One of my favorite applications of this is when the sum is 1/2. When solving 1/x + 1/y = 1/2, the resulting (x, y) give dimensions of a rectangle whose area equals its perimeter, and by extension, when 1/x + 1/y + 1/z = 1/2, the resulting (x, y, z) give dimensions of a box whose volume equals its surface area.
I didn't think I could solve something like this but I think I have, I haven't watched the video yet. Without loss of generality I defined x to be less than or equal to y. For the cases where x
At the beginning you state that m and n should be relatively prime, but at the end it turns that m and n are just posituve integers without any other restrictions.
If you allow m, n to have common divisors there are different triples (k, m, n) which result in the same (x, y, z). Let's assume m, n have a common factor a, e.g. n = a*b, m = a*c. With the final equation in the video you can easily check that (k, m, n) and (k*a^2, b, c) result in the same triple (x, y, z). So, if you want (x, y, z) to have a unique representation in terms of (k, m, n) you require gcd(m,n)=1. Otherwise you don't need this condition.
Make a common denominator (x+y)/xy=1/4 xy=4x+4y xy-4x-4y=0 xy-4x-4y+16=16 x(y-4)-4(y-4)=16 (x-4)(y-4)=16 Check divisors/factors of 16... The rest is trivial
I did work on this a while ago, but I had assumed that 'z' was a known natural number and we're looking for naturals x and y that satisfy 1/x + 1/y = 1/z for this I found that all solutions are of the form (x, y) = (z + f, z + z²/f) where f is a divisor of z² (obviously) leading to a number of solutions equal to number of divisors of z² But hey, this one's pretty neat as well. Kudos.
Hey dude just wanted to say something regarding your announcement ... You don't have to go for views ok? What I mean that when you first opened this channel you probably wanted to share your interest regarding geometry puzzles with other people , right ? So you made your own geometry puzzles to amuse people . If you keep on following your passion then views will automatically increase ! And I personally stayed with your channel bcoz of them ! So my solution is that keep on making videos on geometry puzzles but perhaps make them a bit more interesting , like maybe change your outlining shape to a heart , pentagon and other weird shapes etc . In short be a bit more creative 🙂 And an another piece of advice , Micheal Penn recently opened a Google form to take viewer suggested problems , so why don't you ? Or you can even increase the difficulty of problems ! So that's it from my side and I hope my advices help you in the near future !
Srijan, I agree with you. This channel started with my passion of solving problems that I like and sharing it with the world. It's not all about views, of course. Geometry is not my strength but I like solving puzzles that involve inscribed figures because they are fun imo and also not very hard to solve! If you check out my twitter, you will notice a great number of people that are very strong in geometry. They can basically tackle very hard problems with ease, such as the ones that appear in olympiads. I'm not one of them. I take your advice to heart and will work to come up with more interesting puzzles. Thanks for the feedback!
One of my favorite applications of this is when the sum is 1/2. When solving 1/x + 1/y = 1/2, the resulting (x, y)
give dimensions of a rectangle whose area equals its perimeter, and by extension, when 1/x + 1/y + 1/z = 1/2,
the resulting (x, y, z) give dimensions of a box whose volume equals its surface area.
Wow. This was just beautiful. I had no idea you could use gcd’s like this. Really awesome!
Glad you like it!
HAMZA GCD IS HCF OR HIGHEST COMMON FACTOR
Cool! I have yet to learn how to solve diophatine equations with general solutions. Thanks for the “intro lesson”
Glad you liked it! You mean the video on Diophantine Equations?
ua-cam.com/video/3U3PbYnkkAM/v-deo.html
This a famous diaphontine equation,which came in many mathematics Olympiad
Oh, I did not see this video(ua-cam.com/video/3U3PbYnkkAM/v-deo.html) yet. Thanks!
I didn't think I could solve something like this but I think I have, I haven't watched the video yet. Without loss of generality I defined x to be less than or equal to y. For the cases where x
Great!
This is a famous diophantine equation, which came in many mathematics Olympiad. Thanks for solution
Absolutely! Glad to hear that it helped!
@@SyberMath can you cover Chinese Remainder Theorem , as you explain well
@@SONUKUMAR-vr2jg You mean in a video?
@@SyberMath make a video on Chinese Remainder Theorem
this video is great, subscribed
Glad to hear that! Welcome to the channel!
Loving all your number theory videos recently
Nice to meet you, too!
