What a wonderful treat bumping into your well organized list of categories. Adventure, BASE jumping, Climbing, Art and Design, Animals, Birds and Beer, Oh my!

The Art Of Broken Pieces. Well I can certainly relate to that as I’ve been broken and put back together a few times. ” Kintsugi” Real lacquer, hon urushi dusted with gold powder, kinpun of the best quality used not only to piece together the expensive imported but sadly broken Sung dynasty tea bowls, but to honor even the most humble vessel. This concept is called “Reigi” Respect. All Japanese tea related arts begin and end with reigi. This is life long meditation is called Michi or Do, “The Way”.

I came into climbing from Kendo, “the way of the sword”, back in the early sixties. I loved the picture you posted of what I used to call the Eugene crowd. As I looked at the picture I started to smell deep fried mushrooms, french fries and beer.

The Red Barn was filled with climbers from outdoor programs, with young and old climbers shouting their stories above the overly loud juke box.” I’ve been through the desert on a horse with no name”,spliced together with, ” off width”, “EB” PA, bad pro, ” and did you hear?”he bought it in Yosemite”. Half heard, half drunk stories to be patched together in the future by some that were there and many that were not, including guide book writers and pundits.

For years I’ve read these horror stories of bad bolts, retro bolting, hangdog routes, etc. The advancement in Equipment is wonderful. Better shoes, better clothing, better hardware, cams , bolts and hangers, and biners that are so much lighter and climbing gyms that have improved technique greatly are all good but please remember.

Smith Rocks is not a gym and even if we have a long history with the place, we are not entitled to screw it up for the sake of our egos. At one time some people stamped their initials onto hangers just to claim some dubious first ascent kudos for the guide book.

I loved the skinner butte story. many of us have almost the same story. maybe not the modified flat bed truck, but the quick trip to Eugene, the food co op, skinner butte, putting on the torn up patched, worn out E.B.s and third classing hard routes and leaving before the locals got preachy, hostile and territorial, much like surf beach gangs.

Richard, the great Daimyo Furuta Oribe lived in the late 16th century and early17th c. he did break some things to make whole with kinsugi. Living in a humble house, even a hovel, patched with mud and leaving fallen leaves scattered carefully across the stone path is what is called ” wabi Zumai.” This house is simply a temporary dwelling. This is the way of the samurai and the way of tea. The fukusa, a rag used to wipe the tea utensils and bowl is often made of a patch work of brocade much like like the Buddhist monks robe. The master teacher instructs the student in the oral tradition,” We have no enemies except self. Defeat self, hit clearly! The four diseases are fear, doubt, apprehension and confusion. In traditional kendo, a live sword match is called shinken shobu, it is absolute commitment, life or death, Trad climbing, BASE jumping, Big wave surfing etc. are beyond sport.

If you’ve got, as it seems,an interest in BASE, check out the film, Sunshine Superman, coming out in the fall of 2014. This is the first feature film covering Trad BASE jumping history, from round chutes to wing suits.

Richard, again, thank you for your eclectic taste and obvious artistry.

respectfully, Stephen S. with regards to one of my favorite intellects and climbers, Paul Landrum

]]>Dear Richard,

A person named George pointed out typing errors in mathematical formulas. They do not make the proof wrong. It is, just because while I write in papers I do not make errors. But the similarity in typing made me get mistakes. I am sorry, and here is the corrected one.

1. There is another explanation of a simple proof of Fermat’s last theorem as follows:

X^p + Y^p ?= Z^p (X,Y,Z are integers, p: any prime >2) (1)

2. Let‘s divide (1) by (Z-X)^p, we shall get:

(X/(Z-X))^p +(Y/(Z-X))^p ?= (Z/(Z-X))^p (2)

3. That means we shall have:

X’^p + Y’^p ?= Z’^p and Z’ = X’+1 , with X’ =X/(Z-X), Y’ = Y/(Z-X), Z’ =Z/(Z-X) (3)

4. From (3), we shall have these equivalent forms (4) and (5):

Y’^p ?= pX’^(p-1) + …+pX’ +1 (4)

Y’^p ?= p(-Z’)^(p-1) + …+p(-Z’) +1 (5)

5. Similarly, let’s divide (1) by (Z-Y)^p, we shall get:

(X/(Z-Y))^p +(Y/(Z-Y))^p ?= (Z/(Z-Y))^p (6)

That means we shall have these equivalent forms (7), (8) and (9):

X”^p + Y”^p ?= Z”^p and Z” = Y”+1 , with X” =X/(Z-Y), Y” =Y/(Z-Y), Z” =Z/(Z-Y) (7)

From (7), we shall have:

X”^p ?= pY’’^(p-1) + …+pY’’ +1 (8)

X”^p ?= p(-Z”)^(p-1) + …+p(-Z”) +1 (9)

Since p is a prime that is greater than 2, p is an odd number. Then, in (4), for any X’ we should have only one Y’ (that corresponds with X’) as a solution of (1), (3), (4), (5), if X’ could generate any solution of Fermat’s last theorem in (4).

