Bulgarian Math Olympiad | 1999
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- Опубліковано 5 жов 2020
- We present a solution to a nice number theory problem from the 1999 Bulgarian math olympiad.
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People who paused this video after the presentation of the problem to give it a try on their own are going to be so mad when they find out the second y in the equation was supposed to be squared!
bruh moment
The problem with x^3=y^3+2y+1 can be solved in a similar (slightly harder) way.
You need to distinguish the cases x>y and xy the procedure is similar to the one in the video, meaning that you have y^3+2y+1=x^3≥(y+1)^3=y^3+3y^2+3y+1 => 3y^2+y≥0, which has solution y∈[-1/3,0], so the only integer solution is y=0, for which x=1.
2) if x 3y^2-y+2≤0, which has solution y∈[-2/3,1], meaning that the only integer solutions are y=0 (which we have already considered in point 1) and y=1 (for which y^3+2y+1=4, which is not a cube).
So, the only integer solution to x^3=y^3+2y+1 is (x,y)=(1,0)
Believe me, I am
Based on this video, the solution to the next video’s problem is 17. At least it will be after I change the problem! Hehe we all make silly mistakes in Mathematics.
I'm not mad. I was disappointed in myself for getting it "wrong," and after I read your comment, now I'm not.
I suggest you do a follow-up video showing the solutions for x³ = y³ + 2y + 1 (but be sure the thumbnail for that one has the equation x³ = y³ + 2y² + 1).
I think you end up with 3y^2 + y
Plot both curves on desmos
then the only solution is (x,y)=(1,0)
Finally something from my country. Love from Bulgaria🇧🇬.
Привет
Айдее наште.
@@IoT_ what u write this??
@@garrethutchington1663 what u write this ?? Because I from great country India
Love from Mexico to Bulgaria 🇧🇬
hehe, I have to say that answering a completely different question is an interesting problem solving method :)
😂
Great video, as always! As a Bulgarian, I especially enjoyed this one! Thank you!
Thank you, more problems from Bulgaria in the future please!
hey, just wanted to say that you are one of the best math youtubers if not the best i've encountered. You are super clear and your series in real analysis has been a huge help for me to understand, what kind of properties have the real numbers.
You are an amazing teacher, and your content is way too good, although a huge fan of the vertex algebra playlist it is a tad too hard for me since i am barely starting to study abstract algebra.
keep up the good work!
Love the way you tackled the problem.
Nice little problem and solution. Thanks for sharing!
Great solution, thanks, Michael Penn
This is what I'v been waiting for
Nice to see the approach of finding new constraints with inequatliries. 👍🙏
Intuitively, the gaps between successive cubes grow larger than 2y^2+1.
Wait, wait, wait. Is the equation x³ = y³ + 2y + 1 or x³ = y³ + 2y² + 1? Your solution solves the second equation and the problem on the thumbnail, but the initial statement of the video omits the 2 exponent...
should be x³ = y³ + 2y² + 1 liek in the thunbnail
Angel Mendez-Rivera it is clearly meant to be squared, as that is the entire solution. It's clearly a simple mistake when writing the first part of the question.
@@angelmendez-rivera351 at this moment, as i am writing this comment, i am looking at 5 full thumbnails in the recommended bar to the right of the video:
1) How Much of The Earth Can We See at Once? - a picture of a satellite, the Earth and a cone stretching between them as a form of field of view
2) SynRM | A new giant in the electrical world - a picture of some sort of mechanism related to engineering
3) Why Machines That Bend Are Better - a picture of a flexible-looking structure
4) Why is pi here? And why is it squared? A geometric answer to the Basel problem
5) How High Can We Build? - a depiction of Earth and some supposedly extra tall building that seemingly stretches to outer space (a reference to a space cable or a space elevator)
And as the comment's length grew, more thumbnails popped into view:
6) The Mathematics of Winning Monopoly - a picture of two people sitting at a table with a Monopoly board on it
7) The Ingenious Design of the Aluminum Beverage Can - a picture of a man seated at a table with an aluminum can in front of him
8) Fixed Points - a picture of a punctured, bounded disk with several points drawn, which is a reference to the fixed point theorems, such as Brouwer's fixed point theorem
9) We likely live in on of these 18 universes - an abstract representation of a space with more than 3 dimensions or a non-Euclidian space (hyperbolic, in this case)
10) Visualizing quaternions (4d numbers) with stereographs - a picture which depicts a stereographic projection of the quaternions which, from the looks of it, have the sum of the squares of the imaginary coefficients equal to 1
11) The History of Mathematics and Its Applications - a picture spelling out a part of the title and having an abstract background (to accentuate the fact that it's about mathematics)
snipboard.io/eBHqOy.jpg
snipboard.io/tvKZYu.jpg
Links to screen captures showing what I have described just now for verifications. I suggest you either change your claim or agree with me that what you've said is quite the overgeneralization in the very least.
