Main videos are at: kzread.info/dash/bejne/lK2gqNx7qN23abA.html and kzread.info/dash/bejne/gJeVu6eahrenZMo.html

Proof of the Goldbach conjecture is left as an exercise for the viewer.

@m3sam

2 ай бұрын

😂

Regarding #3, it's funny I had a dream a year ago exactly about that problem. In that dream I thought you'd sometimes need a lot of steps to add and subtract squares until you land on a number you need. I woke up and thought: wow, what an unusual dream, looks like an interesting question; only to be immediately disappointed because I realised all odd numbers are a difference of two consecutive squares, which means you'd never need more than two squares to get any odd number you want, or more than three squares to get any even number (and then there's one exception, number 2, which needs four squares).

If you can solve the first one, try this one, it's a nice easy exercise: Let f(s) = sum 1/n^s, n = 1 to inf, for Re(s) > 1 and let F be the analytic continuation of f to the entire complex plane. Solve the equation F(s) = 0.

@jakobhablitz1429

Жыл бұрын

If you can solve the first share with the world bro. Pretty sure that's the unsolved one!?

@dabluse3497

Жыл бұрын

@@ephemera2 he’s just making a joke chill man 😭😭😭

@ephemera2

Жыл бұрын

@@dabluse3497 I know he's trying to but it makes no sense and the biggest problem in this world is how rare logic is and in turn people listening to those who sound smart but speak nonsense and lies i e politicians. His "joke" is like one of those "what's the difference between" jokes but he only gives a single object for comparing and no punchline because he's trying to tell a joke that he didn't understand but everyone else around him laughed when he first heard it so he tried to memorize it and doesn't even understand that you need two objects to compare the difference. And as you can see someone, aman, wants to understand something he doesn't completely understand at the moment and even try to give it a shot with an answer. People like this are toxic just like Faith Healers psychics and mediums. And now we have a world where half the population believes in astrology and horoscopes and determine what they should and shouldn't think about logically based on social taboos and social norms. In fact, since the entire world is run based on the decisions made and the words spoken by those who try to sound smart whether they are or aren't, so that people listen to garbage instead of thinking, the sway of ignorance has spread so far out of control that our scientific community is affected by it demonizing the truth and propagating what makes money.

@nathangreene3

Жыл бұрын

@@ephemera2 Dude, you might need medication or therapy. This is the comment section of KZread, not a math journal. Lighten up, have some composure, and let other people participate. You're embarrassing yourself in trying to put down someone for misrepresenting the Riemann hypothesis. Do you realize that?

@benjaminflorian6941

Жыл бұрын

@@nathangreene3 I believe the first one is false, but unfortunately the YT comment section is too small to fit the counterexample.

SPOILER ALERT: 1. Open problem. This a famous conjecture called goldbach conjecture which states that all even numbers greater than 2 can be expressed as sum of two (not necessarily distinct) primes. It could not be proven by induction (at least obviously) yet because of the "randomness" (see PS) of primes. 2. Wrong. It works for all n from 0 to 39, but it doesn't work when n=40 where n^2+n+41=1681, which is exactly the square of 41. In fact, for all positive integer d, there is no way that n^2+n+d is prime for all positive integers n since it breaks when n=d-1. Induction doesn't work here because it is hard to relate the consecutive terms with primes relationships. Fun fact: The statement n^2+n+d is prime for nonnegative integers n

@krizeot

Жыл бұрын

Just flip the sings for getting all negative integers :)

@SmileyMPV

Жыл бұрын

direct proof for 3: odd integers: for n=1 we can take n = 1^2 for n≥1 we can take 2n+1 = (n+1)^2-n^2 even integers: for n=2 we can take n = 6^2-5^2-3^2 for n=4 we can take n = 2^2 for n≥3 we can take 2n = n^2-(n-1)^2+1^2 this also shows you only ever need at most three different squares with 2 = 5^2+11^2-12^2 we see that we also only ever need one negative

@Vodboi

Жыл бұрын

For question 2 I just tried replacing n with 41, which then obviously becomes divisible by 41: 41^2+41+41 = 41*(41+1+1) = 41*43. Didn't know that it always breaks for d-1 as well though, thats interresting!

