Monday, August 21, 2017


It was vrey dark here for a little more than two and 1/2 minutes. There was no air movement during that time, and not surprisingly the birds stopped singing. But there were no other noticeable side effects. Certainly no eartquake ---thankfully.


Posted Wednesday, June 21, 2017


The link below is to an image of the paths of the total Solar eclipses of 2017 and 2024. The X happens to mark a spot a few miles from the epicenter of the largest known earthquake ever in the US, the New Madrid quake of 1812. We're told that a large quake, 8 to 9 0n the Richter scale, has happened about every 200 years in this area according to geological evidence.  Will the combined gravitational pull of the earth and sun lined up in a total solar eclipse on August 21, 2017 be enough to trigger the next one? Incidentally, we live about the same distance from the center of the X as the epicenter of the New Madrid Missouri quake of 1812, well within the paths of both solar eclipses! We have a front row seat to view the 2 minute and 40 second maximum eclipse on August 21, and we're only a few miles from the New Madrid fault. Would you be excited, or scared?


Sunday, August 13, 2017


I think that this recent exchange with a friend who has been looking at FLT65 for some time may be of interest to  those who are following the subject of Fermat's Last theorem, which as they may know, is importent to the explanation of quantum phenomena in TDVP. This conversation may be helpful for anyone interested in understanding the logic of the 1965 proof. For this reason, I am posting my latest response here without identification of my friend or his math professor friend.


I spent some time thinking about Prop. P between meetings and events in LA last week. There really should be no confusion about when P is true and when it is false. I will attempt to clarify this while answering your latest comments, because they do reflect what I see as confusion about what FLT65 says and does. I shall also attempt to show you how distinguishing when Prop. P is true and when it is false, and when the identity sign is appropriate, leads to a better understanding of FLT65. 

From your email earlier today:

      Q: Do you believe that Prop P is ever false?

Let’s consider this. Here is Prop P as you stated it:

     P: Equating a polynomial to a constant, for the purpose of finding a  
        specific solution, automatically turns the polynomial itself into a  

This should not be a matter of “belief”, because it is easy to show that P can be either true or false, depending upon the nature of the polynomial, and the circumstances. And, it is an added benefit that the circumstances also allow us to clarify the proper use of the identity symbol ≡.

P is true, if and only if, all variables are specified as constants. For example, we know that, if Y in the FLT equation is an integer, then for any Y there is some integer factor A such that z – a = A. If we should also know that Z and Y are the specific integers Z1 and Y1, then a is determined, and we have: Z1 – a1 A, an identity.

But in FLT65, z is an unknown. Clearly, integers can be assumed for X and Y, and if so, then whether or not z can be an integer is as yet, in the proof, unknown. To assume that it is an integer at the beginning, is premature and leads to circular reasoning. For the polynomial z – a in FLT65, z is a real number, but its specific value is unknown, so when we set the polynomial z –a = A, a known integer factor of Y, P is false, because z and a can take on an infinite number of values. The only requirement imposed by setting z – a = A is that their difference is always A. The value of z can vary, and thus this equation is not an identity.

 I once sent you the judgement of a mathematician who had spent his entire working life as a mathematics professor in a university. You dismissed his judgement that FLT65 was invalid as “something you had seen before.” 

To be clear, I had seen this line of reasoning several times before, and have given it all the serious thought it deserves, so I saw it as the same knee-jerk reaction I’ve seen numerous times, and refuted every time. I felt that you should have recognized the circularity of the argument. But I apologize for my abrupt manner. I should not have been so abrupt, and I certainly should not have been condescending. I’ll copy the math professor’s comment here and try to respond more appropriately.  He said:

“Here is a possible way of putting it that might convince Mr. Close.  In his argument he sets  a = Z - A   and considers the divisor polynomial  g(Z) = Z - a, which he says is a polynomial of degree  1  in Z.  But  g(Z) = Z - a = Z - (Z - A) = A,  which is not of degree  1  but of degree 0.  When the divisor A  is a (nonzero) constant, the polynomial division algorithm over the reals just says there exists a (unique) polynomial  q(Z)  such that  f(Z) = Aq(Z) + 0, where, of course, q(Z) = f(Z)/A.  so there is no  f(a).”

