Please keep making these kind of videos. I cannot express my happiness after finding your channel as you explain techniques for math olympiad. Looking forward to see more videos of this kind from you.
HI! Great work and ty for this video! I want to point out at the last problem, when we have y²+y is congruent to 3 (mod 5), we can rewritr y²+y as y(y+1), therefore those are 2 consecutive numbers, in which ONE of them MUST be an odd and one an even, therefore their product will be even number(not important but, let's note it as 2k, where k is in N), and from congruence to 3 (mod 5), we can rewrite 3 mod 5 as 3+5m, m is in N, so we can obviously see that it will be either odd or even, we eliminate the odd ones (they're useless here), as we're left with evens(8,18,28,38...), so for example we have 8, which can be written as 4*2, and YES is divisible by a even number ofc(4 and 2), BUT it'll never be divisible by two consecutive numbers, because of this: the FIRST number(from right, aka with value 1s) of the product of 2 consecutive numbers, will always go in cycle 2,6,2,0,0 , for exp. 1*2=2, 2*3=6, 3*4=12, 4*5=20, 5*6=30, 6*7=42, so we see that the 1s are: 2,6,2,0,0,2... respectively, SO the product of WHICHEVER 2 consecutive numbers will END up with one of the no. 0,2 or 6, which is DEFINITELY not =8, therefore there aren't any solutions!
I know that this sounds rather complicated than normal, but in mathematical terms (as you already know) this can be written in like 2 lines/rows, but I wanted to give you expanded explanation of my sol., BUT also your solution is pretty damn good!
Yes. I think you are right. I had the same problem. It would be useful if Sybermath verified this so that we are satisfied that we are not overlooking something.
My solutions (of course way faster... AND WITH NO MISTAKES): Problem 1: let's look at the multiples of 23: 23, 46, 69, 92, 115... wait a minute: 115=116-1 and 116=4.29. ALWAYS TRY TO FIND OBVIOUS SOLUTIONS BEFORE STARTING A PROBLEM! Then our equation becomes: 23.x+29.y=4.29-5.23 23.(x+5)=29.(4-y) 23 divides the LHS then 23 divides the RHS then 23 divides 4-y (since 23 doesn't divide 29). So 4-y=23.q where q is an integer. We also see that 29 divivdes x+5. Then x+5=29.p where p is an integer. Our equation becomes: 23.29.p=23.29.q so p=q. The solution for (x,y) are: (29p-5,4-23.p) where p is an integer. Problem 2: again, we're looking for obvious solutions. (3,4) is among the candidates to be tested (because of 3²+4²=5²), and it works: 3²+4²=(3x4-7)². We also see that (-3) and (-4) also work. This gives us the idea to introduce x+y-7 (case x+y=7) and x+y+7 (case x+y=-7) (x+y-7)(x+y+7)=(x+y)²-7²=x²+y²+2xy-49 But x²+y²=(xy-7)²=x²y²-14xy+49 Then (x+y-7)(x+y+7)=x²y²-12xy=xy(xy-12) Let's introduce a=x+y and b=xy-6 Our equation becomes: (a-7)(a+7)=(b-6)(b+6) then a²-49=b²-36 then a²-b²=13 then (a-b)(a+b)=13. The solutions of these equations for integers are: (1,13), (13,1), (-1,-13) and (-13,-1). Since b=(s2-s1)/2, the possible values for b are 6 and -6. If b=-6 then xy=0. And if we suppose (the problem is symetrical) that x=0 then the original equation becomes: y²=7² then y=7 or y=-7. If b=6 then xy=12. a²=13+b²=49 implies a=7 or -7 then x+y=7 or x+y=-7. Let's solve this for x+y=7: x and y are solutions of the equation: u²-7u+12=0. We already know the solutions: 3 and 4. For x+y=-7 our quadratic equation becomes u²+7u+12=0 and we also know the solutions: -3 and -4. Conclusion: the solutions are: (0,7), (0,-7), (7,0), (-7, 0), (3,4), (-3,-4), (4,3), (-4,-3). Problem 3: since all integers are positive, 1/x x>5/3 and since x integer x>=2. Same for y and z. 3/5=1/x+1/y+1/z= min(x,y,z)=2, 3, 4 or 5. Since the problem is symetrical we can suppose than x=min(x,y,z). If x=2: 1/y+1/z=1/10. 