A floor problem from the 1998 IMO shortlist.

[@mxpxorsist]

Note that floor(b) could be zero. You can work around this as follows: note that if a,b is a solution then na, nb for integer n is again a solution. So take n>>0 such that nb > 1, then your argument shows na = nb, therefore a=b.

[@奕無反顧市長]

That I am just an English novice but I can fully understand this video is one of the amazing part of Math.

[@juliankern3528]

There is a small problem at the beginning, because the floor of b is zero if 0

[@zacharyjoseph5522]

But wouldn’t a large enough n make it bigger than 1?

[@linggamusroji227]

@@zacharyjoseph5522 he said a/b=floor(a)/floor(b), but floor(b)=0 for 0

[@pandas896]

You are a fool

[@jkid1134]

Well, Julian, you gonna explain the workaround too? It is not hard to account for this.

[@salim444]

@@linggamusroji227 I think but not so sure. you can pick n' suck that the floor(bn') is not zero and thus from the equation we get also that floor(an') is not zero. you define an' and bn' to be the new number. we follow the same logic and get that they are the same which says an'=bn' => a=b

[@SL-lv8et]

I have a nice number theory problem for you Mr Michael The problem: Find all integers (a,b,c,d) such that 1

[@mach2570]

Can you do a few lectures about how you approach an Olympiad problem, how does that vary through algebra, geometry, combi etc.? That would be very helpful, thanks.

[@Jacob-sj1jp]

Agreed

[@subnow4862]

I'd want this too. I love watching the way he solves problems and I want to do that as well.

[@DavidSavinainen]

11:00 Could you say that floor(-a) = -ceiling(a) where ceiling(a) is the smallest integer larger than a? If so, fl(-a) = -fl(a) implies -ce(a) = -fl(a), which means fl(a) = ce(a) which is only true for integers a? To me, it feels slightly more intuitive than setting a=m+r and then separating r=0 and r=/=0

[@antoniopalacios8160]

I think the key to this beautiful problem is in the 7:51 minute when you say, "Any number between 0 and 1 can be placed between two rational numbers that way." Actually, there are infinite intervals in which you can place r and then modify it to get the interval [0,1) without affecting the fact of its divisibility by p. In other words, your demonstration is not for a specific n, but for an infinite number of n,s. In this way the final result is correct.

[@zygoloid]

The summary at 5:30 contains a mistake. We have not shown that p|[na], only that p|[a]. This unproven claim is used at 9:05, so I don't think this argument is complete.

[@alexeydrobyshevsky3375]

But [an]/[bn] =a/b for all n, so they are all the same number - p/q

[@zygoloid]

@@alexeydrobyshevsky3375 Thanks, I think that is the step missing from the argument: all [na]/[nb] are equal, so are equal to p/q, so applying the same logic by which we determine that p|[a] we can show that p|[na] for all n. But those steps are missing from the argument as presented, and the missing steps are hidden by a board change.

[@sirgog]

Definitely excited to watch this, 98 was my first IMO.

[@linggamusroji227]

Where are a and b? Easy question. There, in the blackboard

[@helloitsme7553]

Hold on , isnt a,b in natural numbers in gwneral also a solution

[@OlivierMIEL]

7:00

[@aadityajha7502]

Nice, you are a great teacher.

[@kamalnehra4295]

Kindly make video on this number problem. And sir I have one more request ,make a patron account so that we can contribute you and you can create more amazing content Show that for any natural number n, between n^2 and (n+ 1)^2 one can find three distinct natural numbers a, b, c such that a^2+ b^2 is divisible by c. (1998 St. Petersburg City Mathematical Olympiad)

[@goodplacetostop2973]

12:46

[@MrRyanroberson1]

when i see this, i actually think of turning it into an unary metric kinda thing, because this is necessarily true: floor(bn)/b = floor(an)/a, and possibly even floor(an)/floor(bn) = a/b

[@thayanithirk1784]

Please do some problems on combinatorics

[@anjaneyasharma322]

a and b such that a and b are decimals with decimal part ending in 5. Just one type

[@leif1075]

You could also have both a and b equal zero right?

[@princeofrain1428]

What's interesting is that as a budding mathematician, I understood intuitively the answers to this. On the other hand, as someone who is not great at problem solving, I couldn't disprove other answers. Seeing the proofs made my brain hurt a little bit, as it felt impossible to even reach those conclusions to me. Any tips?

[@captainsnake8515]

I know that this is a late reply, but my number 1 rule is “don’t start with the proof.” Just try and get an intuitive understanding of the problem you’re working with. Only start with the rigor once you have an outline of the proof.

[@princeofrain1428]

@@captainsnake8515 How do you mean? I don't think I quite understand. Like, ensure I understand what it is I really am trying to prove and the implications of it?

