Posts 29, 30, 33 and 35 turn out to be looking at a cube, not the triangle. While an interesting exercise, it turns out that in trying to see what made it tick, I lost track of what the “it” was.
Posts 29, 30, 33 and 35 turn out to be looking at a cube, not the triangle. While an interesting exercise, it turns out that in trying to see what made it tick, I lost track of what the “it” was.
Might you summarize anything useful you’ve discovered in your posts above?
Yes, it might help if you were able to collect it into a unified, step-by-step proof. Being scattered throughout so many posts makes it very hard to follow you know…
<delete>
Ok. Went looking at some proofs on Pythagorean theorem last night. I ran across this site which inspired this illustration.
This should show the pattern I am seeing. Putting it into the language of mathematica is not exactly my strong suit. It uses 43, 44, 45 developed along the line illustrated in the pdf’s contained in post 33.
FermatsLC-Final_Sheet_5.pdf (25.8 KB)FermatsLC-Final_Sheet_6.pdf (50.3 KB)FermatsLC-Final_Sheet_7.pdf (132 KB)
I’m not sure what value this has, but it appears these three can be restated something like:
X3 , X6, X9, . . . yeilds an equalateral triangle the sides being X(n/3). –>Sheet.5
X4 , X7, X10, . . . yeilds three sides, two of which are X and the longer one being nX. –>Sheet.6
X5 , X8, X11, . . . yeilds an isocoles triangle the sides being (n-1)X with the shorter leg resolving to X. –>Sheet.7
I’m finding juggling the exponents a bit confusing to manage.
dream_weaver, are you going to write a summary of your ideas so that others might understand your thoughts, or are you simply having a conversation with yourself?
I think I just posted a summary in post 46.
I think I just posted a summary in post 47.
I’m the one who wrote post 47, and I don’t see any summary in your recent posts.
Sorry, it should have read post 46.
I think I have the right formulas now.
X3 , X6, X9, . . . yeilds an equalateral triangle the sides being X(n/3). –>Sheet.5
X4 , X7, X10, . . . yeilds three sides, two of which are X((n-1)/3) and the longer one being 4X((n-1)/3). –>Sheet.6
X5 , X8, X11, . . . yeilds an isocoles triangle the sides being 4X((n-2)/3) with the shorter leg resolving to X((n-2)/3). –>Sheet.7
In these cases, X would be an integer, n would be the exponent >3
(a2-b2), (2ab), (a2+b2) is somehow contained in X in all three equations, much like it is here.
In the first equation, it would be getting X(n/3)and get the (a2+b2) deal out of it.
In the second equation, it would be getting X((n-1)/3) out of it.
And the third would come from X((n-2)/3).
The other sheets uploaded regarding this are in post 33 and post 45
Separate the cube as indicated and end up with
1 - b3, 3 - b2 x (a- B) , 3 - b x (a- B) 2, and 1 (a- B) 3
These don’t appear to be able to be reassembled into two cubes of their own.
Those dang-nab emoticons.
The second group breaks down into the following volumes:
1 - b3
3 - b2 x (a- B)
3 - (b x (a- B) 2
1 - (a- B) 3
1 - b2 x 3a
2 - b x (a- B) x 3a
1 - (a- B) 2 x 3a
And the final one yeilds these sub-volumes.
1 - b3
3 - b2 x (a- B)
3 - b x (a- B) 2
1 - (a- B) 3
2 - b2 x 3a
4 - b x (a- B) x 3a
2 - (a- B) 2 x 3a
1 - 3a2 x b
1 - 3a2 x (a- B)
Maybe Fermat was not trying to solve for An+Bn=Cn, but it came as a by-product of applying the Euclidean method to cube space.to derive the a, b, (a- B) relationship and noticed the resulting blocks could not be re-arranged into 2 separate cubes. Considering he is one of the fore-runners to Galileo, Leibniz and Newton, this comes to mind viewing how this can be disassembled so far.
Sorry, it should have read post 46.
I don’t consider post 46 to be a summary of your ideas. There’s no way to know, from that post alone, what you’re talking about.
This is becoming way more complex that I thought it was at first.
The block illustrations are like a three dimensional Euclidean puzzle. The quantity of blocks generated by X blocks to the nth power are assembled into the hexahedrens.
The a, b, a-b relationships are used to subdivide the hexahedrons into the list of blocks described in post 52 and post 53. Does that help any?
The size of the hexahedron would be explained by instead of triangles, a hexahedron X(n/3) x X(n/3) x X(n/3).,
one of X((n-1)/3) x X((n-1)/3) x 4X((n-1)/3)
and finally one of 4X((n-2)/3) x 4X((n-2)/3) x X((n-2)/3).
where X is the number of cubes began with.used in conjuntion with an nth power.
This is becoming way more complex that I thought it was at first.
The block illustrations are like a three dimensional Euclidean puzzle. The quantity of blocks generated by X blocks to the nth power are assembled into the hexahedrens.
The a, b, a-b relationships are used to subdivide the hexahedrons into the list of blocks described in post 52 and post 53. Does that help any?
No, it does not! I think you might get a few people interested in your ideas if you were to write a summary that explained things from scratch. If you don’t do that I expect that you will continue to have a conversation all by yourself. If that’s what you want, then go ahead. But if you want to interest anyone else in your ideas then please write a summary that starts from scratch and assumes the reader knows nothing about your subject. Got it?
Thank-you.
I wish I could summarize it better.
Ok. I’ve modified the last three sheets to discover that the block count in posts 52 and 53 were incorrect. Hopefully this will help tie some more of it together. It’s hard to descibe 3 dimensional object in word, much alone discovering relational formulas between the various aspects.
FermatsLC-Final_Sheet_5.pdf (33.4 KB)FermatsLC-Final_Sheet_6.pdf (72.6 KB)FermatsLC-Final_Sheet_7.pdf (190 KB)
Please keep in mind, I’ve never really attempted anything like this before.
Oh, the block count was correct. I had not originated the diagram with that in mind at the time.