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!
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)
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 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).
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!
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...
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
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!!!! 😄
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.
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.
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
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!
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
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
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
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!!!👍
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
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)
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.)
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.
@@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.
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
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?
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
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
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.
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
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
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
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
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?