@@retired5548 FFS, all he meant was that the thumbnail is not AS IMPORTANT as what he actually solved and verbalized.
Plus it's pretty obvious it was intentional to drive up comment interactions.... Lemmings
@@angelmendez-rivera351 i thought i replied to this earlier...hmm...anyway, here's my thoughts on the matter:
you know i'm fucking with you, right? of course i'll be overly pedantic and overformalize when i see something like this, dude. the joke here was that i've written such a fuckin long reply with all the work and everything as a reply to your two short sentences. oh well
With a similar tecnique, I'll try to solve the first problem: finding couples (x,y) in ZxZ such that x³ = y³+2y+1.
Let's divide into three cases: 1) y = 0; 2) y > 0; 3) y < 0.
1) y = 0 implies x³ = 1, so x = 1 and (x,y) = (1,0) is a solution.
2) y > 0 ==> 2y + 1 > 1 ==> y³+2y+1 > y³+1 > y³ ==> x³ > y³ ==> x > y; since we are in ZxZ we have x >= y+1 ==> x³ >= (y+1)³ ==> y³+2y+1 >= (y+1)³ = y³+3y²+3y+1 ==> 3y²+y y(3y+1) -1/3
You forgotten y^2
(-2,-3) is also a solution .See the question once
Challenging solution sir
@@ramanakv3272 it's not
Hi sr, been watching your daily videos for some time, would like to give a request, how about some euclidean geometry ? the one involving cyclic quadrilaterals, power of point, centers of triangle or angle chasing of some sort. Thank you for the content sr ! keep it up !!
This would actually be really helpful. A lot of contest math, at least where I’m from, puts a large focus on geometry. This channel has always been super helpful for competitive math, so I think leaning into that more with some more geometry videos would be great.
Ok so the equation on the left side of the board is wrong (2y is in reality 2y^2 )
Nice solution, keep it up
I'm very happy to see a problem from my country.
На всички българи - здрасти! 😀
i was solving a different question and thought of taking a break, and i started watching this video, and saw you T-shirt...... THAT FORMULA SOLVED MY QUESTION!!!! 😄
@@mr.knight8967 thanks!!
Without going too deep, my tiger hints are:
when x > 0 , y > 0. x > y. Consider x = y + 1. At some point the gap between successive cubes will overpower the linear term 2y + 1.
consider x =3, y=2. x^3 - y^3 = 27 - 8 = 19. 2y + 1 = 5. Already I know the solution can't involve numbers larger than 3 and 2.
when x < 0, y < 0. x < y. x = y - 1. Again, at some point, the gap between successive cubes will overpower the linear expression.
So, we're down to a few solutions that way.
Totally followed that. Thanks.
Greetings from Bulgaria 👋👋🇧🇬🇧🇬🇧🇬
That problem was so much easier to understand than the ones where you have to use modules to solve. I suppose I don't have a good grasp of modules and its applications since its been so long and never worked with them since over 20 years
Modules, or modulo?
How to come with solution that x>y,I mean only practice can show? Or there is way to solve each kind of NT equations
Piece of cake! (after your excellent explanation)
BEAUTIFUL !
ok i'm confused i thought we were finding x,z so that it satisfy x^3 = y^3+2y+1. Can someone explain what happened.
strongly wish there was some indication added in post that the 2y is supposed to be 2y^2. so perplexing!
Can anyone please give me a tip on how to approach number theory problems?
Genial! Gracias!
Sir,I love your videos and the way you solve problem...But I want a problem on inversion.I think it will be much helpful for us who are interested in geometry
Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120,
what is the value of s(3m)?
Can someone please help me with this?
It came in India. PRMO 2019 25th August Paper
Would love to get your amazing self-erasing blackboard.
Question: why is the letter Z used for integers?
I get that the double bar in the vertical is standard notation for sets of the standard number types, but why not I (Capital I, also written with a double downstroke to distinguish it from unity)?
I have never seen the set of imaginary numbers named, is it maybe reserved for that?
Back in the day when i was learning basic number theory I always got confused because lower case z is so often used in algebra for a complex quantity...
It is clearly too late to change now, but i'd invite you to do a video covering the history of the choices for the names of these infinite sets
Yours hopefully
The question says solve x^3 = y^3 + 2y + 1 but it has been solved for x^3 = y^3 + 2y^2 + 1.
