@@SyberMath but what's in the pare theses doesn't have have be 9oth power jjst because y is a 99th power..so whybdid you saybthat sorry? Just because they are equal does not mean they are the same power..
My sincerest congratualtions on 100,000 subscribers SyberMath! I was waiting for this day I knew it would come because you are a good hearted and kind and nobody can resist your videos they are so good and you explain to us so well and best of all it is free! Thanks for all your precious time making this videos they must have been a lot! I really wish you were my teacher. Actually you ARE my teacher I study with your videos I am preparing for math olympiad next year and I hope to get a good score... maybe I can thanks to you. Maybe i will be a math professor someday, ha ha! Thanks once more SyberMath and may God bless you richley a long and happy life(and also more likes, views, and subscribers) because you deserve it. Best wishes, and also love and praises from YoonHo from South Korea! P. S. This is my second time writing the comment dissapeared all of a sudden and I wrote it the best i can because to make you happy on this nice day:)
HAIKU congratulations on 100K. it warms my ❤ 2 c mathematics so popular on YT. top notch quality (accent + jokes included). may u 1 day win the YT Field's Medal.
Congratulations on 100K! But this equation is very simple bro! Divide both parts by x⁹⁹ and set y/x=z - which should be an integer, since you get 1+z⁹⁹ = x, and from here y = z*x = z*(z⁹⁹+1). Where z is any integer. That is et!!
ez. x⁹⁹(x - 1) = y⁹⁹; if x = 0, 0 = y⁹⁹; y =0. So first solution is x, y = 0. Else: x - 1 = (y/x)⁹⁹. y = nx where n integer. n⁹⁹ = x - 1; n⁹⁹ + 1 = x; n¹⁰⁰ + n = nx = y There we can get any natural n, and get pair of x (n⁹⁹ + 1) and y (nx) for it. So, this equation has ♾️ solution, but we can calculate just x, y = 0; x = 1 and y = 0 (for n = 0); x, y = 2 (for n = 1)
I did it from the other side. substitute y = kx and get x ^ 99 + (kx) ^ 99 = x ^ 100 x ^ 99 + (kx) ^ 99-x ^ 100 = 0 x ^ 99 * (1+ (k) ^ 99-x) = 0 lure x ^ 99 = 0 or 1 + (k) ^ 99-x = 0 if x ^ 99 = 0 then x = 0 and y = 0 from the second equation we have x = 1+ (k) ^ 99 y = kx = k * (1+ (k) ^ 99) = k + k ^ 100 for k = -1 we have x = 0, y = 0, so the above formula gives all real solutions. for integer solutions it suffices to note if 1 is an integer and x is an integer, so k ^ 88 is an integer and a hence k is an integer. So for any integer k we have then x = 1 + (k) ^ 99 y = kx = k * (1+ (k) ^ 99) = k + k ^ 100
We can factor the LHS (x+y)(...a lot of x and y ...) = x^100 As (x+y) | x^100 So y = 0 or x+y = some power of x. If x+y = x^2 we get the (2,2) solution. I should be able to get the general solution from here but it seems to be a dead end 😔
Thank you very much! 🥰💖 ⭐ Suggest → forms.gle/A5bGhTyZqYw937W58 If you need to post a picture of your solution or idea: twitter.com/intent/tweet?text=@SyberMath
x-1 = k^99 ; k integer; why ?? First a= x-1 is an integer... Look at a> 0 (a< 0 and (-1)^99 = -1 and then x0 Take the 99.th root.(x ,y integers ) then y = x * k..... k = 99√a. then k is an integer: why?? you can show: n natural number , n >1, a natural number then n√a is natural number or irrational.... why?? Look at n√a = p/q. or. a = p^n / q^n p, q natural and greatest common divisor (p,q) = 1 ....and show q =1.
Since it's he's searching integer solution, if one side is 99th power of an integer. Other side has to be 99th power of an integer too. And since x99 * (x-1) has to be 99th power of an integer, x-1 has to be 99th power of some integer too. Or x will be irrational
K is more of a general solution, since there are going to be infinite solutions to this eq, so he uses k^99 in order to make a solution pair of x and y in terms of k so it means [all values of x and y that follow it]. In the first place, since y^99 = x^99 * (x-1), and you want this to be rational, only possibility is that x-1 is some power of 99, so x - 1 = k^99, and x = k^99 + 1.
