Indian though, it is like half African. Hardest math question in African, what would that be?

@@krishna-vb4ok India is the place where people literally worship cow.

@@AbirInsights Hindus do not worship cows, but a majority refuse to eat all meat, including beef. Cows play a central role in the history of Indian husbandry and culture (indeed, it is perhaps impossible to have a complex society without large domestic animals), and dairy products like milk and fresh cheese are staples of their diet. So the refusal to slaughter cows (which offer more alive in the long run) is seen as a first step toward Hindu vegetarianism, which is based on a principle of non-violence and belief in the soul of every living thing. So over time we have seen cows increasingly venerated in Hindu society, and they are important in certain Hindu celebrations. But no Hindu believes cows are divine in any sense or "better than people" or anything like that, and they do not worship them. Other Indian religions like Jainism also promote vegetarianism and have varying levels of respect for cows, but none of them include worship of cows. The special place cows have in Indian society is perhaps in some ways analogous to the special place dogs have in Western society, where people typically find the slaughter of dogs for meat as far more repugnant than of other animals. The reasons are different (cows aren't pets in the same way as dogs), but the result is the same.

@@EebstertheGreat Thanks for that much needed clarification

@@AbirInsights I didn't know people has this kind of myth about India set in their mind.

Put a big flag on the thumbnail and watch the nationalists arrive.

@@Banzybanz no, he hid the square term. I just came to look at the question, but ended up watching the whole video. If I had seen the question, I wouldn't click on it because I knew the answer already, but didn't knew the proof.

@@stv3qbhxjnmmqbw835 he didn't hid the square term instead of writing "square" he symbolized square as ■

@Tangent of circle. Nationalist have no place in the modern world,All the Extremist party are nationalist,Free Kashmir

@Tangent of circle. Lol you have to be trolling?Nobody with alt right views would come open and tell that their Extrimist if you are really a nationalist then you can take the middle finger of mine,Stupid nationalist modick suppoter

Thanks Michael for responding

Exactly this one is a little tougher than the general one

Usually 8th and 4th (last problems per day) are ether takes half of time or none depending on can you solve it or not I think there just above 4 hours per day so like 1,5 hours on solving and 0,5 on writing or viseversa. Anderstending can you solve it or not is the hardest and most important part, and if you can solve it you should know if you can find a way to solve it in rational time. You don't have to solve everything you just need to solve most.

Also tbh it pretty easy problem, I think there must be 5 problems per day and it is 3rd per second day (basically last is coffin hard and 2nd from the end is just tough, so you should just take others as free ones) it also seems that 2^0+3^1=2^2 is a solution too, but I may missed the term in technical English that is for natural numbers and another for whole numbers starting from 0 In second case there probably family of roots where power of 2 is just 0 so we need to insure that there is fixed amount of this solutions (2^0 is 1 base 4, that may cause something in second part of solving to go wrong, basically can reduce amount of points for number by half) I think it easyly fixable with this solving, even if I am right, but most likely I just missed a word in problem (such missunderstanding cost me a lot..)

I actually wrong m, n are natural, so it is absolutely clear done number. I just focused on my solution a bit to much

I got lots of requests for this problem. Everyone seems to really like it!

@@MichaelPennMath Yea, I really enjoyed how you get to do the same thing over and over to solve the problem :)

That's how I figured it!

Or he could stop at 16:30 by noticing no two powers of 3 are separated by 2, besides 3^0 and 3^1. The solution in the hint is the smallest solution.

@@erikb3799 and only solution

That's how I solved it! Great approach

Yup. That is exactly how I solved it too.

I'll put it on the list.

Bom ver um companheiro brasileiro aqui kkkk fiz a obm ano passado mas real essa tava difícil

Bom saber q n sou o único

Tive dificuldade até pra entender essa questão no dia

Boa sugestão parceiro, vou esperar ansioso pela solução dessa questão

Well, actually, if we look at the wording used on the official test paper, it says 'positive integers' only, not natural numbers(you can check on AOPS), really unsure why michael changed the wording

Can't agree with you Daniel. I studied always that the set of natural numbers start from 1 and go on and on.