This argument completely misses the point of FLT65 by assuming the definition a = Z – A which is not the case in FLT65. His statement that I set a = Z – A is false. You will not find this anywhere in FLT65. Z is the unknown, the dependent variable. Z – a is defined as a polynomial of the 1st degree, meaning that the value of Z depends on the value of A, an integer factor of Y and the value of a, which is an integer if there is an integer solution of the FLT equation, which is as yet, in the FLT65 chain of logic, unknown. By assuming that Z is an integer by definition, the conclusion that Z –a is a constant, and therefore of 0 degree, is of course, circular reasoning, as mentioned above, and stated in my more abrupt response. By the way, I meant no disrespect to your professor friend by dismissing his comments. A university professor known as a number theory expert made the same mistake, but he quickly acknowledged that it was circular reasoning when I pointed it out to him.

Thank you for your statement of what you see as a disproof of FLT65. It enables me to better understand why you kept coming up with propositions that had nothing to do with FLT65. I think other critics may have had this same misconception about how FLT65 goes about proving FLT.

Your statement has the logic of FLT65 completely backward:

 “The equation of constants, f(Z1) = Ap = (Z1 - a)p, does not imply that the variable (Z - a) is a variable factor of the variable f(Z). 

I agree! However, what FLT65 says is the converse: the fact that f(z) cannot contain z – a as polynomial factor for real number values of the variables, implies that, if there were an integer solution for the Fermat equation, then there would be an integer version of f(z)/(z – a) = q(z) + f(a) where f(a) would equal zero, and that would violate the ‘if and only if’ condition of the division algorithm for polynomials.

It appears to me that the confusion comes from considering Z to be a specific integer before it is known whether z can be an integer or not. Maybe this will be clearer if we go step-by-step:

Dividing f(z), a polynomial of degree p-1 by z - a, a 1st degree polynomial, we have unique polynomials q(z) and f(a) of degree less than p such that:

(zp-1 + zp-2x + zp-3x2 +•••+ xp-1)/(z-a) = q(z) + f(a)/(z-a). Multiplying through by z-a, we have:
f(z) = (zp-1 + zp-2x + zp-3x2 +•••+ xp-1) = q(z)(z-a) + f(a).

From this we see that the polynomial f(z) is factorable into two polynomial factors, q(z) and z-a, if and only if f(a) = 0. But f(a) = ap-1 + ap-2x + ap-3x2 +•••+ xp-1, which cannot equal zero because a and x are positive for any integer solution of the FLT equation.

Therefore, the FLT equation factor polynomial f(z) cannot be factored into two polynomials of degree less than p, one of which is z - a.

If there is an integer solution, then with the term-by-term substitution of the integer variables into the variable polynomial f(z), we obtain the variable polynomial f(Z) for any integer solution of the Fermat equation. But, for any integer solution, f(Z) = (Z-a)p, where Z-a = A, a single integer factor of Yp. So now we have:
The hypothetical integer polynomial f(Z) = (Zp-1 + Zp-2X + Zp-3X2 +•••+ Zp-1) = q(Z)(Z-a) + f(a) = (Z-a)p.

By inspection of this integer polynomial equation we see that f(a) contains Z - a as a factor, and although we don’t know what the specific values of a and X are for an integer solution, we know that they are positive constants. So f(a) = M(Z-a) where M is an integer constant, and we have f(Z) ) = (Z-a)p, = (Zp-1 + Zp-2X + Zp-3X2 +•••+ Zp-1) = q(Z)(Z-a) + M(Z-a), from which we have: f(Z) = [q(Z) + M](Z-a).

And q(Z) + M is a variable polynomial in Z of degree less than p, call it q1(Z). Thus for hypothetical integer solutions of the FLT equation, we have the variable polynomial f(Z) = q1(Z)(Z-a). But this is a violation of corollary III of the division algorithm, which tells us that the variable polynomial f(Z) cannot be divided into the factors Z - a and another polynomial of degree less than p. The only way we can avoid this contradiction is for z to be an irrational real, not an integer. 

I have demonstrated in at least three different ways, including Fermat’s favorite method of proof, infinite descent, that if we ignore this contradiction, and assume that two of the three variables x, y and z are integers, and solve the FLT equation for the third variable, then that third variable cannot be an integer.