@@synaestheziac That only applies for y = [polynomial]. In this case this is not really a polynomial since its y^99. Moreover, you technically have infinite solutions to diophantine equations like ax^2-by^2 = 1 [instead of 2]. Thing is that they're finding integer solutions of the graph itself and not when the graph intersects another at a certain point. First one is bound to have infinite solutions [if it happens to]
let's try: x^99 + y^99 = x^100 (*) could be rewritten as: (x^99)(x-1) = y^99 ; let's have k an integer and suppose y=kx. then: (x^99)(x-1) = (k^99)(x^99) let's suppose X# 0 . therefore we can simplify the equation to get: (x-1) = k^99 so I think we can infer that x is an integer (obviously). next we can say that (from (*)) that y^99 is an integer as well. now let's determine whether y is an integer: as we had y=kx, we can write x=y/k . substituting in (*) we get: (y^99)/(k^99) + (y^99) = (y^100)/(k^100) . now multiplying each member by (k^100) and simplifying, wet get: k(y^99) + (k^100)(y^99) = y^100 and after factoring and reduction, we obtain: y = k(1+k^99) which is obviously an integer as well.
I think you over complicated it. Once you conclude that k and x are integers, you can immediately conclude that y = k·x = k·(k^99 + 1) is an integer without having to substitute into the original equation.
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
Congratulations, Syber!! 🎉
Thank you!!! 💖🥰💖
First! Congratulations on 100k!! :) I started to watch your videos last year and they are amazing, you can learn lots of thing. Keep us good work!
Thank you!!! 💖🥰
@@SyberMath but what's in the pare theses doesn't have have be 9oth power jjst because y is a 99th power..so whybdid you saybthat sorry? Just because they are equal does not mean they are the same power..
Congratulations, Syber! Deserved!
Thank you!!! 💖🥰
100k congrats sir!!
hope you hit 500k by the end of 2022
you mean 2022?
@@Jha-s-kitchen ohh yep
Thank you!!! 💖🥰
@@pranjalsrivastava3343 i hope he reaches ∞ or ℵ (aleph null)
My sincerest congratualtions on 100,000 subscribers SyberMath! I was waiting for this day I knew it would come because you are a good hearted and kind and nobody can resist your videos they are so good and you explain to us so well and best of all it is free! Thanks for all your precious time making this videos they must have been a lot! I really wish you were my teacher. Actually you ARE my teacher I study with your videos I am preparing for math olympiad next year and I hope to get a good score... maybe I can thanks to you. Maybe i will be a math professor someday, ha ha! Thanks once more SyberMath and may God bless you richley a long and happy life(and also more likes, views, and subscribers) because you deserve it. Best wishes, and also love and praises from YoonHo from South Korea!
P. S. This is my second time writing the comment dissapeared all of a sudden and I wrote it the best i can because to make you happy on this nice day:)
my typing is slow i typed in about 20 minutes
Thank you very much for the kind words!!! 💖🥰💖
I wish you the very best of everything!
@@SyberMathno problem and thanks!
@@SyberMath pin me plz😍😍
HAIKU
congratulations on 100K.
it warms my ❤ 2 c mathematics so popular on YT.
top notch quality (accent + jokes included).
may u 1 day win the YT Field's Medal.
Wow, thank you! 🤩🥰
100k followers well deserved. Problems pitched at a good level, frequent updates and of course a clear solution (or two, or three..).
Thank you! 🥰
Congrats on 100k, Syber - ur the best!
Thanks a ton! 💖🥰
Great vid! Congrats on 100k
Thank you!!! 💖🥰
@Owls Math Hi sir! Good to see you here!
@@SuperYoonHo Hey YoonHo! We know whats good and watch all the same youtube channels 🤣🤣
Congratulation on 100k Subscribers SyberMath!!!
Thank you so much 😀🥰💖
Congratulations on 100K! But this equation is very simple bro! Divide both parts by x⁹⁹ and set y/x=z - which should be an integer, since you get 1+z⁹⁹ = x, and from here y = z*x = z*(z⁹⁹+1). Where z is any integer. That is et!!
