a prime circle
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- Опубліковано 31 гру 2024
- We look at a nice problem from a Finnish math contest.
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![What is the intersection of all open intervals containing [0,1]?](/img/n.gif)
In the part 8:41 there's a fallacy.
The factorization by x+1 only holds when n+1 is odd(n is even), whereas if n is odd, there's no particular prime that divides all 2^(n+1)+1. (It is clear by the example; Fermat Primes, form of 2^2^n+1)
However, we can show that n should be even. If n is odd, 2^(n+1)+1 is congruent with 2 on modulo 3. A perfect square, (p^m)^2 should be congurent with 0 or 1 on modulo 3. Hence it contradicts, so n should be even.
i was just about to comment that !
thank you for explaining why n is even
4:07 He should've moved p^2m to the right side which would avoid the error and lead to a quicker solution.
Mersenne primes are of the form 2^n - 1 not 2^n + 1
x^n - 1 = (x-1)*p(x) for some polynomial p(x), but if x=2 then (x-1) = 1 then the mersenne number is not necessarily prime
What he showed in the video is accurate (but yes, only for odd n)
@@rohitg1529 yep thanks for the point!
I was meant to say Fermat Primes, by the way. But since there exists some primes, there are no common factor that divides them all
we are getting closer and closer to the Fermat's Last Theorem
😂 The journey was Single for Wiles Andrew 💔🤫
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Yeah, like having walked the first mile of a thousand miles.
Holy Diophantus, professor, what a way to hide a 3-4-5 triangle!
Thank you!
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At 11:25, a simpler argument would be to ask, which two powers of 2 differ by 2? Why 2^2 and 2^1, course :)
In fact that trick can be used as early as 7:30 and it would shorten the video by a few minutes.
@@replicaacliper I came here to make this exact comment. Well done, mathematicians of UA-cam!
I noticed right off that this is a Pythagorean triple with a and b both having one prime factor, if multiple times. As all PTs except one have at least one multi-prime composite leg, it's gotta be that triple.
8:21 the factorization is true only if n is even (and thus n+1 is odd).
You could've factored p^2m-1 into (p^m+1)(p^m-1), after that since they must be both powers of 2 and they differ by 2, they can only be 2 and 4, so p^m=3 and so p=3 m=1. Then 2^(m+1)=3^2-1, so m=2. Subbing back into the original equation you get r=5.
At 7:20 you could've cut out a step by factoring the difference of squares right then and there. You'd get that p^m is trapped between two powers of 2 and thus p^m=3.
Idea for a livestream: solving contest problems that you're seeing for the first time. I would love to see the process in action!
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8:20 You are assuming that n is even(also the plus/minus 1 should just be +1 but whatever). I wonder if we can justify that assumption.
You didnt show that n+1 must be odd in order to do that factorization at 8:41
4:38 Good Place To Be
13:37 Good Place To Stop
Leet
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@@Mystery_Biscuits You’re a man of culture as well
I started with the fact that if a^2+b^2=c^2, with a, b and c integers, then there are integers x, y, and z such that a = x^2-y^2, b = 2xy and c = x^2 + y^2. It is easy to see that z = 1 in this problem. Since a or b has to be a power of 2, it can only be b, so x = 2^s and y = 2^t. From there on it is easy.
A strange corollary to this result - really, to the fact that there is one singular solution - is that no power of 2 besides 4 can be a leg of a Pythagorean triple where the other leg is a prime number.
I'm not sure I understand what's being said around 8:41
We factor 1 + 2^(n+1) into 3*N for some integer N
Take the case n = 3 : then 2^(n+1) = 2^4 = 16
However 16+1 cannot be factored as a multiple of 3
So i'm not sure I understand why we can factor it that way
Right. It only works for odd exponents, so for even n. You could do:
2^(n + 1) = p^(2m) - 1 = (p^m - 1)(p^m + 1)
Now, it is impossible that both p^m - 1 and p^m + 1 are divisible by 4. Therefore,
p^m - 1 = 2
This immediately gives you p = 3 and m = 1.
Very nice Michael you're doing a great job.I love this kind of things which you can evaluate the radius of circle with insufficient information .keep going man 👍👍
2:38 - Why is r odd?
I got schooled watching the video, but even more so by reading the comments. I'm really glad that this much free educational material exists on UA-cam. Thank you to everyone here!
This problem is very trivial due to the wording since it implies upfront that there is only one solution. 3-4-5 triangle comes to mind immediately where 4 is 2^2. If the problem said find all solutions then what you did was a solid method.