Michael, your idea is nice. My first idea was
1) x>0,y>0 then x^3=z^3-(z^2+z-2) where z=y+1, only for z=1 Is there solution, because for z in R\((-1+sqrt5)/2;(-1-sqrt5)/2) there is y
Amazing
For fun:
0 "and so on and so forth",
1 "go ahead and ...",
1 "great",
and the "and that's a good place to stop" at the end.
Solving the problem from the thumbnail of the video. Nice!
Playing the video and finding that the problem is actually different and there was a mistake on the thumbnail. Oh, come on! Screw it, I am watching the video anyway.
It turns out the thumbnail was correct and so was my solution. Nice!
On the left it says y not y^2.
I did almost the same, but I rewrote it as:
x^3-y^3 = 2y^2+1
So a difference between two cubes have to be something postive (so x>y).
I then said that y = (x-1) must be the value of y for which the LHS is the smallest (while it also is the value for which the RHS has the greatest value, before dipping into the negative values, but at that point it is already far behind the LHS and it cannot keep up with something cubed, since for negative values we get x^3+y^3), so if we want some kind of solution *that* at least has to be smaller or equal to the RHS.
Calculating 2(x-1)^2+1 and x^3-(x-1)^3 and setting the inequality gives the second order polynomial:
x^2+x-2 leq 0
which means x in [-2,1]. Looking through those values for x gives the same three solutions as shown above.
The method you showed was way better, though :) thanks for the puzzle.
How did you find out that y=(x-1) is the smallest value of y for which the LHS is smallest?
@@gervasiociampin2062 We know that x^3-y^3 geq 1. This means that x>y (strictly) so the next highest number we can give to y is x-1 which is also the number that will minimize the LFS since it is just two numbers subtracted from each other.
Wasn't the original question x^3 =y^3+2y+1? But then you wrote the second polynomial in second degree in the beginning of the solution.
I have tried like this x^3-y^3=2y+1 then (x-y)(x^2+xy+y^2)=2y+1
You can cleraly see x and y have diffrent parity so l ts says x-y is even and x^2+xy+y^2 is odd
Hello Michael! I want to share with you my solution to the problem which is not so straightforward but a little bit easier. What I did was taking the x³ to the other side and consider that equation a polynomial equation: y³+2y²+(1-x³)=0. Then I used the integer roots theorem saying that:
If this polynomial has integer solutions, the solutions will divide exactly the steady value, which in this case is 1-x³. We know that 1-x³=(1-x)(1+x+x²) so the possible integer solutions are 1,-1,1-x³,x³-1,1-x,x-1,1+x+x²,-1-x-x². So, if we place those values in y we get equations in respect of x which we solve seperately. This way, we find integer values of x which we place at the original equations so as to get values of y too. If we do it like this we will get the same pairs as you did!!!👍
Holy moly! This is so damn close to the problem v^2 = u^3 + 2u + 1 😮
And then I realised the problem in the video is y^3 + 2y^2 + 1.
I think someone replied to the problem “Find all integers u such that u^3 + 2u + 1 is a perfect square.”, and he/she found that the only solutions are u = 1, u = 0 and u = 8.
Adam Romanov I think you have seen this in the comments section of Michael’s video on a mathematical problem from Turkey. (Find all primes p such that “some expression” equals some square number.)
You could've just differentiated the right hand side and placed it greater than 0 ( x3 is strictly increasing)
How do you proceed?
Can someone explain me how from x3=y3+2y+1 we went to x3=y3+2y2+1?
6:44 Tova e dobro myasto za spirane
Woah , i thought I was the first one to watch but I saw ur comment lol
2:33 is great!
@Adam Romanov naistina e dobro mqsto za spirane
Homework involving elliptic curves: find all rational solutions to this equation.
Great problem with a great solution! Wait, what is the original problem? 😁
Sorry the hint that f(x) = x^3 is strictly increasing is misleading. It decreases between 0 and 1 and -1 and 0. I think the hint is meant to say that it’s strictly increasing for the natural numbers. This is implied in the problem statement but again, the hint is misleading
Nice
don't forget about the square on the y of the equation like the thumbnail
I believe (x,y)=(1,0) is the only solution to the problem without the square.
X^3=x x^3=y^3+2y+1 x is variables symbol y is stables symbol 0x,0y +1 is sum check...
how do you know that x is greater than or equal to y+1 at 2:45 ? x is greater than y does not directly indicate the next inequality that you wrote.
Because if x=4 and y=3 x>y but y+1=4 then the inequality goes x>=y
So is it 2y or 2y^2 ?
probably 2y^2, because if it would be 2y, we would get that y has to be between -1/3 and 0, and thus y would have to be 0, which would be untypical for this kind of olymiadquestions
Thumbnail says it's squared
@@angelmendez-rivera351 i wish there were more black, women and lgbt in mathematics
@@angelmendez-rivera351 yeah I don't see a problem with giving a fact about the thumbnail without avoiding the "content" or saying it doesn't matter🤔😕
2y^2 >= 3y^2 + 3y , what am i missing here, how is this right
Is he a professor at Penn State? If not, he should be.