By the equivalence between X’^p + Y’^p ?= Z’^p (3) and X”^p + Y”^p ?= Z”^p (7), we can deduce a result, that for any X” in (8), we should have only one Y” (that corresponds with X” ) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.

X” cannot generate any solution of Fermat’s last theorem, because we have illogical mathematical deductions, for examples, as follows:

i) In (8), if an X”1 could generate any solution of Fermat’s last theorem, there had to be at least two values Y”1 and Y”2 or at most (p-1) values Y”1, Y”2,…, Y”(p-1), that were solutions generated by X”, of Fermat’s last theorem. (Please note the even number (p-1) of pY”^(p-1) in (8)). But we already have a condition stated above, that for any X” we should have only one Y” (that corresponds with X”) as a solution of (1),(7),(8),(9), if X” could generate any solution Fermat’s last theorem. Fermat’s last theorem is simply proved!

ii) With X”^p + Y”^p ?= Z”^p, if an X”1 could generate any solution of Fermat’s last theorem, there had to be correspondingly one Y” and one Z” that were solutions generated by X”, of Fermat’s last theorem. But let’s look at (8) and (9), we must have Y” = -Z”. This is impossible by further logical reasoning such as, for example: We should have : X”^p + Y”^p ?= Z”^p , then X”^p ?= 2Z”^p or (X”/Z”)^p ?= 2. The equal sign, in (X”/Z”)^p ?= 2, is impossible. Fermat’s last theorem is simply again proved, with the connection to the concept of (X”/Z”)^p ?= 2. Is it interesting?

]]>A simple proof of Beal’s conjecture:

Now let’s start with A^x +B^y ?= C^z (1)

A,B,C are coprimes and x,y,z are natural numbers >2

Then we have: A^x + B^x + B^y –B^x –C^z ?=0 (2)

(A+B-B)^x +B^x + B^y-B^x –C^z ?=0 (3)

A^x + (B+A-A)^x +By-B^x – C^z ?=0 (4)

a) x is an odd number:

Then we get:

B^y-B^x –C^z + (A+B)^x -x(A+B)^(x-1).B+…+ x(A+B).B^(x-1) ?=0 (5)

B^y-B^x –C^z + (A+B)^x -x(A+B)^(x-1).A+…+ x(A+B).A^(x-1) ?=0 (6))

Let’s take any set of three coprime actual numbers A1, B1, C1 and partially replace A, B, C with A1, B1,C1 as follows:

B1^y-B1^x –C1^z + (A1+B1)^x -x(A1+B1)^(x-1).B+…+ x(A1+B1).B^(x-1) ?=0 (7)

B1^y-B1^x –C1^z + (A1+B1)^x -x(A1+B1)^(x-1).A+…+ x(A1+B1).A^(x-1) ?=0 (8)

Please note the common mathematical structure among (7) and (8). Both equations (7) and (8) are from an equation as follows:

B1^y-B1^x –C1^z + (A1+B1)^x -x(A1+B1)^(x-1).u+…+ x(A1+B1).u^(x-1) ?=0 (9)

Now we can easily see that the equation (9) has not any integer solution, because if there had been a solution U1 in (9), we had to get at least a set of three number U1 = A1 = B1 to replace U,A,B in (7),(8), (9) by U1, A1, B1. The replacement to get any solution with the condition U1 = A1 = B1 is impossible, with A1 and B1 are coprimes. Please note that if A=1, we then will get at least a triplet of U=A=B ?= 1, and the conjecture is also proved with 1+1 ?= C^z.

In other words, When A1 and B1 are coprime, either A1 > B1 or A1 < B1, then we can never get U1 = A1 = B1 to replace U, A,B in (7),(8),(9) to get any solution.

b) x is an even number: We can use a similar approach as the above approach, to prove the case x is even.

Fermat's last theorem can be proved by this method and some other simple proofs.

