What this links to is units in the ring Z[rootk]. Every unit has norm 1 and thus is a solution to the equation. Clearly units form a multiplicative group.
But now the most important part: how do you find that "first solution" - and more importantly, enure that it is the first (i.e. smallest)? Guess and check sounds very involved for equations with non-trivial coefficients, e.g. x²-73y² = 1217
That notebook was the least asian thing in the video.I can confirm that every competitive exam aspirant prefers longbooks(400pgs+) for exam prep because it is easier to maintain and search old things.
Squaring 9+4*5^(1/2) gives the pair {161, 72}. (I was waiting for you to give that pair, but I had to figure it out myself.) You can also use this method to give rational approximations of the square root of 5: 9/4, then 160/72=20/9...
so what? this is quite obvious since x^2=(-x)^2 for all reals. we are looking for the solution where x and y have the smallest absolute values they could have and we take the positive one for simplicity's sake. we say that the first solution (x1,y1) is the one for which |x1|
@ yeah I’m just saying these are solutions too that weren’t mentioned since it was asked if these are all the solutions in the video, wasn’t a major point.
If "f" is a differentiable function on the interval [0, 1], with the following boundary conditions: f(0) = 0 f(1) = 1 Then, find the minimum value of the integral: ∫ from 0 to 1 of (f'(x))² dx.
![Seungmin "그렇게, 천천히, 우리(As we are)" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/kAzmhLHePqU/mqdefault.jpg)
What are the tens and the units digits of 7^7^7?
Solution ruclips.net/video/_OAFyKZWwy8/видео.htmlsi=npITrvCxthrMc_QN
What this links to is units in the ring Z[rootk]. Every unit has norm 1 and thus is a solution to the equation. Clearly units form a multiplicative group.
But now the most important part: how do you find that "first solution" - and more importantly, enure that it is the first (i.e. smallest)? Guess and check sounds very involved for equations with non-trivial coefficients, e.g. x²-73y² = 1217
Exactly, how do you even know a solution exists for all nonsquare k?
You can use continued fractions to find the smallest solution and then build the others from that
@@SyberMathi don't it is so obvious how to use continued fractioj for that
Great! Please do make that video showing there are no other solutions!
That notebook was the least asian thing in the video.I can confirm that every competitive exam aspirant prefers longbooks(400pgs+) for exam prep because it is easier to maintain and search old things.
Squaring 9+4*5^(1/2) gives the pair {161, 72}. (I was waiting for you to give that pair, but I had to figure it out myself.)
You can also use this method to give rational approximations of the square root of 5: 9/4, then 160/72=20/9...
Log Ladders, Lemmermeyer's Product
Nice to meet you ❤ Tomorrow morning I have mathematics exam wish me success
Same here
good luck!
Good luck! Calc exam?
@ of course bro ,thank you
good luck,,
Wouldn’t negating x values also give solutions so we also have all the reflected values
so what? this is quite obvious since x^2=(-x)^2 for all reals. we are looking for the solution where x and y have the smallest absolute values they could have and we take the positive one for simplicity's sake. we say that the first solution (x1,y1) is the one for which |x1|
@ yeah I’m just saying these are solutions too that weren’t mentioned since it was asked if these are all the solutions in the video, wasn’t a major point.
Happy new year y'all!! 🎇🎇🎇🎆🎆🎆
I want to see the proof! Also, can this be extended to when the right hand side is composite?
Very well explained 😊
Amazing video! Happy new year to everyone. 🎉🎆
OK, now how about doing this for x^2 - 61y^2 = 1 ?
Nice proof!
Very interesting!
And for that matter, is there a simple way of finding all integer solutions for elliptic curves of the form y^2 = x^3 - x + r^2 ?
1:32 later is when? give me timecode so i can watch and understand every step not just seing something and forgeting about it
I may not understand the math completely but the video is amazing to watch
The pell equation x^2-2y^2=+-1 gives approximation of square root 2
Secant ,tangent/k are all of the solutions.
張耕宇學長好！
你好⊙ω⊙😊
The thumbnail is so cool
In my classes, we solve this by having the solution be a continued fraction and just write answer in continued fraction format
I dont understand anything, I just watch these to fall asleep
The Chinese are very smart thank you
Sorry, I really need comments for this gentleman.
🎉🎉🎉
Can you find a room with more echo? 😂
im in 9th grade, I cant even do basic calculus
it's not important for you, don't do it
"Do not worry about your difficulties in Mathematics; I can assure you mine are still greater"
- Einsteine
@@prabhakarsingh6821 thank you i will use this quote later every there and where
존나잘생겼네
Zamn
peak
ruclips.net/video/uqwC41RDPyg/видео.html
If "f" is a differentiable function on the interval [0, 1], with the following boundary conditions:
f(0) = 0
f(1) = 1
Then, find the minimum value of the integral:
∫ from 0 to 1 of (f'(x))² dx.
1
I will find god
X=i^4 is solution of sqrt(x)=-1
its not
Sqrt(i^4)=i^2=-1
@nouarislimani sqrt(i^4) = sqrt(1) = 1, youre thing is wrong cuz whenever you do sqrt(x^2) its always equal to |x|
In mathematics we have a definition- √(x)² = |x|. Here it's √i⁴ = √(i²)² = √(-1)² = |-1| = 1. So this isn't possible.
Also with that logic in mind you can say that (-1)² is a solution too.
He has good knowledge of mathematics
I had never learned till 10
im too fast
wow, that's so cool
Happy New Year🎉❤ 🎂🎊💐👑💵💵💵💵💵💵🧠🧠🧠🧠🧠🧠📐(^o^) to MM is MATHEMATICS MASTER