Thank you!!! 💖
If you divide by x , you have to check what happens when x=0
Congratulations!
ez. x⁹⁹(x - 1) = y⁹⁹; if x = 0, 0 = y⁹⁹; y =0. So first solution is x, y = 0. Else: x - 1 = (y/x)⁹⁹. y = nx where n integer. n⁹⁹ = x - 1; n⁹⁹ + 1 = x; n¹⁰⁰ + n = nx = y There we can get any natural n, and get pair of x (n⁹⁹ + 1) and y (nx) for it. So, this equation has ♾️ solution, but we can calculate just x, y = 0; x = 1 and y = 0 (for n = 0); x, y = 2 (for n = 1)
תותח אתה
Congratulations on 100K Subscribers!!!
Thank you!!! 😍🤗
I did it from the other side.
substitute y = kx and get
x ^ 99 + (kx) ^ 99 = x ^ 100
x ^ 99 + (kx) ^ 99-x ^ 100 = 0
x ^ 99 * (1+ (k) ^ 99-x) = 0 lure
x ^ 99 = 0 or 1 + (k) ^ 99-x = 0
if x ^ 99 = 0 then x = 0 and y = 0
from the second equation we have
x = 1+ (k) ^ 99
y = kx = k * (1+ (k) ^ 99) = k + k ^ 100
for k = -1 we have x = 0, y = 0, so the above formula gives all real solutions.
for integer solutions it suffices to note if
1 is an integer and x is an integer, so k ^ 88 is an integer and a hence k is an integer.
So for any integer k we have then
x = 1 + (k) ^ 99
y = kx = k * (1+ (k) ^ 99) = k + k ^ 100
Oh, my congrats with big 10OK!
Thank you very much! 🥰💖
He’s spilling the t!!! Congrats on 100k!!
Thank you! 😁 🥰💖
Well, I'll tell you that I had to learn English to be able to follow you, I'm from the Dominican Republic
Wow! Nice to meet you! Greetings from the United States! 😍💖
Mabrook on 100K Syber.
Shukran!
@@SyberMath you deserve it you been going strong since the beginning.
Congratulations, sir!!!!
Thank you!!! 💖🥰
Yes, this time u, (not the letter u) made the difference! Congrats!
Keep up making these nice videos
Thanks! 😃🥰💖
Congratulations on 100k 🎉👏 following u since 2k sub afair
I appreciate it! 💖🥰
Congratulations on 100k.
Thank you so much 😀🥰
Congratulations 🙏
Thank you!!! 💖🥰
Congrats!!
Thank you very much! 🥰💖
Congrats syber !
Thank you! 💖🥰
congratulations
Thank you!!! 🥰
Congrats sir
Thanks for video
Np. Thank you!!! 💖🥰
@Pranav Mali IN
Hello and good day Pranav bro
@@SuperYoonHo same to u my loving bro have a great sunday
@@pranavamali05 Thanks bro and you too
@@SuperYoonHo hello master bro how are u didn't see u from many days
Congrats for 100k
Thank you!!! 💖🥰
We can factor the LHS
(x+y)(...a lot of x and y ...) = x^100
As (x+y) | x^100
So y = 0 or x+y = some power of x.
If x+y = x^2 we get the (2,2) solution.
I should be able to get the general solution from here but it seems to be a dead end 😔
i think to say that x+y | x^100 means x+y is a power of x is only true if x were a prime number
take x=6, then x+y | 6^100 so we could have 2 etc
Should use Roman numberal C instead of constant k for 100
Congrats on 100k, Syber.
By the way, I have some math problems I want to see you solve on your channel. Is there a way I can send them to you?
:^(
Thank you very much! 🥰💖
⭐ Suggest → forms.gle/A5bGhTyZqYw937W58
If you need to post a picture of your solution or idea:
twitter.com/intent/tweet?text=@SyberMath
There seem to be many people who like to hear you writing ✍️.
I don't get it
축하 !!! congratuations!!!
고맙습니다!!! 💖
@@SyberMath Wow can you speak korean?
우와 SyberMath 대단해요
Nice equation and solving.
P.D.: Yes, you neeeded some cup of tea xD
Thanks! 😁
@@SyberMath aah ok... Sorry the joke of tea
101k subscribers now. Including myself.
YAY!!
🥳🥰
Great .Super 👍👍👍
😊🥰
x-1 = k^99 ; k integer; why ??
First a= x-1 is an integer...
Look at a> 0 (a< 0 and (-1)^99 = -1 and then x0
Take the 99.th root.(x ,y integers )
then y = x * k.....
k = 99√a. then k is an integer: why?? you can show:
n natural number , n >1, a natural number then n√a is natural number or irrational....
why??
