Oh dear: this instalment was intended to be a routine exposition of the special case when p=7 of the reasoning contained in the second part of Blog 2 (2022) which introduced the exciting new elementary approach to Fermat’s lost theorem. The point of focusing onto this special case was to illuminate the core argument, which can be understood much more easily when we are dealing with a specific value of p. Unfortunately a problem has emerged at the last moment.
The Main Case of Fermat’s Enigma concerns the possibility that his (Fermat’s) equation harbours a contradiction when Tab > 0 (mod p). An illumination of the core argument here was set out in the previous Blog.
But work on the Minor Case when Tab = 0 (mod p) has, alas, revealed an overlooked glitch in the original Blog. The reasoning of proposition (39) of Blog 2 was that Y p– pLpbp-3 necessarily had the factor A. But if it had the factor A, it would —as a freestanding specific unknown natural number— have the value zero when A was replaced by 0. Thus the expression Y p– pLpbp-3 which we arrive-at when A=0 cannot be zero —it was argued— because it is the difference between a pth power and p times a pth power. This seemed at the time to be rock solid, and the presumption that it was valid remained until this week (May 29th 2022). But, as part of the build-up to the present Blog it was re-worked carefully, taking into account the neglected fact that when A = 0 the effect on Y and L is to turn them away from being natural numbers.
[When A=0, Y becomes b4/7V which is an irrational number because V becomes the 7th root of b4/117649. L too becomes an irrational number because it, too, is defined in terms of V. (In Blog 2 Y and L were used to denote the specific unknown numbers which obtained when the As in their specific defining expressions were replaced by zeros.) The result of this was, regrettably, overlooked in Blog 2. But now, using the correct expressions for Y and L instead, it can be clearly seen that the expression above (Y p– pLpbp-3) —which was supposedly an expression composed of natural numbers— ends up being equal to b4 – b4! This shows unmistakably that Y p– pLpbp-3 is zero, and that Y p– pLpbp-3 does have the factor A! With the advantages of hindsight it is obvious that that, if one takes away the one and only non-A term in a polynomial, the result will be that what remains has the factor A.]
This glitch does not invalidate the reasoning in the main case, when Tab > 0 (mod p) in any way. The value of Tab could be either a multiple of p or not. The nominal “probability” that it is not —disregarding the unlikely possibility that it is necessarily a multiple of p— is, of course, p times the “probability” that it is. Also the possibility that a value of Tab might have the factor p a huge number of times (p-1 times) when p is large (i.e. in the majority of cases, bearing in mind that p might be as large as you like) is very remote indeed. So we are dealing in the glitch-ridden case with a much, much rarer situation. But a repair will , of course, be needed, and it is a case of “back to the drawing board”.
The same strategy underlying the main case when Tab > 0 (mod p) will be applied in the repair exposition, which has, fortunately, already been provisionally identified. It will be carefully checked-over and set out in the July Blog. This strategy is that it is necessary (I) to conceptualise clearly in detail how the situation that Tab = 0 (mod p) might happen. This will then (II) imply results which lead to a clear contradiction.
A Glitch Review Careful followers of this series of Blogs will have already noticed that a succession of glitches occurred when the extraordinary Enigma discovered by Richard Beetham —the author’s former teacher in 1946— was picked up in the Mathematical Gazette in 2020. The Beetham Enigma was actually much more difficult to resolve than this current hitch in dealing with Fermat’s Enigma. This arose from the fact that there was no apparent reason why the Beetham Enigma was so difficult. The author is now rather old, and he has outlived virtually all his former mathematical friends —who might otherwise have been able to check his reasoning with unbiased eyes. It took him more than two years —during which he explored more than 300 possible elementary constructions in vain— to put together a strategy for tackling the Beetham Enigma. Several of these suggested paths of reasoning which were aired in these Blogs but subsequently turned out to be glitched. The solution was to base the reasoning on a strategy which amplified an unobvious overall symmetry in the configuration. There is actually also a minor glitch in the final Beetham Blog (the point marked E in the last diagram is not the same as the point marked E in the earlier diagrams) but this has no effect on the reasoning, so it is not significant.
CHRISTOPHER ORMELL June 1st 2022