1/y y>=10, and same for z. 1/10=1/y+1/z= z=110. y=12: 1/z=2/120=1/60 => z=60. y=13: 1/z=3/130: no solution. y=14: 1/z=4/140=1/35 => z=35. y=15: 1/z=5/150=1/30 => z=30. y=16: 1/z=6/160: so solution. y=17: 1/z=7/170: no solution. y=18: 1/z=8/180: no solution. y=19: 1/z=9/190: no solution. y=20: 1/z=10/200=1/20. If x=3, 1/x+1/y=3/5-1/3=4/15. 1/y y>15/4 => y>=4, same for z. min(y,z)= 1/z=1/60 => z=60. y=5: 1/5+1/z=4/15 => 1/z=1/15 => z=15. y=6: 1/6+1/z=4/15 => 1/z=1/10 => z=10. y=7: 1/7+1/z=4/15 => 1/z=13/105: no solution. If x=4, 1/x+1/y=3/5-1/4=7/20: y>=20/7 then y>=3, same for z. But x=min(x,y,z) then y>=4, same for z. min(y,z) 1/z=1/10 => z=10. y=5: 1/5+1/z=7/20 => 1/z=3/20: no solution. If x=5, 1/y+1/z=2/5 and x=min(x,y,z) forces x=y=z=5. Problem 4: (x-y)^3+(y-z)^3+(z-x)^3=30 (x-z)^3=[(x-y)+(y-z)]^3=(x-y)^3+(y-z)^3+3.(x-y)(y-z).[(x-y)+(y-z)]. Let's introduce a=x-y and b=y-z. (a+b)^3=a^3+b^3+3ab(a+b). But the original equation can also be written: a^3+b^3-(a+b)^3=30. Then our equation becomes: -3ab(a+b)=30 => ab(a+b)=-10. ab divides 10: if |ab|=1 then |a|=|b|=1 and |a+b|=
Even though there were some inaccuracies while solving some questions, the methods and the ideas are sitll applicable and very useful, it was great to see them being applied....
19:51 that is an example for the beauty of logic (im always surprised how human mind can reduce a huge problem and solved !!!!!!!!! Cheers for humanity 🍻)
Maybe if you could make a kind of "map of diophantine equations" in regards of solution methods used/required? What if you can train a 'neural network' to solve these problems?
@@nnsnumbersandnotesunlimite7368 It looks maybe possible... very difficult reading. Maybe if using some type of classification is available. But no guaranteea if they could be easily classified into different groups according to solution method required. It can be too hard to find somecases the exact solution method by reading.
Thank you for posting. I'm curious as to why several online Diophantine equation solvers show the general solution to 23x + 29y = 1 as x = -5 + 29k, y = 4 - 23k where your solution is x = -5 - 29k and y = 4 + 23k. Any insight would be appreciated.
@@SyberMath how did you get 6y equals 1 mod 23..just because 6 isnthe remainder when diving 29 by 23 but you didnt really justify that..i dont see why anyone would think of that..
@@SyberMath but 8 equals 3 mod 5 because 8 divided by 5 gives you remainder 3 not negative 2...i get 10 is a multiple of 5 but that would men that -2 mod 5 equals 3 mod 5...see what I mean?
I hope you don't object to me making just a rather general rambly post. It's the day after you posted this video and it's early in the morning for me. I woke up thinking about the last problem you demonstrated where you reduced mod 5. I thought what would happen if the final step did not show no solutions. Now I am interested in the psychology of how problems are constructed. For that point of view if this approach is going to give you anything very definite then the only outcome is no solutions. Other outcomes might give you something but also another problem to solve. After all reducing mod 5 gives the same left hand side regardless of x (even irrational).
I appreciate your comment! If the outcome is that there are some solutions, there may or may not be a way of finding them. Sometimes it's easy to find an upper and/or lower bound and then proceed. Sometimes, there will be infinitely many solutions and we may find some relationships. It all depends.