Depends on how you interpret the wording. I always assume that you still have to prove that there's exactly one solution.
8:10 wrong. This factorization works only for odd n+1
I agree
n+1 is odd it is 3
The results is actually a trivial consequence of the Catalan conjecture (now a theorem) that the only solution of x^a - y^b = 1 for x,y,a,b being positive integers is 3^2 - 2^3 = 1
Proving the Catalan conjecture is another matter though :P
I found myself thinking of the usual hint that such problems have very few solutions, often just one. We have one solution by inspection. Is it unique?
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A satisfying ending.
Question: For positive integer n, let S(n) denote the sum of the digits of n.
E.g S(24) = 2 + 4 = 6
S(92) = 9 + 2 = 11
S(200) = 2 + 0 + 0 = 2
Find the smallest positive integer satisfying, S(n) = S(n + 864) = 20
Source: Aime 2015 I Problems/Problem 8
The disdain for the idea that 0 is a natural number is so strong that he prefers to call the set of odd numbers 2Z + 1 instead of 2N + 1.
Actually he doesn't have any disdain for the idea. In fact, he cares so little that he writes N often because he's just lazy, but has said many times that he doesn't care whether or not 0 is an element of N.
@@PubicGore I've seen him say as much before, but I don't buy it.
The old difference of squares trick !
Perhaps one of the hardest geometry problems you're doing so far... 😍😍😍
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@@quickyummy8120 Thanking you, Boss 🙏 🙌 ♥
It's not really a geometry problem. It's a number theory problem dressed up as geometry.
13:22 "That Finnishes the solution to this problem"
I immediately thought about the Pythagorean triple
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And that's a good place to like
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Toward the end you started skipping a bit. I think you should have shown how you got x=1 and y=2, as well as actually plugging things in to get n=2.
For the former, you can note that 2^(y-x) - 1 must be odd, and thus is equal to 1. So 2^x=2, and thus x=1. Plug that back in, and you get 2^(y-1) - 1 = 1, so 2^(y-1) = 2. Thus y-1=1 and y=2.
Plugging in for the other you get that 3^2 - 1 = 2^(n+1) => 8 = 2^(n+1) => n+1=3 => n=2.
3-4-5 triangle. This is the first Math Olympiad question I have solved myself!
torille
Wait this isnt a review of highly acclaimed album "Soundtracks for the Blind" by Swans
Extraordinaire !!
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The requirement that r be odd is superfluous. If r were even, then both p^m and q^n would need to be even, which would require p=q=2, which is impossible.
p and q could both be odd, but there are no solutions for that case either.
Absolutely beautiful problem, where evennness and primeness of the number lets you discard all of the bullshit and just have fun.
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The intro is sick
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Finland👍
I thought to myself "it must be a pythagorean triplet", so could it be 3,4,5 5,12,13 7,24,25 8,15,17.... I checked whether 3,4,5 satisfied the conditions and it did.... Solved in 15 seconds!!!
YOU MUST SIMPLIFY MATHS OF ALL PHYSICS
If 3,4,5 is the only Pythagorean triple where both legs are powers of primes and the hypotenuse is odd, that should be enough to prove the problem without all the number theory
Yeah and how do you prove that?!
@@JoGurk some of the many known principles of Pythagorean triples are as follows:
1. Exactly one of a, b is divisible by 2 (is even), but never c.
2. Exactly one of a, b is divisible by 3, but never c.
3. Exactly one of a, b is divisible by 4, but never c (because c is never even).
4. Exactly one of a, b, c is divisible by 5.
The 3, 4, 5 triple is the only set of numbers that satisfies these principles and the requirement that both legs be powers of primes.
@@GabeKorgood alright, yeah. Sorry :D
The radius is also prime. 😁
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واصل.قناة رائعة.
Молим Вас прескочите тривијалне ствари у доказима а обратите пажњу на на идеју задатка. Поздрав Милан Ребић, Србија.
I try to do this myself and when I reach the point at 7:20 I just make use of the Catalan conjecture since I dont know what to do even I know the answer lol
To me it was obviously the 3,4,5 triangle from the start, but perhaps that's not the point.
You still need to determine that that solution is unique.
@@TJStellmach
Fair enough, thanks. :)
у отца пустая квартира, это значит не тждома
Any Swans fans here just for the thumbnail?
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WRONG!!! YOU MADE THE MISTAKE OF FACTORISING 2ⁿ+1!!! IT ONLY WORKS WHEN n IS ODD!!!
на тебе р
Second