Pham Duc Sinh thaotrangtvt3@gmail.com

]]>A simple proof of Beal’s conjecture:

Now let’s start with A^x +B^y ?= C^z (1)

A,B,C are coprimes and x,y,z are natural numbers >2

Then we have: A^x + B^x + B^y –B^x –C^z ?=0 (2)

(A+B-B)^x +B^x + B^y-B^x –C^z ?=0 (3)

A^x + (B+A-A)^x +By-B^x – C^z ?=0 (4)

a) x is an odd number:

Then we get:

B^y-B^x –C^z + (A+B)^x -x(A+B)^(x-1).B+…+ x(A+B).B^(x-1) ?=0 (5)

B^y-B^x –C^z + (A+B)^x -x(A+B)^(x-1).A+…+ x(A+B).A^(x-1) ?=0 (6))

Let’s take any set of three coprime actual numbers A1, B1, C1 and partially replace A, B, C with A1, B1,C1 as follows:

B1^y-B1^x –C1^z + (A1+B1)^x -x(A1+B1)^(x-1).B+…+ x(A1+B1).B^(x-1) ?=0 (7)

B1^y-B1^x –C1^z + (A1+B1)^x -x(A1+B1)^(x-1).A+…+ x(A1+B1).A^(x-1) ?=0 (8)

Please note the common mathematical structure among (7) and (8). Both equations (7) and (8) are from an equation as follows:

B1^y-B1^x –C1^z + (A+B)^x -x(A+B)^(x-1).u+…+ x(A+B).u^(x-1) ?=0 (9)

Now we can easily see that the equation (9) has not any integer solution, because if there had been a solution U1 in (9), we had to get at least a set of three number U1 = A1 = B1 to replace U,A,B in (7),(8), (9) by U1, A1, B1. The replacement to get any solution with the condition U1 = A1 = B1 is impossible, with A1 and B1 are coprimes. Please note that if A=1, we then will get at least a triplet of U=A=B ?= 1, and the conjecture is also proved with 1+1 ?= C^z.

In other words, When A1 and B1 are coprime, either A1 > B1 or A1 < B1, then we can never get U1 = A1 = B1 to replace U, A,B in (7),(8),(9) to get any solution.

b) x is an even number: We can use a similar approach as the above approach, to prove the case x is even.

Fermat's last theorem can be proved by this method and some other simple proofs.

Pham Duc Sinh thaotrangtvt3@gmail.com

]]>Now let’s start with A^x +B^y ?= C^z (1)

A,B,C are coprimes and x,y,z are natural numbers >2

Then we have: A^x + B^x + B^y –B^x –C^z ?=0 (2)

(A+B-B)^x +B^x + B^y-B^x –C^z ?=0 (3)

A^x + (B+A-A)^x +By-B^x – C^z ?=0 (4)

a) x is an odd number:

Then we get:

B^y-B^x –C^z + (A+B)^x -x(A+B)^(x-1).B+…+ x(A+B).B^(x-1) ?=0 (5)

B^y-B^x –C^z + (A+B)^x -x(A+B)^(x-1).A+…+ x(A+B).A^(x-1) ?=0 (6))

Let’s take any set of three coprime actual numbers A1, B1, C1 and partially replace A, B, C with A1, B1,C1 as follows:

B1^y-B1^x –C1^z + (A1+B1)^x -x(A1+B1)^(x-1).B+…+ x(A1+B1).B^(x-1) ?=0 (7)

B1^y-B1^x –C1^z + (A+B)^x -x(A+B)^(x-1).A+…+ x(A+B).A^(x-1) ?=0 (8)

Please note the common mathematical structure among (7) and (8). Both equations (7) and (8) are from an equation as follows:

B1^y-B1^x –C1^z + (A+B)^x -x(A+B)^(x-1).u+…+ x(A+B).u^(x-1) ?=0 (9)

Now we can easily see that the equation (9) has not any integer solution, because if there had been a solution U1 in (9), we had to get at least a set of three number U1 = A1 = B1 to replace U,A,B in (7),(8), (9) by U1, A1, B1. The replacement to get any solution with the condition U1 = A1 = B1 is impossible, with A1 and B1 are coprimes. Please note that if A=1, we then will get at least a triplet of U=A=B ?= 1, and the conjecture is also proved with 1+1 ?= C^z.

In other words, When A1 and B1 are coprime, either A1 > B1 or A1 < B1, then we can never get U1 = A1 = B1 to replace U, A,B in (7),(8),(9) to get any solution.

b) x is an even number: We can use a similar approach as the above approach, to prove the case x is even.

Fermat's last theorem can be proved by this method and some other simple proofs.

Pham Duc Sinh thaotrangtvt3@gmail.com

]]>