@@leif1075 Good point! You can but it will be harder. Consider the following: y^2+y-2=0 (mod6) which implies (y-1)(y+2)=0 (mod6). Then by using y-1=0 (mod6), we can write y=6k+1, k is an integer. Substitute into the original equation: 6^x=(6k+1)^2+6k+1-2. Simplifying, we get 6^{x-1}=3n(2n+1). Since 2n+1 is always odd and 3 is odd, we must have 2^{x-1}=n and 3(2n+1)=3^{x-1} from which we find 3^{x-2}-2^x=1 and this equation has no integer solutions. The other case should be similar.
@@SyberMath å måth softwåre thåt knows åll the formulås to win everytime in business,economy,sports,wår,poletics, ånd knows how to find gåps ås in mårket ideås thåt solves problems or creåtes innovåtive ideås. every business cån use it, every sports teåm, ånd it could solve wår ånd fåmine . usåble on micro-meso-måcro scåle.
@@leif1075 While solving for a Diophantine Equation, it's important to be able to express one variable in terms of the other especially for using divisibility criteria!
@@leif1075 You're right. I skipped a step there, trying to keep the video a little shorter. You can take an extra step and write 10y/(y-10)=(10y-100+100)/y-10=(10(y-10))/y-10+100/(y-10)=10+100/(y-10)
i had to skip your video because you talk too fast. sometimes you eat your words. Just remeber that we your audience come here to learn. It feels like you are explaining it to someone that already gets it. Thanks!
Interesting video! I'm a high school student starting number theory and this video abundantly help me! Thanks!
Glad it was helpful!
Please keep making these kind of videos. I cannot express my happiness after finding your channel as you explain techniques for math olympiad. Looking forward to see more videos of this kind from you.
Happy to hear that! Thank you!
I’ve started college and I want to learn some number theory in advance, so thank you for this awesome video. Gonna like it and download it.
Happy to help!
HI! Great work and ty for this video! I want to point out at the last problem, when we have y²+y is congruent to 3 (mod 5), we can rewritr y²+y as y(y+1), therefore those are 2 consecutive numbers, in which ONE of them MUST be an odd and one an even, therefore their product will be even number(not important but, let's note it as 2k, where k is in N), and from congruence to 3 (mod 5), we can rewrite 3 mod 5 as 3+5m, m is in N, so we can obviously see that it will be either odd or even, we eliminate the odd ones (they're useless here), as we're left with evens(8,18,28,38...), so for example we have 8, which can be written as 4*2, and YES is divisible by a even number ofc(4 and 2), BUT it'll never be divisible by two consecutive numbers, because of this: the FIRST number(from right, aka with value 1s) of the product of 2 consecutive numbers, will always go in cycle 2,6,2,0,0 , for exp. 1*2=2, 2*3=6, 3*4=12, 4*5=20, 5*6=30, 6*7=42, so we see that the 1s are: 2,6,2,0,0,2... respectively, SO the product of WHICHEVER 2 consecutive numbers will END up with one of the no. 0,2 or 6, which is DEFINITELY not =8, therefore there aren't any solutions!
I know that this sounds rather complicated than normal, but in mathematical terms (as you already know) this can be written in like 2 lines/rows, but I wanted to give you expanded explanation of my sol., BUT also your solution is pretty damn good!
More on Diophantine equations Mr Gauss: czcams.com/video/6rjoO4K_XuI/video.html
Wow 🤯
Your solution was so cool
Nice work man 👍
Hello
In the problem where you gave the solution with inequalities you stated that 2
Yes. I think you are right. I had the same problem. It would be useful if Sybermath verified this so that we are satisfied that we are not overlooking something.
I was also thinking the same, wish syber maths checks it
My solutions (of course way faster... AND WITH NO MISTAKES):
Problem 1: let's look at the multiples of 23: 23, 46, 69, 92, 115... wait a minute: 115=116-1 and 116=4.29. ALWAYS TRY TO FIND OBVIOUS SOLUTIONS BEFORE STARTING A PROBLEM!
Then our equation becomes: 23.x+29.y=4.29-5.23 23.(x+5)=29.(4-y)
23 divides the LHS then 23 divides the RHS then 23 divides 4-y (since 23 doesn't divide 29). So 4-y=23.q where q is an integer.
We also see that 29 divivdes x+5. Then x+5=29.p where p is an integer.