Tech Toppers i'm with Daniel, set of natural numbers N starts at 0, and set N* starts at 1. I think that we shouldn't use the terms natural numbers, because each country has their different definition for that.

The standard ISO definition has 0 as a natural number. The set theoretic construction of the natural numbers requires 0. Almost all modern number theorists regard 0 as natural. The Indian examiners who set the paper said ‘positive integers’ . Yet here we are again,

@@exacerbated no , I was taught zero is no natural numbers

as an engineer, i think it is in fact more elegant to solve it using the same thing again and again

It's insane to solve it in an exam. Who would have the sanity levels to really believe that this approach is going somewhere and spend precious time of the exam? If there is no elegant solution, I guess no one would invest time to try it.

@@michaildokianakis5725 If it was asked in olympiad in INMO or International Maths Olympiad then they will have 3 hours for 6 questions

@@michaildokianakis5725 I think the solution presented is pretty similar to what most people sitting this who manage to solve it would come up with. If you have some experience at doing Olympiad questions, one of the main skills you pick up is the skill of being able to quickly tell if a particular approach is likely to work. In this case I think if you're about imo bronze level then it's pretty easy to tell this kind of method would work, without even having to get out some paper to make some notes. Personally if I was doing this question, I'd do it the following way, which is similar to the way in the video but just a little different (this proof is pretty close to what I'd write for an Olympiad, only thing I'd change is remove the comments in brackets and add a little detail about how to trace back the steps). First let x^2 = 2^m + 3^n. If m = 0 then by Mihailescu's theorem, this has no solutions if n > 1. (If you don't want to use Mihailescu's, then instead the factorisation (x - 1)(x + 1) = 3^n can be used). Then you just need to check whether n = 0 or 1 are solutions, and we find that the only solution for m = 0 is 2^0 + 3^1 = 4. If m = 1 then we get x^2 = 2 + 3^n. There are no solutions for n > 0 because this would imply x^2 is 2 mod 3. Clearly n = 0 is also not a solution. Therefore there are no solutions for m = 1. The last case is when m is greater than or equal to 2. In this case by using mod 4 we find that n is even, so let n = 2a. Then we can write the equation as x^2 = 2^m + 3^2a => (x - 3^a)(x + 3^a) = 2^m. Then let x - 3^a = 2^y and x + 3^a = 2^z. By subtracting these we then find that 2(3^a) = 2^z - 2^y and then dividing both sides by two gives 3^a = 2^(z-1) - 2^(y-1). Note that y = 1 otherwise the right hand side is even, whilst the left is clearly odd. Therefore this equation can be written as 3^a = 2^(z-1) - 1. If a = 0 then we get the solution 3^0 = 2^(2-1) - 1 and then tracing back the steps, this yields the solution 2^3 + 3^0 = 9. If a = 1 then this gives the solution 3 = 2^(3-1) - 1 and tracing back some steps gives the solution 2^4 + 3^2 = 25. If a > 1 then the equation has no solutions by Mihailescu's theorem (again you can avoid Mihailescu's if you want by using a factorisation). So overall it's not too hard to come up with and not too hard to realise beforehand that the method will work. It involves a lot of very standard tricks which make it pretty easy to do if you know them. One trick is to deal with small cases first in order to use some modulo arithmetic and the other main trick is using difference of two squares factorisation a lot. Using Mihailescu's theorem is also handy to make the written proof a little bit shorter, but not knowing it probably wouldn't hinder you too much in finding a proof. Overall if you have a fair bit of experience then a question like this might take a few minutes to solve, and then a little while longer to write up a proof, maybe 15-20 minutes, but in most Olympiads this is a very reasonable amount of time, and in fact it probably wouldn't be a problem to spend a while longer than that, even an hour or two working on the question probably would be fine in an olympiad. In the international mathematical Olympiad, you get 4h30 to solve 3 questions, but a significant proportion of people only solve 1 question.