I don’t consider adding such demonstrations to FLT65 to be necessary because the contradiction f(a) ≠ 0 versus f(a) = R = 0 is sufficient to prove FLT by itself. This contradiction is valid because it is obtained by applying the division algorithm and corollaries to variable polynomials, not constants, and a single contradiction is sufficient to prove there can be no integer solutions for zp – xp = yp.

I believe the whole confusion for most critics arises from assuming that the division algorithm and corollaries are inappropriately applied to constants. They think this because they jump to the conclusion that all three variables must be treated as integer constants from the beginning of the proof. Thanks to you, this confusion has been made clear with the analysis of Prop P! 

Once you see that P can be true or false, depending upon the nature of the polynomial, and that in FLT65, f(Z) and Z – a are still polynomials of the variable Z, even though for a hypothetical integer solution, they are equal to the constant integer factors of Yp, the logic of FLT65 becomes clear, and the search for counter examples and counter propositions becomes unnecessary and irrelevant.

Edward R. Close  August 12, 2017

Friday, August 4, 2017


in 1637, Pierre de Fermat, a judge at the French Parliament of Toulouse, wrote in the margin of a book on Diophantine equations that he had devised a marvelous proof that there are no positive whole number solutions to the equation xn + yn = zn for n greater than two. His proof, for n = 4 is known, but his general proof for all n greater than two was never found.

It is an interesting aside that Fermat was not a professional mathematician. He did not publish his findings, he simply conveyed them in letters to other mathematicians, and thus was considered an amateur. The less than modest French mathematician and philosopher Rene Descartes tried to discredit Fermat by proclaiming that he was “a troublemaker who owed his reputation to a few lucky guesses”. However, in one dispute after another, e.g. their derivations of the sine law for the refraction of light, Fermat proved to be right and Descartes wrong. While Descartes clearly considered himself to be the superior intellect of the day, a comparison of their works reveals the fact that Fermat was the better scientist and mathematician of the two.

Descartes’ arrogance shines out in the following statement: “‎I hope that posterity will judge me kindly, not only as to the things which I have explained, but also to those which I have intentionally omitted so as to leave to others the pleasure of discovery.” Implying that he could have explained much more. This is analogous to a classmate of mine who liked to say “I’m not conceited, I’m actually twice as smart as I say I am!”

Peter Bernstein, in his book Against the Gods, states that Fermat "was a mathematician of rare power. He was an independent inventor of analytical geometry, he contributed to the early development of calculus, he did research on the weight of the earth, and he worked on light refraction and optics. In the course of what turned out to be an extended correspondence with the Pascal, he made a significant contribution to the theory of probability. But Fermat's crowning achievement was in the theory of numbers."

Regarding Fermat's work, Sir Isaac Newton wrote that his own early ideas about calculus came directly from "Fermat's way of drawing tangents."
Speaking of Fermat's work in number theory, Mathematician Andre Weil says that: "what we possess of his methods for dealing with curves of genus 1 is remarkably coherent; it is still the foundation for the modern theory of such curves. It naturally falls into two parts; the first one ... may conveniently be termed a method of ascent, in contrast with the method of descent which is rightly regarded as Fermat's own. … With his gift for number relations and his ability to find proofs for many of his theorems, Fermat essentially created the modern theory of numbers.

Never-the-less, popular history has treated Rene Descartes very well, while ignoring Fermat. The name Descartes is well known to every student of mathematics and science, while Fermat’s name is virtually unknown except for in relation to Fermat’s Last Theorem, -and most modern mathematicians openly doubt that he actually proved it! Why? Because for more than 300 years, professional mathematicians tried to find a proof, and failed.

The power of Fermat’s math lies in his method of infinite descent, and the principle of “efficient purpose in nature”, and in 1964-5, while teaching high school mathematics, using the same simple principles used by Pierre de Fermat, I produced a proof which I first submitted to a professional mathematician in 1966. Because my proof was completed in 1965, I call it FLT65.

To see FLT65 as it was submitted to the first reviewer on January 25 1966, copy the link below and paste it into your web browser.

This link will take you to a lengthy discussion of FLT65. To see the original submitted version, scroll down to Appendix C.

I have submitted FLT65 to more than 50 mathematicians over the years, and only a few of them who responded actually offered any mathematical arguments attempting to disprove FLT65. And while a precious few have admitted it, none of them were able to produce a valid refutation of FLT65. Because I believed it would eventually be recognized as a valid proof, I have carefully documented the submittals and responses.