Our equation becomes: 23.29.p=23.29.q so p=q.
The solution for (x,y) are: (29p-5,4-23.p) where p is an integer.
Problem 2: again, we're looking for obvious solutions. (3,4) is among the candidates to be tested (because of 3²+4²=5²), and it works: 3²+4²=(3x4-7)². We also see that (-3) and (-4) also work.
This gives us the idea to introduce x+y-7 (case x+y=7) and x+y+7 (case x+y=-7)
(x+y-7)(x+y+7)=(x+y)²-7²=x²+y²+2xy-49
But x²+y²=(xy-7)²=x²y²-14xy+49
Then (x+y-7)(x+y+7)=x²y²-12xy=xy(xy-12)
Let's introduce a=x+y and b=xy-6
Our equation becomes: (a-7)(a+7)=(b-6)(b+6) then a²-49=b²-36 then a²-b²=13 then (a-b)(a+b)=13.
The solutions of these equations for integers are: (1,13), (13,1), (-1,-13) and (-13,-1).
Since b=(s2-s1)/2, the possible values for b are 6 and -6.
If b=-6 then xy=0. And if we suppose (the problem is symetrical) that x=0 then the original equation becomes: y²=7² then y=7 or y=-7.
If b=6 then xy=12. a²=13+b²=49 implies a=7 or -7 then x+y=7 or x+y=-7.
Let's solve this for x+y=7: x and y are solutions of the equation: u²-7u+12=0. We already know the solutions: 3 and 4.
For x+y=-7 our quadratic equation becomes u²+7u+12=0 and we also know the solutions: -3 and -4.
Conclusion: the solutions are: (0,7), (0,-7), (7,0), (-7, 0), (3,4), (-3,-4), (4,3), (-4,-3).
Problem 3: since all integers are positive, 1/x x>5/3 and since x integer x>=2. Same for y and z.
3/5=1/x+1/y+1/z= min(x,y,z)=2, 3, 4 or 5. Since the problem is symetrical we can suppose than x=min(x,y,z).
If x=2: 1/y+1/z=1/10. 1/y y>=10, and same for z. 1/10=1/y+1/z= z=110.
y=12: 1/z=2/120=1/60 => z=60.
y=13: 1/z=3/130: no solution.
y=14: 1/z=4/140=1/35 => z=35.
y=15: 1/z=5/150=1/30 => z=30.
y=16: 1/z=6/160: so solution.
y=17: 1/z=7/170: no solution.
y=18: 1/z=8/180: no solution.
y=19: 1/z=9/190: no solution.
y=20: 1/z=10/200=1/20.
If x=3, 1/x+1/y=3/5-1/3=4/15.
1/y y>15/4 => y>=4, same for z. min(y,z)= 1/z=1/60 => z=60.
y=5: 1/5+1/z=4/15 => 1/z=1/15 => z=15.
y=6: 1/6+1/z=4/15 => 1/z=1/10 => z=10.
y=7: 1/7+1/z=4/15 => 1/z=13/105: no solution.
If x=4, 1/x+1/y=3/5-1/4=7/20: y>=20/7 then y>=3, same for z. But x=min(x,y,z) then y>=4, same for z. min(y,z) 1/z=1/10 => z=10.
y=5: 1/5+1/z=7/20 => 1/z=3/20: no solution.
If x=5, 1/y+1/z=2/5 and x=min(x,y,z) forces x=y=z=5.
Problem 4:
(x-y)^3+(y-z)^3+(z-x)^3=30
(x-z)^3=[(x-y)+(y-z)]^3=(x-y)^3+(y-z)^3+3.(x-y)(y-z).[(x-y)+(y-z)].
Let's introduce a=x-y and b=y-z.
(a+b)^3=a^3+b^3+3ab(a+b).
But the original equation can also be written: a^3+b^3-(a+b)^3=30.
Then our equation becomes: -3ab(a+b)=30 => ab(a+b)=-10.
ab divides 10: if |ab|=1 then |a|=|b|=1 and |a+b|=
Wow!!!
Helpful! Thanks!
Glad it was helpful!
Very intersting information thank u!!!
Uf a>b (-a
Very good
Very Nicely Explained. Thanks a Lot , Sir.
You are most welcome!