Look at n√a = p/q. or. a = p^n / q^n
p, q natural and greatest common divisor (p,q) = 1 ....and show q =1.
Awesome, so the y value is just the x value multiplied by the integer k ^^
Congratz tho, love your Humble and interestimg videos, keep em coming!! (:
Despite that u have a problem with "TEA"🤣, The main question is: WHERE IS THE GRAPH😡
And CONGRATULATIONS ON 100K (do a video regarding the 100k😂)
😁 Thank you very much! 🥰💖
Ok I've got an interesting problem for u maybe it's too easy for u but please try.
a³³+b³³+c³³=3d¹¹ find integer solutions
That looks hard! 😁😜
@@SyberMath hmm yeah
Ans: (1,1,1,1)
Everyone: But can you prove that's the only solution!!!
@@mcwulf25 haha i can't😅
If (a,b,c,d) is a solution then (x *a,x*b,x*c, x^3 *d ) is a solution...
genial 100k
💖
تهانينا اليك
شكرًا لك! 💖
x = y = 2 or 0
x = 1, y = 0
(X,Y):(2,2).
Oops - I found (0,0) and (1,0) and did not see the infinite family at all! Nice problem.
Thank you!!! 🥰
You are amazing person.it's not t but you need tea 🤣😂😂🤣
I do!!! Thanks! 😁😍💖
I don’t understand the K^99, can you give the theory of it(write it down)?so i can search and follow you.
Since it's he's searching integer solution, if one side is 99th power of an integer. Other side has to be 99th power of an integer too.
And since x99 * (x-1) has to be 99th power of an integer, x-1 has to be 99th power of some integer too. Or x will be irrational
K is more of a general solution, since there are going to be infinite solutions to this eq, so he uses k^99 in order to make a solution pair of x and y in terms of k so it means [all values of x and y that follow it]. In the first place, since y^99 = x^99 * (x-1), and you want this to be rational, only possibility is that x-1 is some power of 99, so x - 1 = k^99, and x = k^99 + 1.
y^99 = x^99(x - 1)
x - 1 = (y^99)/(x^99) = (y/x)^99
Let y/x = k (k is an integer)
x - 1 = k^99
x = k^99 + 1
There are infinitely many solutions to a polynomial equation? I thought it had to be bounded by something involving the degrees…
@@synaestheziac That only applies for y = [polynomial]. In this case this is not really a polynomial since its y^99. Moreover, you technically have infinite solutions to diophantine equations like ax^2-by^2 = 1 [instead of 2].
Thing is that they're finding integer solutions of the graph itself and not when the graph intersects another at a certain point. First one is bound to have infinite solutions [if it happens to]
good
Thanks
Nice problem for celebrating 100k subs🎆😃💯🍻🥂
Thanks! 😃🥰💖🥳
X=1 and y=0
Rainbowlike button
x=y=2
That's it?
2,2.....1,0
SYBER U DID IT 1E5 SUBSCRIBERS YOU FINALLY DID IT
Yess! With your support, of course! 🥰
Stop yelling in all caps.
let's try: x^99 + y^99 = x^100 (*) could be rewritten as: (x^99)(x-1) = y^99 ; let's have k an integer and suppose y=kx. then: (x^99)(x-1) = (k^99)(x^99)
let's suppose X# 0 . therefore we can simplify the equation to get: (x-1) = k^99 so I think we can infer that x is an integer (obviously). next we can say that (from (*)) that y^99 is an integer as well.
now let's determine whether y is an integer: as we had y=kx, we can write x=y/k . substituting in (*) we get: (y^99)/(k^99) + (y^99) = (y^100)/(k^100) . now multiplying each member by (k^100) and simplifying, wet get: k(y^99) + (k^100)(y^99) = y^100 and after factoring and reduction, we obtain:
y = k(1+k^99) which is obviously an integer as well.
I think you over complicated it. Once you conclude that k and x are integers, you can immediately conclude that y = k·x = k·(k^99 + 1) is an integer without having to substitute into the original equation.
@@angelmendez-rivera351 Thank you Angel; I did it right away so I didn't take time to think more about it, but you're right!
@@angelmendez-rivera351 Hi good to see ya
100K中有我一份力😄
你确实有份！🥰
@@SyberMath 哇! 涼爽的!
Congratulations !!!
Thank you so much, Srijan!!! 😀🥰💖