You might well ask: if no one has refuted FLT65, why hasn’t it been accepted? The answer to this is an interesting story by itself. To learn the details of the history of the odyssey of FLT65, copy the link below and paste it into your web browser.

I believe that the time for Fermat to become a household name has come, because his simple methods of Diophantine analysis are totally appropriate and exactly what is needed for application to quantum physics. It is time to go beyond the calculus of Newton and Leibniz and apply the Calculus of Distinctions to TRUE quantum equivalence units to produce a better description of multi-dimensional reality. See details in the posts of TDVP and

Because of excessive academic specialization, and the fact that modern thinking has lost its metaphysical basis in Infinite Intelligence, science has gone astray and adopted atheistic materialism as its metaphysical basis. The result is a world society that is morally adrift. This lack of meaning and purpose is so dangerous that, if not corrected, it could spell the demise of the human race. It is time to revisit the simple infinite descent and principle of “efficient purpose in nature”, of Pierre de Fermat.

Edward R. Close    August 4, 2017

Wednesday, August 2, 2017



A friend pointed out to me that I spend a lot of time and effort these days defending my 1965 proof of Fermat’s Last Theorem (FLT65) against hypotheses and propositions that are designed to show flaws in FLT65’s logic. My friend suggested that instead of being on the defensive, perhaps it is time for me to take a different approach. The only legitimate questions raised by critics seem to involve, in one way or another, questioning the legitimacy of the application of the division algorithm and corollary III to integers in FLT65. So let’s have a look at how the division algorithm is applied in FLT65.

Division is one of the four fundamental operations of mathematics, and the division algorithm describes the operation of division for polynomials (algebraic summations with multiple terms). The following are examples of algebraic polynomials:
f(z) = z3 – a3; g(y) = y2 + by + c; h(x) = x4 – 2x2 + 5x + 13; and q(X) = 12X
A general expression representing an algebraic polynomial in x, of degree n, is given by;
f(x) = axn + bxn-1 + cxn-2 + …+ sx + k, where there can be any number of terms, and a, b, c. …s, and k are constants that can be either positive, negative or zero.

The first part of FLT65 shows that for any polynomial f(x) over the field of real numbers, there exist unique polynomials, q(x) and g(x), such that f(x)/g(x) = q(x) + r(x)/g(x). This is nothing more than a statement of the division algorithm for polynomials. I also provided proof of the three corollaries of the division algorithm used in the FLT65 proof, including corollary I, which says that if f(x) and g(x) contain a common factor, r(x) contains it also; corollary II, which says that the remainder, when f(x) is divided by z-a is f(a); and corollary III that says that a polynomial f(x) of degree greater than 1, is divisible by the polynomial x-a, if and only if, f(a)=0. FLT65 also contains proof that if FLT is true for n = p, when p is a prime number greater than 2, then it is true for all n.

In FLT65, if x, y and z are integers, then, for a comprehensive proof, it is sufficient that they are relatively prime integers. For FLT65, y is chosen as a variable that does not contain p, and f(z) is defined as the larger factor of the right-hand side of the equation below, the Fermat equation:
yp = zp – xp = (z-x)(zp-1 + zp-2x + zp-3x2 +•••+ xp-1)
It is shown that z–x, and f(z), as factors of yp, can be considered to be relatively prime. Thus f(z) = Ap, and A is an integer factor of y.

It is noted in FLT65 that the division algorithm and corollaries hold when the terms and coefficients of the polynomials are integers because the integers are elements of the field of real numbers. Also note that there are no restrictions on the application of the division algorithm and its corollaries with regard to the values of the coefficients, or the number of terms in the polynomial to which they are applied. This means that the statement of the division algorithm, f(x)/g(x) = q(x) + r(x)/g(x), is true whether f(x) and g(x) are polynomials of many terms or single terms.

Applying the algorithm and corollaries to the polynomial f(z), and defining g(z) as z-a, we have:
(zp-1 + zp-2x + zp-3x2 +•••+ xp-1)/(z-a) = q(z) + f(a)/(z-a). Multiplying through by z-a, we have:
(zp-1 + zp-2x + zp-3x2 +•••+ xp-1) = q(z)(z-a) + f(a).
From this it is clear that the polynomial f(z) is factorable into the two polynomial factors q(z) and z-a, if f(a) = 0.