Even though there were some inaccuracies while solving some questions, the methods and the ideas are sitll applicable and very useful, it was great to see them being applied....
Excellent!
19:51 that is an example for the beauty of logic (im always surprised how human mind can reduce a huge problem and solved !!!!!!!!! Cheers for humanity 🍻)
I totally agree!
Tony haddad please stop commenting! You have received too many hearts from syber!
Solve : Find all positive integers x , y and z with z odd , which satisfy the equation :
2018^x =100^y +1918^z
I thought you would also explain also with euclid algorithm for linear diophantine equation
Absolutely awesome. You really should make more videos like this
Thank you for the kind words! 🥰
That was great - I'll refer to this whenever I encounter another Diophantine challenge.
This video rocks
this is awesome, thank you so much! :D
Glad you like it!
Imho ab(a+b)=-10 looks better. Overall the video is very helpful. Thanks!
Np
Just discovered your channel. Loved the content. Keep up the good work.
Welcome aboard! Thanks for watching!
Thanks!
💖 Oh, thank you!!! 💖
Love your channel. Please keep the interesting problems coming!
@@acesovernines Thanks! Will do! 💖
Very interesting, great job bro!
Thank you!!! 🤩
Is exponent = 5, x=1/y=1, z=2 [1, 1,1]--->Origin A valid solution? 3:55
The COEFFICIENTS ARE 1, 1, 2🎉
Please make more videos like this teaching something
Good idea!
Maybe if you could make a kind of "map of diophantine equations" in regards of solution methods used/required?
What if you can train a 'neural network' to solve these problems?
Interesting questions! That would be really cool!
More on diophantine equations : czcams.com/video/6rjoO4K_XuI/video.html
@@nnsnumbersandnotesunlimite7368
It looks maybe possible... very difficult reading.
Maybe if using some type of classification is available. But no guaranteea if they could be easily classified into different groups according to solution method required.
It can be too hard to find somecases the exact solution method by reading.
Making denominator nigative in the third equation,we can valed solutions
Thank you very much!
You're welcome!
Thank you for posting. I'm curious as to why several online Diophantine equation solvers show the general solution to 23x + 29y = 1 as x = -5 + 29k, y = 4 - 23k where your solution is x = -5 - 29k and y = 4 + 23k. Any insight would be appreciated.
They both generate the same set of solutions. Just replace k with -k.
@@MrDowntownjbrown Thank you.
Hello, which program do you use to write your demos?
Hi. I use notability
How often do you use the phrase “Without Loss of Generality” that you created an acronym for it that you remember before the phrase xd
you are GOD...!!!!! thnx for this vdo...
Thank you so much ❤
You're welcome 😊❤
Thx for this video, you got a new subscriber
Thanks for the sub! 💖😊
😃Great work!
Thank you so much 😀
@@SyberMath I really hope you can respond to my.mod 5 question when you can because Im very curious. Thanks
@@SyberMath how did you get 6y equals 1 mod 23..just because 6 isnthe remainder when diving 29 by 23 but you didnt really justify that..i dont see why anyone would think of that..
Problem #3
19:47 You said that values of x and y will lie between 2 and 5. So how come the solutions of y and z are greater than 5. Please Help
There was an error in the video that wasn't picked up.
2
@@CrYou575 Thanks
That's exactly what I was thinking about
@@CrYou575 Nice catch!
It will be a nice video 😍
Hope so!
Those were very helpful
I'm glad!
another greeat explination!
How many coffee cups did you drink before recording the video, man? You speak as fast as the wind!
Seriously bro 😅
Sorry, I was trying to keep the video shorter!
Sir, can you still use the SFFT in problem #3?
No b/c you have three variables
yup sir but when you have that 1/y + 1/z = 1/10 maybe you can use that ??
@@roviedimaano575 Absolutely!
He replied all the comments except this one
Not only this one. But it's okay
21:37 I think its 4/15 not 1/15. Still a great video nonetheless :)
That's right! :) Thanks
Great info! Thanks 😀
Absolutely! Thanks for watching.
@@SyberMath but 8 equals 3 mod 5 because 8 divided by 5 gives you remainder 3 not negative 2...i get 10 is a multiple of 5 but that would men that -2 mod 5 equals 3 mod 5...see what I mean?