@pieffe8

Жыл бұрын

@@SmileyMPV You are supposed to sum/subtract squares starting from 1. So you cannot say generate 4 by taking 2^2, because the sum doesn't contain 1^2. I had this problem at a national math olympiad competition many, many years ago. I could not solve it, but I was definitely not trained for it.

@SmileyMPV

Жыл бұрын

@@pieffe8 I guess that does make the problem more interesting. But as far as I can tell, this was not explained in the video.

Zvedelina is awesome!)

First one sounds like Goldbach's or similar, unproven Second fails on 41 Third is proven true because, - 1 is a perfect square - 2 = 16 - 9 - 4 - 1 - 3 = 4 - 1 - 4 is a perfect square - all odd naturals [edit: greater than 1] are differences of consecutive perfect squares - all of the odds which are greater than 3 do not need to depend on 1 - all evens greater than 4 can just be made by adding 1 to those odds

we can never have too much of dr. stankova

The polynomial question is brilliant - it turns out there's a mathematical reason why n²+n+41 is prime for every n from 1 to 40. Its not just a coincidence! Even more interesting, the trick used to make this work is NOT generalisable, this polynomial is the largest such able to use this trick. If anyone knows more about this I'd love to know. Obviously one can find polynomials which are prime for arbitrarily many inputs, but I'm more interested in number theoretical derivations. EDIT: Holds for 0

@n0tting

Жыл бұрын

For n² + an + b to produce a prime, starting with n=0, b must be a prime, that's all I can tell. n² -79n + 1601 produces 80 primes for the consecutive values 0 ≤ n ≤ 79.

@Czeckie

Жыл бұрын

what's the mathematical reason for the polynomial to give primes? (just fyi, it doesn't work for n=40)

@gammaknife167

Жыл бұрын

@@Czeckie Ah thank you! Apologies, I should have said prime for 0

@Czeckie

Жыл бұрын

@@gammaknife167 thanks! this sounds interesting. I won't force you to give proof in yt comments. It seems to me that with this setup it should be an ideal norm argument but I'm way too rusty to work it out.

Best cliffhanger since Lost season 1 finale

What a lovely series of videos :) Always grateful for Numberphile and the amazing presenters

If you decide Goldbach, it will be by showing something about prime gaps coming from opposite ends of 2 and 2N. Specifically, you'll be showing whether, given the set of primes P from 2 to N, there can be some N such that a gap is always found at 2N-P. This is not patently ridiculous, because primes are denser nearer the origin. Whoever solves it will probably end up showing that, if we're covered with the first K primes, then we're also covered with K+1 primes, where the Kth prime is taken to be 2N+1. So, whatever they do will no doubt involve lemmas about arithmetic progressions. Maybe someone studies something about prime gaps, and notices that the maximum gap size is (or isn't) dictated by our ability to cover the intervening even numbers with Goldbach pairs? I suspect the reason we haven't solved it is only that we haven't figured out the right intermediate questions to ask.

SPOILER: The first one is the Goldbach Conjecture, so it's the unknown one, and I'm sure that if it could be done by induction, someone would have done that 100+ years ago. I think that the "gazillion" that it has been checked up to is something times 10^18, last time I heard about it. The middle one is provably false, because 41^2+41+41 is divisible (obviously) by 41. So, by the process of elimination, that means the last one has to be the true one that is provable by induction -- but I haven't tried to do it.

• 3rd prob. is definitely True because square numbers (except 0) can be written as a sum of sequential odd numbers b/c (k+1)² = k² + (2k +1). Ex: 1² = 0² + 2*0 + 1 = 0² + 1 = 1 2² = 1² + 2*1 + 1 = 1² + 3 = 1 + 3 3² = 2² + 2*2 + 1 = 2² + 5 = 1 + 3 + 5 4² = 3² + 2*3 + 1 = 3² + 7 = 1 + 3 + 5 + 7 ... So , if n is odd we just need to choose the right square numbers and subtract them (except for 1 which is trivial 1 = 1²). Pick n, equate to 2k + 1, i.e. n = 2k + 1 and solve for k. Then n = (k+1)² - k². If n is even then n-1 and n+1 are odd. Find the square numbers that subtracted give n-1 or n+1, and then add or subtract 1² respectively. For instance, n=11 = 6² - 5² and n=13 = 7² - 6². To do n=12 we either add 1² to 11 or subtract 1² to 13. The "hard" one to do is n=2, but that was shown in the video how it can be done. • 2nd prob. is definitely False. For instance for n=41 the all expression is divisible by 41, therefore is not a prime number. • So, the 1st prob. must be the one we don't know if its true or not.