We also know that, if there is an integer solution to the equation zp – xp = yp, then yp is an integer containing the integer factor f(Z) = Ap. = (Z-a)p. This gives us two equations:

1)         (zp-1 + zp-2x + zp-3x2 +•••+ xp-1) = q(z)(z-a) + f(a), an algebraic polynomial equation and
2)         (Zp-1 + Zp-2X + Zp-3X2 +•••+ Xp-1) = q(Z)(Z-a) + f(a), the same equation with all integer terms. Equation #2 exists if there is an integer solution (X,Y,Z) for zp – xp = yp.
For an integer solution, if one exists, equation #2 can be reduced to an equation of single integer terms:
3)         Ap = Q·A + R, which implies that the integer f(Z) = Ap is equal to the product of the two factors Q and A, if and only if R = 0.

Critics have suggested that there is no correspondence between the polynomial factors of equation #1 and the integer factors of equation #3, in which case, f(a) 0 does not necessarily imply that R 0, and the fact that the polynomial f(z) cannot contain the polynomial z-a as a factor, has no bearing on whether the integer that f(Z) reduces to can contain the integer value that (Z-a) reduces to, or not.

So the real question whose answer will resolve the disagreement about the validity of FLT65 is:

What is the nature of the relationship between the polynomial factors of equation #1 and the integer factors of equation #3?

In FLT65, Z-a is defined as A, a factor of Yp, and f(Z) = Ap, and A is an integer if there is an integer solution. By inspection of equations #2 & 3 above, we see that there is a one-to-one relationship between the four expressions of the two equations. By direct substitution of single-integer values for X, Z and a, assuming an integer solution of the FLT equation, for any prime degree, p, the polynomial factor (Zp-1 + Zp-2X + Zp-3X2 +•••+ Xp-1) yields Ap, the polynomial factor q(Z) yields Q, (Z-a) yields A, and f(a) yields R. This means that R = f(a) ≠ 0, and we have a contradiction:

For there to be an integer solution to the FLT equation, Yp =ApBp =f(Z)Bpf(Z) =Ap and this in light of equation #3 implies R has to be zero. But because of the form of f(z), with all coefficients equal to unity except the constant term, which is zero, f(a), which yields R, cannot be zero. And thus the fact that f(a) 0 for equation #1, implies that R 0 in equation #3, and this contradiction, featured in FLT65, is sufficient to prove there are no integer solutions for the FLT equation.

Edward R. Close August 3, 2017

Monday, July 31, 2017



The big bang theory has been popular with the general public for more than half a century, but contrary to popular belief, the simplistic idea of a universe flying apart from a single explosion of an infinitely dense mathematical singularity (a dimensionless point) billions of years ago has never had unanimous support among physicists and cosmologists. The most obvious problem was that a straight-forward regression, or running backward in time in accordance with the general theory of relativity resulted in having stars that were older than the universe. This was side-stepped by Alan Guth’s rapid early inflation theory in 1981. There were other problems, and in 1989, I applied the Calculus of Distinctions to the red-shift constant-light-speed expanding-universe big bang and found that there were unresolved contradictions in the theory, indicating that the universe could not have originated in a singularity explosion. I also concluded that the universe has always existed in some form, and will always exist. I sent my findings to Stephen Hawking in 1989 and published them in 1990, but they were generally ignored by mainstream science. Professor Hawking responded that my ideas were interesting, but he didn’t have time to pursue them, and he had a problem with my assertion that time, like space, has to be three dimensional.
I am not the only one who had (has) problems with the big bang theory. The latest ripple comes from the Laser Interferometer Gravitational-wave Observatory (LIGO). See the Daily Galaxy site:
From the article: “Correlated noise in the two LIGO gravitational-wave detectors may provide evidence that the universe is governed by conformal cyclic cosmology (CCC) which assumes that the universe consists of a succession of [big-bang] aeons, says Roger Penrose of the University of Oxford. Penrose proposes that the apparent noise is actually a real signal of gravitational waves generated by the decay of hypothetical dark-matter particles predicted by CCC. Penrose argues that a significant amount of this noise could be a signal of astrophysical or cosmological origin – and specifically CCC.
Last month, physicists at the Niels Bohr Institute, writes Hamish Johnston, editor of, pointed out that some of the noise in the two LIGO detectors appears to be correlated – with a delay that corresponds to the time it takes for a gravitational wave to travel the more than 3000 km between the [recording] instruments [of the interferometer].

