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.
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...
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
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: ua-cam.com/video/e8IOmQnnlNk/v-deo.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,,
Amazing video! Happy new year to everyone. 🎉🎆
Great! Please do make that video showing there are no other solutions!
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...
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.
I may not understand the math completely but the video is amazing to watch
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
The conjugate come from the fact that its the other root to the minimal polynomials, which are if degree 2 in this case.
Very interesting!
OK, now how about doing this for x^2 - 61y^2 = 1 ?
Nice proof!
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.
I dont understand anything, I just watch these to fall asleep
What are the tens and the units digits of 7^7^7?
Solution ua-cam.com/video/_OAFyKZWwy8/v-deo.htmlsi=npITrvCxthrMc_QN
Log Ladders, Lemmermeyer's Product
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 pell equation x^2-2y^2=+-1 gives approximation of square root 2
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 ?
❤❤
I want to see the proof! Also, can this be extended to when the right hand side is composite?
Is undefined > ∞?
So how do you find the "first" solution (9,4)? What technique/method is used?
Just use our Asian power😂
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
The thumbnail is so cool
Happy new year y'all!! 🎇🎇🎇🎆🎆🎆
The Chinese are very smart thank you
Hey can you please solve this inequality for me?
sqrt(8 + 2x - x ^ 2) > 6 - 3x
🎉🎉🎉
Sorry, I really need comments for this gentleman.
Secant ,tangent/k are all of the solutions.
1:32 later is when? give me timecode so i can watch and understand every step not just seing something and forgeting about it
K = 61 の場合を解決しました。
K = 61 no baai o kaiketsu shimashita.
Can we solve this using graph?
It is a hyperbola after all.............
I solved the case where K= 61!
Wow
張耕宇學長好！
你好⊙ω⊙😊
Can you find a room with more echo? 😂
Resolvi o caso em que K = 61 ?
Please solve this
I didn't find the solution anywhere.
Integrate (x^x)(x^(x^x))(1/x+ln(x)+(ln(x))^2)
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
100 differential equations...... Pls tq
peak
Very well explained 😊
Zamn
ua-cam.com/video/uqwC41RDPyg/v-deo.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.
wow, that's so cool
I will find god
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
im too fast
존나잘생겼네
Happy New Year🎉❤ 🎂🎊💐👑💵💵💵💵💵💵🧠🧠🧠🧠🧠🧠📐(^o^) to MM is MATHEMATICS MASTER