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.
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.
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
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.
i have a question that comes in an olypiad in my country and i weren't be able to solve it a,b ,c are positive intgers an hey are not eqaul to 0 prove that if (a/b)+(b/c)+(c/a) is an integer that mean that a*b*c is a cube
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.
中文版: th-cam.com/video/e8IOmQnnlNk/w-d-xo.html
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.
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,,
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.
Great! Please do make that video showing there are no other solutions!
I may not understand the math completely but the video is amazing to watch
Amazing video! Happy new year to everyone. 🎉🎆
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
In my classes, we solve this by having the solution be a continued fraction and just write answer in continued fraction format
Can you explain how we can use continued fractions @Nikkikkikkiz? Kinda confused on how they can be used here.
The conjugate come from the fact that its the other root to the minimal polynomials, which are if degree 2 in this case.
I dont understand anything, I just watch these to fall asleep
Log Ladders, Lemmermeyer's Product
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...
OK, now how about doing this for x^2 - 61y^2 = 1 ?
Sorry, I really need comments for this gentleman.
The Chinese are very smart thank you
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.
K = 61 の場合を解決しました。
K = 61 no baai o kaiketsu shimashita.
The pell equation x^2-2y^2=+-1 gives approximation of square root 2
The thumbnail is so cool
Very interesting!
I want to see the proof! Also, can this be extended to when the right hand side is composite?
Nice proof!
Happy new year y'all!! 🎇🎇🎇🎆🎆🎆
Is undefined > ∞?
Can you find a room with more echo? 😂
So how do you find the "first" solution (9,4)? What technique/method is used?
What are the tens and the units digits of 7^7^7?
Solution th-cam.com/video/_OAFyKZWwy8/w-d-xo.htmlsi=npITrvCxthrMc_QN
I solved the case where K= 61!
1:32 later is when? give me timecode so i can watch and understand every step not just seing something and forgeting about it
❤❤
🎉🎉🎉
張耕宇學長好！
你好⊙ω⊙😊
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 ?
Secant ,tangent/k are all of the solutions.
Can we solve this using graph?
It is a hyperbola after all.............
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
Resolvi o caso em que K = 61 ?
Hey can you please solve this inequality for me?
sqrt(8 + 2x - x ^ 2) > 6 - 3x
i have a question that comes in an olypiad in my country and i weren't be able to solve it
a,b ,c are positive intgers an hey are not eqaul to 0 prove that if (a/b)+(b/c)+(c/a) is an integer that mean that a*b*c is a cube
100 differential equations...... Pls tq
peak
He has good knowledge of mathematics
I had never learned till 10
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
Zamn
Very well explained 😊
I will find god
th-cam.com/video/uqwC41RDPyg/w-d-xo.html
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.
im too fast
존나잘생겼네
wow, that's so cool
Happy New Year🎉❤ 🎂🎊💐👑💵💵💵💵💵💵🧠🧠🧠🧠🧠🧠📐(^o^) to MM is MATHEMATICS MASTER