First proposed over a decade ago by Penrose, CCC assumes that each aeon begins with a big bang and proceeds into an unending future in which the universe expands at an accelerating rate. As this expansion becomes infinitely large, Penrose argues that it can be transformed back into the next big bang.”
This is a huge departure from the popular big bang theory and abandons the idea of an absolute beginning or end. After more than 27 years, Roger Penrose, co-author of the Penrose-Hawking Mathematical Singularity Theorems, appears to be agreeing with some of my 1989 findings.

Edward R. Close, July 31, 2017

Thursday, July 27, 2017


Link to the Power Point used in the second part of the St. Louis presentation Saturday July 22, 2017

The surface of the cube has 54 colored faces: 9 red, 9 orange, 9 blue, 9 green, 9 yellow, and 9 white. This makes 9x6 = 54 total. The cube appears to be divided up into 27 small cubes, but they are attached in the center so that they aren’t able to show all of their faces. If they were all colored on all 6 sides, there would be 27x6 = 162 colored faces, and there would be more possible configurations than all the visible stars in the night sky! Fortunately, they are only colored on 1, 2 or 3 sides.

There are 6 center pieces that never move relative to each other, with only 1 colored face. There are 12 edge pieces with 2 colored faces, and 8 corner pieces with 3 colored faces. This makes 6x1 + 12x2 + 8x3 = 6 + 24 + 24 = 54 colored faces, and that means that the number of different configurations possible is:

 N = [(1!x61)/6]x[(8!x38)/3]x[(12!x212)/4] =[(8!x38)x(12!x212)]/12

Fortunately, I didn’t have to work this formula out from scratch. It uses the concepts of combinations and permutations, for which mathematicians, with nothing better to do, have worked out formulae. Determining the number of all possible combinations and permutations is part of a broader mathematical subject known as combinatorics. If this sort of thing interests you, you can look it up on the internet. Carrying out these multiplications and divisions, we get the number of non-redundant configurations to be: N = 43,252,003,274,489,856,000.

This is still a very huge number of different configurations. But take heart! There are algorithms (sequential series of rotations) that reduce the number of necessary rotations to a reasonable number on a path through this maze of possibilities to a solved cube. By “solved”, of course we mean having all nine faces of each color neatly arranged on each of the 6 sides of the cube. Each side is “mono-chromatic”, i.e. all one color. 

In order to develop algorithms (formulas) for solving the cube, we must have a notation or code for describing all of the rotations needed.

Facing any given side of the cube, the front, or facing side is indicated by the label F. the left side is labeled L, right side R, upper side: U, back side: B, and bottom side: D. The bottom side is labeled D for “down” to avoid confusion with B for “back”. Further, for algorithms describing 90-degree (quarter-turn) rotations in the nine planes simulated by the cube: F, L, R, U, D, and B, denote clockwise rotations, and F’, L’, R’, U’, D’, and B’, denote counter-clockwise rotations. (Clockwise and counter-clockwise are defined as if you were facing the side being rotated.) It is good to practice each of these quarter-turn rotations, saying “L” when you turn the left side clockwise, and “L-prime” for L’, when you turn the left side counterclockwise, etc., until they become second nature.

Trying to learn how to solve the cube by reading written instructions alone, is like trying to learn to play a violin by reading. It is possible, but reading is no substitute for a good instructor. I will have more complete instructions and pictures in the book, but just having the algorithms may give you a start. Once you get started, you may even find some algorithms of your own.

Start by holding the scrambled cube with the red center square up. Look for the 4 two-faced edge pieces with one red face, and move them up, one at time, into the top layer with the red face up. Each of the 4 will be in place when its other face (not red) is lined up with its color in the middle layer below it. This will give you a big red “X” that will be your reference frame for the rest of the solution.

Next, find the 4 corner pieces that have one of their 3 colored faces red, and rotate them into place with the red side up and the other two colors matched up with the same color center pieces of the two sides. You will be able to rotate some of these into place without disturbing the red “X”, but when it is necessary to disturb it, you must restore it before going on to the next corner piece. When this is done, you will have completed one layer of the cube. No algorithms are needed to this point. Most people can get this far by trial and error. But it is best to practice doing this until it becomes routine, because the more efficient you become at this routine, the easier the next step will be.

