"tweaking" the Fermat last theorem equation
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- Опубліковано 16 жов 2024
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Wait wait wait, you're just gonna say that there are no solutions for n ≥ 3 without a proof? What, is the proof too wide to fit on the board?
The proof is too complicated . look at Wikipedia to get an idea
en.wikipedia.org/wiki/Fermat%27s_Last_Theorem
@@Ahmed-Youcef1959it can be hard to tell in a comment, but I'm 90% sure they were making a joke
😂
Don't be alarmed, he does have a proof, but the file size of *that* video is too large for UA-cam to handle.
@@thenateman27 Guess you can push your estimate to 99.9999… 😂
2^n + 2^n = 2^(n+1), gg
Usually x, y are intended to be different integers. But I like this one anyways.
@@versacebroccoli7238 no they aren't. This is a common misconception among non-mathematicians and one of the first things you learn when starting to study mathematics at a university level. If it is not specified that x and y are different, then they might be the same.
Oh hell na jigsaw
im tweakin brah
I like Fermat and all his consequences
10:25
Thanks for that.
Lol
With luck and more power to you.
hoping for more videos.
Since in the example 81 is a perfect square of a perfect square, with a=1 and b=2 :
x^3 + y^3 =z^5 (z = 81)
x^3 + y^3 =z^2 (z = 9^5)
x^3 + y^3 =z^4 (z = 3^5)
But with a=2 and b=3:
x^3 + y^3 =z^5 (z = 1225 = 35^2)
x^3 + y^3 =z^2 (z = 35^5)
Probabely there is still a lot to do with this equality ...
Great video
There is a lot of beauty in this and the presentation of it.
Sublime Michael, perfectly sublime 🌟
OR: Never underestimate the Fundamental Theorem of Arithmetic
Which, come to think of it, must mean there are some interesting modular arithmetic insights no?
EDIT: hmmm .... @ 5:29 nv - mu = 1 is uncannily like the determinant of a 2 by 2 matrix with values n, m, u, v arranged in the most suitable order.
I could swear I read something saying that x^l+y^m=z^n was a Clay Millennium Problem but I can't find it now.
Beal's conjecture, slightly different
Why won't you show the cases where l and m aren't equal or where gcd (l,m,n)=2 ?
Gcd (l,m,n)=2 has infinite solutions and those are the pythagorean triples
What about Fermat's Lost Theorem?
😏
I did an Excel sheet and there are several problems with the formula I noticed
1/ (x^m+y^m=z^n) for m=3 and n=5, The values a=1 and b=4 gives x=274625, y=1098500 and z=4225, and it doesn't work
2/ for m=3 and n=5, you said we need to the find values u and v such as nv-mu=1. You found the smallest set u=3 and v=2. But other sets that respect this equality such as u=8 and v=5 gives values for x, y and z that don't work for any values a and b I picked
3/ I thought just the smallest values u and v made it work; But for m=5 and n=7, the set u=4 and v=3 doesn't give correct values for x,y and z for any values a and b I picked
I believe you're running into the problem that Excel isn't computing exact integer answers -- at some point, it's switching over to floating point. After all, z^5 = 4225^5 is approx 1.34E+18. That's going to be a nineteen digit number.
@@WRSomsky Thanks. It works on Wolfram Alpha
I knew some researchers who would not touch Excel for anything serious. Basis:
Excel algorithms are not made public and so compromises independent peer group reviews. Essential for any serious researcher.
and
Add 1/3 * 10^19 to 1/3 * 10^(-19) then do some complicated stuff that messes decimal point accuracy.
Computers as a rule do not do real real numbers and are always are reduced to a limited subset of quotients and powers
L. Auby?
Michael, it's rare I'm captivated by a problem. Haven't watched the video yet but the thumbnail alone led me down some super fun garden paths. Now I will see what you intend.
0:23 that is not entirely correct because a=b=c=0 is a solution. The correct statement should be "There are no positive integer solutions"
remember that anglos think that N*=N, despite R*=/=R, Q*=/=Q and C*=/=C
@@Crusademashup-zz6ls well that's because subtraction is defined in all of these but not in N
@@UA-camBS So what ? He is not talking about binary operations, only about the definition of a natural number.
@@Crusademashup-zz6ls They also confuse monogenous groups and cyclic groups
just say non-trivial sols don't exist.