In the comments, like ..wait, “there’s more of you?!”
Yay I finally got one
Hi everyone
For x^3=y^3+2y+1
add -y^3 to all sides and we get that
(x-y)(x^2+xy+y^2)=2y+1
If y=0, then x is 1. Assume y>0, then x > y (notice that x^2+xy+y^2 is always positive if one of x and y is nonzero)
In this case, x-y is at least 1, but x^2+xy+y^2>2y+1 (since y is at least 1 and x is at least 2).
So there is no solution for positive x and y.
Assume y1. So x^2+xy+y^2 is greater than the absolute value of 2y-1.
Therefore there is no solution for this last case.
The only solution is (0,1)
Use hit & trial method
Only (1,0) satisfied x^3=y^3+2y+1
You put in a square in the y term that wasn’t in the original problem.
That's pretty elegant. I shudder when I see degree-three diophantines, because the solutions can get quite crazy, even for the most innocent-looking equations
Example: a/(b+c)+b/(a+c)+c/(a+b) = 4, find the smallest positive integer solutions for a, b, c
Looks innocent, but without googling you won't find the solution if you're not well-versed in the theory of elliptic curves.
Good
Hardly rocket science. When y = 0 , x in this equation must be 1. And if you take y^3 + 2y down to 0 you must start with -2 or the first one. The third solution is a bit of a number fluke. But if you can do 2/3 of this sitting on a train feeling bored......... do they have a lot of tunnels in Bulgaria?
Pause at 1:37 if you want to solve the problem he really intended.
Any one please answer this question
1)Prove that there exists distinct positive integers x,y such that x+j divides y+j where j=1,2,3,.......,n
2)if for some positive integers x,y, x+j divides y+j for all positive integers show that x=y.
This is question from inmo 1996
I feel like number 2 is true intuitively but I'm not very good at number theory so I'm having a hard time proving it to myself.
Hmm ok no prob
1) x+j|y+j, also x+j|x+j.
=> x+j|y+j-(x+j) => x+j|y-x, set y=x+c where c is an integer.
=> x+j|c, for all j in{1,2,...,n}. Let c = (x+n)!/x! = (x+1)(x+2)...(x+n) then c divides all x+j, for j in{1,2,...,n}.
so for any non-negative x and y=x+(x+n)!/x!, x+j|y+j for j in{1,2,...,n}
2) Suppose for a contradiction that x does not equal y. then y=x+c where c is a non-zero integer.
From 1): x+j|c for all positive integers j.
if x
In the second part of your question, did you mean for all positive integers j?
Thank you Thomas strokes
I love to solve NT EQUATIONS
But these days I am practicing Geometry bcoz I am going weak and weak in Geometry
First I thought it was y^3+2y+1, which probably doesn't have any solutions but 1 and 0. But clearly you just had typo on the left of the board and it was supposed to be y^3+2y^2+1
Is it a typo when it appears on a chalkboard?
I would be tempted to call it a "chalko" or a "write-o" but then nobody would know what i meant...
@@trueriver1950 good one 😁
(1,0),(-2,-3)
Just put x=1 & y=0 what Messy here!
But why does 3.366666 work?
Haha made u calculate it XDD
YOU CAN SOLVE WHAT YOU THING TINKERING THE QUESTION AND YOU WILL COME WITH FLYING COLOR IN OLYMPIADS!!! ARE THEY FOOL????
dude very Seaxyyy Question
We proved that these values are integer solutions to the equation, but are the steps sufficient to demonstrate that these are ALL the integer solutions? Wouldn't we need to provide an extra step in a formal setting proving there are no others?
Nope the equation y(y+3)
I see, but, we arrived at that conclusion via means of inequality which is not as strong as means of equality. Is that approach still sufficient?
@@Div1nePiece That is the easy thing about inequality involved with INTEGERS. y(y+3)
I think he forgot y^2.
You wrote the problem wrong ?
6:44 Esse é um bom lugar para parar
И это отличное место, чтобы остановиться.
Bruh bruh this is harddd I am 7th grade
Pendyala education and entertainment channel👌
TOTAL NON SENSE HE IS SUPPOSING SOMETHING WHICH IS NOT AS PER QUESTION . IS THIS A METHOD FOR SOLVING A MATHS PROBLEM
Leider falsch. Sorry the solution is WRONG! Compare the left side of the blackbord with the right one ...
You failed. You solved the wrong problem.