@adamqazsedc

Жыл бұрын

The first prob is Goldbach's conjecture, one of the most famous open question in number theory because of its involvement with Euler

Spoilers for #3, a much simpler direct proof Because every square is a sum of sequencal odd numbers (eg. 1+3+5+7 =16, which can be proven graphically), you can find two neighbouring squares such that their difference is any arbitrary odd number (eg. 16-9 =7). You can add or subtract 1 to get to its adjacent even number. You can then construct the special case of 2 as shown in the video. Therefore, you can construct every positive integer from squares.

@MrRyanroberson1

Жыл бұрын

the proof was that you could get every number from using ALL squares from 0 to n

I can't prove that every even number is the sum of two odd primes, but I can prove the inverse: That every even prime is the sum of two odd numbers.

@KRiku1000

Жыл бұрын

1 + 1 = 2

@MichaelWarman

Жыл бұрын

What? The sum of any two odd numbers will be even, yet no even numbers (except 2 are prime), therefore no primes are the sum of two odd numbers (except 2).

@AnOldGreyDog

Жыл бұрын

@@MichaelWarman Precisely - _every_ even prime is the sum of two odd numbers. The fact that there is only one even prime makes it easy to check. This crops up in Gödel, Escher, Bach by Douglas R. Hofstadter.

@PhilBagels

Жыл бұрын

@@AnOldGreyDog Yep. That's where I stole it from!

@AnOldGreyDog

Жыл бұрын

@@PhilBagels a brilliant book. Something worth stealing on nearly every page.

n² + n + 41 is the false one by counterexample n = 41

@Hiltok

Жыл бұрын

The first counter example is 40. But for 1 to 39 it works like a charm.

Is the problem with squares to express a number as a sum or difference of squares starting at 1, or of any squares? We know already that any positive integer is a sum of four squares, some of which can be zero.

Or does #3 mean you must use all squares, 1, 4, 9, 16, etc,.. up to some point, and can't just pick whatever squares you want? I.e. you can't get 15 simply with 64-49?

@msolec2000

Жыл бұрын

Yeah, and also -1+16

this one is easy 2n= p1+p2: no one knows //// x^2+x+41: false //// +,- distinct perfect squares: true

Aunt Zvedelina Petros 😂

n^2 + n + p can give you arbitrary long sequence of primes but never an infinite sequnce. Prove that.

The one that's true and proofable, can proof it using deduction?

@Hiltok

Жыл бұрын

The provable one is amenable to an Inductive proof. [Note that Zvedelina warned not to spend too long on the first problem. That's the one that's a conjecture.]

I think the higher the even number , the many more ways there are to add to primes to sum to it , but the proof ?....no. Also I can see immediately the next problem fails for 41, but I haven't checked all the lower numbers. Third one is definitely true by induction, I see it ,but I won't embarrass myself by trying to write it down here.

1.) ??? 2.) False by counter-example: *P(41) = 41 * 43* is not prime 3.) True by induction below ----------------------------------- _Proof 3.)_ (by induction) _Induction hypothesis:_ Any *n ∈ ℕ* can be written as a signed sum of distinct squares. _Base cases:_ *1 ⩽ n ⩽ 4* *1 = 1^2* *2 = -1^2 - 2^2 - 3^2 + 4^2 //given at **5:51* *3 = -1^2 + 2^2* *4 = 2^2* _Induction step:_ *n ⇝ n + 1* Assume the hypothesis is true for a fixed *n ⩾ 4* . For any *k ∈ ℕ* we can rewrite *n + 1* as *n + 1 = (n - 3) + k^2 - (k + 1)^2 - (k + 2)^2 + (k + 3)^2* Notice *n - 3 ∈ ℕ* , so by induction we can rewrite *n - 3* as a signed sum of distinct squares. Choose *k^2* to be larger than the biggest square in that signed sum, and the equation above rewrites *n + 1* as the signed sum of distinct squares *q.e.d.*

@SmirnovSB

Жыл бұрын

Represent four as *4 = −1² −2² + 3²* and you can say that: Any *n* ∈ ℕ can be written as a signed sum of _consecutive_ distinct squares from *1²* up. It's much nicer imo.