and the same goes for when n=0, another possible solution is m=3,n=0 (2^3+3^0=8+1=9=3^2)

but you know that 's cheating isn't it

@@zouhairnouri6863 it's intuition as the person mentioned.. since we can't prove fermat's last theorem in the exam.. I think it's safe to assume that each component on the RHS can't be of power>2.. and from the video Pythageorean triplets visualised by 3blue1brown(awesome channel btw).. one can easily say one possible pair of m&n is possible..

These are solutions of you consider 0 a natural number. I take the more standard definition that N is made up of positive integers.

@@MichaelPennMath Standard in the US ;o) In Europe, we have ℕ*={1,2,3... and ℕ={0,1,2...

That is very interesting and explains a lot when I see talks by Europeans. Anyway, the problem originally read "all positive integers". I shortened it to \in\N to take up less room on the board. Also, how do you get the \mathbb{N} ie the blackboard bold N in the comment?

@@MichaelPennMath This is a unicode symbol. I have a list of them on my phone ready to be copied/pasted. Here are a few: Fractions ⅟ ½ ⅓ ⅕ ⅙ ⅛ ⅔ ⅖ ⅚ ⅜ ¾ ⅗ ⅝ ⅞ ⅘ Die ⚀ ⚁ ⚂ ⚃ ⚄ ⚅ Chess ♔ ♕ ♖ ♗ ♘ ♙ ♚ ♛ ♜ ♝ ♞ ♟ Parallelograms ▱ ▰ ⏥ ⏢ Mathbb 𝕒 𝕓 𝕔 𝕕 𝕖 𝕗 𝕘 𝕙 𝕚 𝕛 𝕜 𝕝 𝕞 𝕟 𝕠 𝕡 𝕢 𝕣 𝕤 𝕥 𝕦 𝕧 𝕨 𝕩 𝕪 𝕫 𝔸 𝔹 ℂ 𝔻 𝔼 𝔽 𝔾 ℍ 𝕀 𝕁 𝕂 𝕃 𝕄 ℕ 𝕆 ℙ ℚ ℝ 𝕊 𝕋 𝕌 𝕍 𝕎 𝕏 𝕐 ℤ 𝟙 𝟚 𝟛 𝟜 𝟝 𝟞 𝟟 𝟠 𝟡 𝟘 You can easily find them (and many, many more) online, the thing is that websites and apps do not necessarily support all of them. I never used some of the ones I just copied/pasted on the YT comments section, so some funny characters might be displayed instead.

And for math formulas, I also have a list of things I wrote in the past: ∂f/∂t = ∂²f/∂x² ∀ x ∈ ℝ ∫ e⁻ᵗ dt

He just tell us the "mathematician's way" that can be used on the general problems

I think this is also a valid way, in fact its pretty elegant

Chabbi!!! Remember me?

Wooww!! Can't believe we met here. Chabbi, why did you disappear all of a sudden?

@@saminhaque13-52 what ??

@@steveleekyon1212 -_- I'm with u on plus lol

@@saminhaque13-52 what do u mean by disappear?

Quadratic residues and quadratic reiprocity are way beyond IMO maths.

Polacy, złączmy się tutaj xD

no siema

I've put it on my list!

From PFTB? Seems like I have seen this somewhere before

@@shantanunene4389, yeah, i think i've seen it on Brilliant

Where is it from

Also, I think there's a typo. No number of the above form can be written as 2 cos(2 k π/m)

The set of natural numbers is {1, 2, 3, 4, ....}, the set of whole numbers is {0, 1, 2, 3, 4, ....}, so if this problem were using the set of whole numbers instead of natural numbers than there would be two solutions.

@@dclrk8331 I would prefer such a clear definition for natural numbers, but you shouldn't be so sure. There are different standards, see: mathworld.wolfram.com/NaturalNumber.html

Good observation.

