the limitations on k itself.

Limitations of k

When writing the search software algorithm, I dealt with the list of prime factors of the factorials, c! and b!. Dividing c! by b! was a matter of subtracting the prime exponent list of b! from the prime exponent list of c!, starting at the index for the c! list - 2 (i.e. for k=2) and moving down no more than half the c! index. That is because b! cannot be any larger than the square root of c! which is around the c index divided by 2. However, a much bigger limit on k comes from the position of c in the list of prime numbers.

Impact of Prime Number Gaps

An area of intense study in number theory is the so called “prime gap” issue. The list of prime numbers contains the only even prime at 2, then goes through the successive odd numbers, but skips some causing “gaps” such as between 7 and 11 because 9 is not prime. The maximum length of these gaps becomes larger as the prime list goes on such that for large numbers the maximum is on the order of:

\( ln(p_n)\)

This impacts the search algorithm because every time increasing c makes it equal the next prime number, the prime factorization list for c! grows by adding c as the top prime factor with exponent 1. That means that the first b! list that is 2 index steps below the c index will be shorter in length, and the subtraction will not eliminate the top prime factor, but will create a zero in the exponent position what was the top prime for b!, so losing NZE and having no possibility to be a!. Furthermore, even if k advances to lower values of b!, having passed the top prime of c! (namely c) on to the proposed a! would mean that a! would be greater than b! thus a > b, which is a contradiction. So, there can be no solution to the ABC equation if c is a prime number.

What if c is a prime number plus one? Then when we look at the k = 2 case of c(c-1) we would have (c-1) be that prime number, which would end up in the product, so again, no solution. But, if c were two numbers above a prime (and in a prime gap bigger than 2), the situation is okay for c(c-1) because both of those factors are not prime, however k = 3 would still be in trouble.

This is a central concept of my proof: the maximum value of k is limited by the size of the prime gap in the neighborhood of c. If c is large enough, this does not matter for the left side limits, but for large c the growing prime gaps allow bigger values of k, so bigger values of the NZE limit based on right side k. Is this a problem as c gets very large? The answer is no, and here is why:

The size of the prime gaps in the neighborhood of c are growing as ln(c) and the right side NZE limit is growing as k^1.5 so given that the prime gap is the maximum k around c, the limit on k grows at:

\(0.5(ln(c))^{1.5}\)

This behavior is mathematically confirmed by the derivative of the function:

\( \frac{d}{dc}[0.5(\ln(c))^{1.5}] = \frac{0.75(\ln(c))^{0.5}}{c} \)

As c increases, this derivative approaches zero, while a linear function maintains constant slope, proving that the right side NZE limit grows more slowly than any linear function for sufficiently large c. Thus, given that I have eliminated all possible solutions above c = 10 up to a large value of c, (through computer search up to c = 3,000,000 where k can get up to 221), and at a large value of c, as c grows c will remain forever above the right side NZE limit, there will be no more solutions other than (6, 7, 10). This concludes my proof. The next step is to formalize this for submission to a number theory journal.

P.S. I gave this post to Gemini at NotebookLM and it made an explanation podcast.