Turn the cube over with the red side down, and proceed to the next step. You only have the 4 edge pieces to worry about in the second layer, because the 4 center pieces are part of your extended “X”. And, only one algorithm is needed to complete the second, or middle layer, but you may think of it as two, because it has a left-hand and right-hand version. Find the edge piece in the third layer that needs to be in the second layer and line one of its colored faces up with the same color center piece. Then use one of the algorithms below to put it into place. Use the right-hand version if the piece being moved into position is facing you, and the left-hand version if the piece is on the left face.

          Right-hand: U’L’ULF’LFL’ or Left-hand: URU’R’FR’F’R

The nice thing about these algorithms is that while moving the correct edge pieces into place in the middle layer, even though pieces in the first layer are moved, they are returned to their correct locations by the time you complete the algorithm. It took me about a week to work these out the first time, and I never wrote them down until recently. They may be confusing at first, but with practice, they become second nature.
Using the right- or left-hand algorithms above from 4 to 8 times should complete the second layer while preserving the first layer. When the first two layers are complete, you are ready to move on to the third, and most difficult layer.

Continue to hold the cube with the red side down. Only three more algorithms (with mirror-image versions) are needed to complete the solving of the cube. But they may have to be used multiple times while watching for specific TRIADIC patterns in the orange faces on the top of the cube. All three are designed to move the faces in the third (top) layer while preserving the two layers already completed. It took me much longer to discover them. The first one is designed to get the top layer arranged for the second one, and the second one is designed to get the top corners in place, and the third one is designed to move the edge pieces of the top layer clockwise or counterclockwise into place to complete the cube.

There are 8 possible patterns for the orange faces in the top layer when the first two layers are complete. Fortunately, you don’t have to know what they are (even though you will get to know them through practicing the algorithms given here). You just need to apply the algorithms below repeatedly, until you obtain a “backwards L” composed of three orange squares in the upper left-hand quadrant of the top face of the cube.
When that orange “backward L” appears, use the “mirror-image” algorithm below:
Now, check to see if the corner pieces are in place. Don’t worry if they are not turned correctly, just check to see whether the corner piece with, e.g., white, blue and orange faces, can be lined up with the white and blue sides, while the yellow, green and orange corner piece is lined up with the yellow and green sides, the yellow, blue and orange with the yellow and blue sides, and the white, green and orange with the white and green sides. Chances are good that they will not all four line up. Use the algorithm below repeatedly, until they do.


This is the most difficult step by far, and sharp focus, concentration, diligent practice and intuition are needed. With practice, you will intuitively find your way through this step.

If, when you line one corner up, none of the others are in place, use the algorithm above and check again. If two of the corners line up, but the others don’t, line the two that are correct up on your left and perform the algorithm again. Keep performing the algorithm until all four corners are in place. When they are in place, there are 3 possibilities: either 1) the cube is complete; 2) you have a large backward L triadic pattern on the top of the cube with three of the corners having colors other than orange facing up, and one orange face on the left side; or 3) a large backward L with the orange face of the lower left corner piece facing to the front. If you have #2, then, keeping the red side of the cube down, turn the cube to the right, so that the orange face of the corner piece on the lower limb of the backward L is facing you, and perform the following algorithm:
If you have #3, don’t turn the cube, and perform the mirror-image algorithm below:  
(Note that the 2 after the U or U’ means perform that quarter-turn rotation twice.)
At this point, you may have a completed cube, or you may have all the orange faces up, but with the edge pieces of the upper layer in the wrong positions. Use the following algorithm to move them clockwise around the top layer, or its mirror-image to move them counterclockwise, until you have a side-wise monochromatic completed cube.
Clockwise: R2UFB’R2F’BUR2; Counterclockwise: R’2U’F’BR’2F’BU’R’2
Congratulations! If you’ve made it successfully through to this stage, you have will have solved the cube. With practice, you will become more and more proficient at avoiding unnecessary rotations, and you may eventually be able to complete the cube from any scramble in less than one minute! HAVE FUN!

Ed Close July 28, 2017

Wednesday, July 26, 2017



This closed conference for CARE Instructors was not recorded. But a link to the powerpoints has been requested by participants and others who were unable to attend. Please click on "Care Presentation" below.

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