Lior orz

@@II_xD_II How do you keep finding me smh

What is 'n' here? Is it true for all natural numbers? And what do you mean by 'ordering'? Strictly ascending / Strictly descending or either? Because if you consider trivial case n = 1, n^2 + 1 is 2 persons and n + 1 is also 2 persons.... So for any given 2 people, we can always have a subset of 2 people (the subset is same as the superset) and any ordering of age of the two people will also give ordering of the height, but it can be EITHER ascending OR descending.... Can you throw in some specificity?

@@050138 yep natural numbers and either ascending or descending.

They just need to be ordered in some way.

@Adam Romanov Wow! Didn't know such a theorem exists! I have exactly followed the steps backwards as you mentioned.... And arrived at the theorem statement on the lines of what you wrote.... Which I had been trying to prove since last two days.... The thing I am trying to prove goes like this.... Given a sequence of (n^2 + 1) numbers which are increasing.... Irrespective of whatever order you may arrange the numbers of this sequence, you will still have (n + 1) of the numbers either strictly increasing or decreasing.... It will be either increasing or decreasing, not both and it will depend totally on the way the terms of original sequence is jumbled

@Adam Romanov thanks for your reply. It's from Scuola Galileiana Unipd Admission Test. (Italian university math undergrad admission test)

..... "tangent to circumcircle of triangle HBC interesects omega....." But tangent at which point?

I like this problem. I will give it a go. btw: I have a collaborator who is originally from Romania!

I think the general term would be 2^n*p+2^{n}-1, now settings n=k(p-1) implies the whole term is divisible by p.

Very interesting question! Was able to solve it using Fermat's Little theorem.... More over, that 'i' for which it is true is the prime 'p' itself, for all odd primes p.... Let's represent nth term as a[n] Since a[1] = p and a[n] = 2*a[n - 1] + 1 we have a[2] = 2*a[1] + 1 = 2p + 1 a[3] = 2*a[2] + 1 = 2*(2p + 1) + 1 = 4p + 3 So we can see that in the nth term, the coefficient of p will be 2 multiplied (n - 1) times and the constant term will be sum of all the powers of 2 from 0 to (n - 2), or a[n] = 2^(n - 1)*p + (2^(n - 2) + ...... + 2 + 1) or a[n] = 2^(n - 1)*p + [2^(n - 1) - 1] or a[n] = 2^(n - 1)*(p + 1) - 1 or generally or a[i] = 2^(i - 1)*(p + 1) - 1 Now, Fermat's Little theorem states that if 'p' is a prime and 'a' is a natural number relatively prime to p (gcd of (a, p) = 0), then a^(p - 1) = 1(mod p) So for every odd prime (all primes except 2), as 2 is relatively prime to p, we have 2^(p - 1) = 1(mod p) We also have (p + 1) = 1(mod p) obviously Multiplying the two congruences, 2^(p - 1)*(p + 1) = 1(mod p) which means 2^(p - 1)*(p + 1) - 1 is divisible by p But 2^(p - 1)*(p + 1) - 1 is nothing but a[p] So we proved that for all odd primes p, a[p] is divisible by p and hence not a prime If p is 2, then we have a[1] = 2 and a[2] = 2*2 + 1 = 5 which is an odd prime, for which we already have proven that there will be an i such that a[i] is not prime QED

@@angelmendez-rivera351 Thank you! ☺️

Yes, once you know that both exponents are even it is clear that you have a primitive pythagorean triple and can use the parameterization you mentioned. This is a very nice and potentially shorter solution. I didn't approach it this way to keep the argument more self contained.

@@MichaelPennMath if I were you I would have used the same way😂. In our junior summer and winter MOP camps, we generally use mod and factorization while solving diophantine equations just like this. Also in competitions and olympiads, the official solutions are based on this method!

Of course there is also 3^1+2^0