[@aryadebchatterjee5028]

@@princeofrain1428 What he is trying to say is if a certain problem/proof is given then what you should do is try to understand what you are dealing with like the implication of it and if 'you suppose it to be true then what basic axioms or theorems can you break down into and once that is done and you have explored all venues then just write the proof by starting with the axioms and basically write all the steps in exploration in a more compact way and this is the methodology professional mathematicians use too. Sometimes you develop a sense of familiarity with the subject in question and then guestimate your way up. BTW fellow budding mathematician here, I am 14 and learning multivariate calc. , number theory, combinatorics, linear and abstract algebra, and real analysis and currently preparing for the Indian national mathematical olympiad right now what is your stage?

[@matematicaspanish8301]

In 8:42, why can you refer to the same original n as the n you know exists for r to be betwen 1/n and 2/n? Feels like you know there exists a natural number n for r to lie there, but you can't asume it will be the same n from the original problem.

[@toriknorth3324]

Since the end result was a contradiction he only needed to check that a single value of n failed to make the original equation true.

[@zzz942]

Because it is said to be true for any n

[@ittaloceara]

where can I found a prove of this?

[@antoniopalacios8160]

@@toriknorth3324 In reality, for that specific n, in addition to the contradiction, he obtains the trivial results from which he already started.

[@skylardeslypere9909]

Why can you just take the general eqn and set n = 1? Can you prove that there are no sols for n>1?

[@skylardeslypere9909]

Then again, I think this entire thing might've been general but I'm not sure

[@toriknorth3324]

There have to be solutions for every n, but in particular there has to be a solution for n = 1. He wrote out _a_ in a form that made it a solution for n = 1, then further constrained _a_ so that it would also be a solution for all n > 1.

[@Daniel-nl3ug]

Another way of phrasing the question would be "Suppose that a * floor(bn) = b * floor(an) is true for every possible natural number n. What are the possible values of a and b?"

[@skylardeslypere9909]

@@Daniel-nl3ug but that's not the question? It says find ALL a,b for ALL n, so if he finds all a,b for n=1, why wouldn't there be more (or less) a,b for n>1?

[@_Ytreza_]

@@skylardeslypere9909 The question is "Find all a and b such that a * floor(bn) = b * floor(an) for all n in N" The important part is "for all n in N", the equation must be true for n = 0, 1, 2 etc and with the same a and b (the ones that you have to find) If you find some solutions for n = 1, they may not work for other values of n so unless they are obvious solutions (like a b integers in this case) you have not solved the problem. But it's still useful to look at a special case (like n = 1) because you know for sure that all values that fail for n = 1 are not solutions

[@pandas896]

isn't a=b also solution.

[@OlivierMIEL]

10:10

[@atreyamajumdar9836]

I am floored!

[@thephysicistcuber175]

Reupload?

[@davidchapper4777]

Hi @Michael Penn, love your videos. I’ve recently stumbled upon a problem (I invented it, don’t know if its original). Let a and b be integers, and n be a natural number greater than 0, find all solutions to: a^(n+1) + a^n + 1 = b^2. I could only solve the case where n = 1, and been stuck on the case where n = 2 for 3 days. My best regards, David.

[@leastsignificantbit5069]

I'd say that n cannot be even (if we want to obtain non trivial solutions), if the n is even: n+1 is odd and it means we cannot factor LHS in such way that it is a perfect square. But that is true only for cases when a is not a perfect square itself. And is also not divisible by perfect square.

[@leastsignificantbit5069]

Trivial solutions are ofc (a, b) = (0,1) and (0,-1) and N being any natural number. Edit: after pondering some time, I stumbled across idea of using Vieta Jumping (it gave me all the solutions I have found so far and some new). I think there is a chance that it covers all the possible solutions to this problem.

[@davidchapper4777]

Least Significant Bit yeah, that’s the problem I found when trying to solve a^3 + a^2 + 1 = b^2

[@davidchapper4777]

I tried to factor like this: a^2(a+1) = (b-1)(b+1) but it didn’t get me anywhere

[@davidchapper4777]

SPOILERS: The only solutions for a^3 + a^2 + 1 = b^2 are: (a,b) = (0, +-1) , (-1, +-1) and (4, +-9), But cant seem to prove why there aren’t anymore solutions. One thing is trivial, b is odd.

[@FisicoNuclearCuantico]

@Michael Penn Best regards. :D

[@cobokobo2115]

another floor . Fractional part function problem homemade

[@ramk4004]

Thank you sir

[@U9191-e6s]

What if n is not equal to 1 !!

[@nikhilkushwah3933]

Hi

[@MrArmas555]

++

[@ertyrtyui469]

n€N*