@@leif1075 that is correct
Problem 3 how about x=2, y=35, z=14?, i know y>z but 1/2+1/35+1/14=3/5
It works.
For x=2
y-10=25
y=35
z=10+4=14
x 2 y 14 z 35 so y
nice
Make more diophantine I like
Number throery
Me too!
Nice
Thanks!
I hope you don't object to me making just a rather general rambly post. It's the day after you posted this video and it's early in the morning for me. I woke up thinking about the last problem you demonstrated where you reduced mod 5. I thought what would happen if the final step did not show no solutions. Now I am interested in the psychology of how problems are constructed. For that point of view if this approach is going to give you anything very definite then the only outcome is no solutions. Other outcomes might give you something but also another problem to solve. After all reducing mod 5 gives the same left hand side regardless of x (even irrational).
I appreciate your comment! If the outcome is that there are some solutions, there may or may not be a way of finding them. Sometimes it's easy to find an upper and/or lower bound and then proceed. Sometimes, there will be infinitely many solutions and we may find some relationships. It all depends.
@@SyberMath For the last problem, why wouldn't you do mod 6 since it's a multiple of 6?? Isn't that the most logical and intelligent choice??
@@leif1075 Good point! You can but it will be harder. Consider the following:
y^2+y-2=0 (mod6) which implies (y-1)(y+2)=0 (mod6). Then by using y-1=0 (mod6), we can write y=6k+1, k is an integer. Substitute into the original equation: 6^x=(6k+1)^2+6k+1-2. Simplifying, we get 6^{x-1}=3n(2n+1). Since 2n+1 is always odd and 3 is odd, we must have 2^{x-1}=n and 3(2n+1)=3^{x-1} from which we find 3^{x-2}-2^x=1 and this equation has no integer solutions. The other case should be similar.
2
18:26 "not all three at the same time" well if all three are 5... but yeah they can't all be 2. 2 be, or not 2 be. :D
Toby or not Toby! 😁
would you work on å project thåt is very revolutionål ? åre u for hire ?
What kind of project?
@@SyberMath å måth softwåre thåt knows åll the formulås to win everytime in business,economy,sports,wår,poletics, ånd knows how to find gåps ås in mårket ideås thåt solves problems or creåtes innovåtive ideås. every business cån use it, every sports teåm, ånd it could solve wår ånd fåmine . usåble on micro-meso-måcro scåle.
A kid said to his math teacher: To show you how good I am at fractions, I only did half my homework!
5 out of 4 people have trouble with fractions! 😁
@@SyberMath Why would anyone think of doing that at 20:18??
@@SyberMath Also, at 20:13, when you solve for z you get z = 10y/(y-10) , not the funky expression you got..
@@leif1075 While solving for a Diophantine Equation, it's important to be able to express one variable in terms of the other especially for using divisibility criteria!
@@leif1075 You're right. I skipped a step there, trying to keep the video a little shorter. You can take an extra step and write 10y/(y-10)=(10y-100+100)/y-10=(10(y-10))/y-10+100/(y-10)=10+100/(y-10)
Problem 4: (x - y)^3 + (y - z)^3 + (z - x)^3 = 30 ==> (x - y)^3 + (y - z)^3 + (z - x)^3 = 3·(x - y)·(y - z)·(z - x) ==> 3·(x - y)·(y - z)·(z - x) = 30 ==> (x - y)·(y - z)·(z - x) = 10
No sum of three factors of 10 is equal to 0, so this has no integer solutions.
Nice!
1995 is "recently"? - OK!
What do you mean?
@@SyberMath That I would call Wiles proof from 1995 not "recently" - but that's surely no problem!
@@keinKlarname I see. I don't know why I called that recent, either! 😁
@@SyberMath Btw, I've learned already a lot from just seeing a few of your videos - thanks a lot for this!
You're very welcome!
More on Diophantine equations : czcams.com/video/6rjoO4K_XuI/video.html
i had to skip your video because you talk too fast. sometimes you eat your words. Just remeber that we your audience come here to learn. It feels like you are explaining it to someone that already gets it. Thanks!
Noted
a³+b⁴=c⁵ is what made me click. Not gonna lie.