Glad you like them!
At the beginning you state that m and n should be relatively prime, but at the end it turns that m and n are just posituve integers without any other restrictions.
m and n are relatively prime but x, y and z do not have to be. We wrote x,y, and z in terms of m,n and k.
If you allow m, n to have common divisors there are different triples (k, m, n) which result in the same (x, y, z). Let's assume m, n have a common factor a, e.g. n = a*b, m = a*c. With the final equation in the video you can easily check that (k, m, n) and (k*a^2, b, c) result in the same triple (x, y, z). So, if you want (x, y, z) to have a unique representation in terms of (k, m, n) you require gcd(m,n)=1. Otherwise you don't need this condition.
Easy problem in one way, difficult in another. Nice!
The simplicity of its look and the parametric solutions we have to use!
Beautiful sir 💐.. Keep it up 👍
Thank you, I will!
I did not quite understand the reason behind the gcd(mn, m+n) =1...
You mean why it is true?
gcd(n, m)= 1 => gcd(n, m+n) = gcd(m, m+n) = 1 => gcd(n*m, m+n) | 1 => gcd(nm, m+n) = 1
@@Relrax Nice work!
6:33 and there gcd must equal 1 for (m,n)
This what i want , cool !!!!!!!!!
Glad you like it!
Can you find the integer solution of 1/x + 1/y= ¼ please sir solve it
Make a common denominator
(x+y)/xy=1/4
xy=4x+4y
xy-4x-4y=0
xy-4x-4y+16=16
x(y-4)-4(y-4)=16
(x-4)(y-4)=16
Check divisors/factors of 16...
The rest is trivial
@@SyberMath woww thanks for the reply and the answer it was unexpected
Easy way is 1/4=1/5+1/20
1/z=1/(z+1)+1/z(z+1) for integer, z>1
Awesome!
Glad you think so!
Nice problem!! But almost any integers m, n, k work, right? Not only the positive ones. The only constraints will be m, n, k =/= 0 and m + n =/= 0.
Sure! You can just negate everything. I just wanted to keep it simple and restrict it to positives.
I did work on this a while ago, but I had assumed that 'z' was a known natural number and we're looking for naturals x and y that satisfy 1/x + 1/y = 1/z
for this I found that all solutions are of the form (x, y) = (z + f, z + z²/f) where f is a divisor of z² (obviously) leading to a number of solutions equal to number of divisors of z²
But hey, this one's pretty neat as well. Kudos.
Good! Thanks!
m, n, k in N-{0} but gcd(m,n)= 1. Isn't it ? ok I see it's for all positive integer : great job ! Bravo
Thank you!
1/(2n) + 1/(2n) =1/n
There are infinite solutions of this so it has inf solns
They are not the only solutions
👍 Дякую.
нема проблем
Going to subscribe you
Welcome to the channel!
so fun
Glad you like it!
(x,y,z)=(4,4,2)
Sure. Are there other solutions?
So easy Anyone could tell easily that this equation has infinite solution
Yes but finding them is fun, don't you think?
@@SyberMath yes indeed
Hey dude just wanted to say something regarding your announcement ...
You don't have to go for views ok?
What I mean that when you first opened this channel you probably wanted to share your interest regarding geometry puzzles with other people , right ? So you made your own geometry puzzles to amuse people . If you keep on following your passion then views will automatically increase ! And I personally stayed with your channel bcoz of them ! So my solution is that keep on making videos on geometry puzzles but perhaps make them a bit more interesting , like maybe change your outlining shape to a heart , pentagon and other weird shapes etc . In short be a bit more creative 🙂
And an another piece of advice , Micheal Penn recently opened a Google form to take viewer suggested problems , so why don't you ? Or you can even increase the difficulty of problems ! So that's it from my side and I hope my advices help you in the near future !
Srijan, I agree with you. This channel started with my passion of solving problems that I like and sharing it with the world. It's not all about views, of course. Geometry is not my strength but I like solving puzzles that involve inscribed figures because they are fun imo and also not very hard to solve! If you check out my twitter, you will notice a great number of people that are very strong in geometry. They can basically tackle very hard problems with ease, such as the ones that appear in olympiads. I'm not one of them. I take your advice to heart and will work to come up with more interesting puzzles. Thanks for the feedback!
@@SyberMath Yeah thanks for taking my feedback with a positive mindset !
@@srijanbhowmick9570 Of course!
can i be of any help?