Croatian Mathematical Olympiad | 2005 Q11.1

[@avivavital3769]

There are 2 little mistakes in the last board regarding the case that k=k. 2. When regarding the case of k=2 in the last line, you didn't plug it in correctly to the right hand side, the equation should be: 2l!=2+l!+m! => l!=m!+2, reducing that mod m, gives that 2 is congruent to 0 mod m which can only work for m=2, which is possible according to the first mistake. Plugging m=k=2 into the original equation gives: 2l!=2+l!+2 => l!=4 but there is no solution here.

[@demenion3521]

the first one is not really a mistake as m=k would give the same kind of equation that he got for the case k=l before.

[@demenion3521]

@David Schmitz but you can still use the same argument. if m=k, then reducing the equation mod l+1 also gives either 0=2 or -1=2 mod l+1. neither of these give a new solution

[@suryanshsharma2451]

@@demenion3521 yaa

[@antoniopalacios8160]

@@demenion3521 actually, he would have to check the k=m

[@alexey_burkov]

@@antoniopalacios8160 I think you are not right a little. As soon as k 2=0(mod m) -> m=2 And k=2, k=m and it is possible on this stage. Now we must check the equation with this numbers, and find out that l! must be equal to 4...

[@adrianfrauca8118]

The second part of Wilson's theorem has an extra case, which is n = p^2 for a prime p. That is proved by noting that 1 < p < 2p < p^2-1 (except for n = 2^2 = 4) and thus p^2 | 2p^2 | (p^2-1)!

[@NegativeAccelerate]

When you procrastinate maths homework to watch more complicated maths 😎

[@timurpryadilin8830]

17:10, it should actually be 2l!=2+l!+m!, which gives l! =m! +2, but this is impossible

[@hans-juergenbrasch3683]

No need to employ Wilson etc. By symmetry assume k=< l For k=k. If m>k, then all terms except the addend 1 are divisible by k+1>1, a contradiction. So m=k and l!=2+l!/k! requiring k>1. Similarly to before, this yields the contradiction k+1>2 divides the addend 2. Therefore, we must have k=l and get l!=2+m!/l! and therefore, l>2 and m>=l which yields m>l. Let d=min(m-l,l), then d! divides l! and m!/l! (note, that (m-l)! divides m!/l! using the binomial coefficient definition). So d! divides 2 and hence d=1 or d=2 smaller than l, so d=m-l. d=1 gives l!=3+l. Division by l>2 requires l=3 and this yields a solution k=l=3, m=4. d=2 gives l!=2+(l+1)(l+2)= l^2+3l+4. Division by l>2 yields l=4 which does not solve the equation.

[@prag9582]

Nice!

[@madcapprof]

The condition of n=p^2 for some odd prime p, also needs to be solved in the Wilson's thrm corollary.

[@timurpryadilin8830]

if p>2, then p < p^2 and 2p< p^2, and thir multiplication gives 0 mod p^2

[@madcapprof]

@@timurpryadilin8830 My comment was a suggestion for including this case. I know the proof and wasn't asking a question.

[@ittaloceara]

why.

[@sammyboy7094]

Because numbers of the form n=p^2 don’t have two divisors a,b such that 2

[@goodplacetostop2973]

18:30 and by the way, how would you say “That’s a good place to stop” in Croatian ?

[@thapakaji8579]

ok

[@birdbeakbeardneck3617]

the vid got 9 likes u got 5

[@thapakaji8579]

@@birdbeakbeardneck3617 he does the work of god dude...

[@marting5019]

"To je dobro mjesto za stati." greetings from Zagreb :)

[@lexyeevee]

4:35 this isn't possible if n is a square of a prime p, e.g. 9. the same idea does still work because p is necessarily greater than 2, so both p and 2p appear in (n - 1)! - which is the same reason 4 /doesn't/ work

[@zeravam]

9 is not prime

[@lexyeevee]

@@zeravam ...yes, it's the square of a prime

[@agfd5659]

Lol, like a minute after he had explained that proof, it clicked for me that he missed that and I decided to scroll into the comments to find yours.

[@jimschneider799]

As Aviv Avital pointed out, m is not necessarily strictly larger than k. Also, the equation that results from dividing through by l!, which is k! = 1 + k!/l! + m!/l!, implies that 1 < k! < 3, since 0 < k!/l!

[@backyard282]

Croat here, thanks for featuring my country :)

[@ajakowski]

I am not a Croat, but I got annoyed when I saw Abramović misspelled :)

[@Dubravko_Sabolic]

Man, you are practically a Croat. And you make some very good videos... Greetings from Zagreb. :-)

[@udic01]

13:57 m does not have to be strictly bigger than k. It could be equal to k (and the proof still stays the same)

[@SlidellRobotics]

for k=l, m!/k! being whole means m>=k, not m>k. It turns out in both cases where he uses the fact it doesn't matter. The first was the k! l! < .. < k! l!, which was still proven. The other was for k=2, where m=1 is actually less than k.

[@robertgerbicz]

You really don't need Wilson's theorem to prove that k!=2+m!/k! has no solution (other than k=3,m=4). If k>=3 and m>=k+3 then it has no solution because the left side is divisible by 3, but the right side gives 2 as a residue mod 3. That gives us few cases to check: k

[@Goku_is_my_idol]

Yes thats the way i did it too. k!=2+(k+1)(k+2) or k!=2+(k+1) This is for k>=3 Next put mod3 for both sides In first case we gotta have (k+1)(k+2)=1(mod3) which is impossible And next k!=k+3 for k=3 Which is the only solution We can easily check for k=1 and k=2 which gives us no solutions (I dont know wilsons theorem. If i had known maybe i would have used it. But this ques doesnt really require that)

[@raghaviyer3097]

@@Goku_is_my_idol me too!

[@ribozyme2899]

7:30 This is also possible without Wilson's theorem. Case m2k: m!>(2k)!>k!^2=2k!+m!, so 0>k!, which is a contradiction. (the only exception is m=1 and k=0, which isn't a solution)

[@lucassandleris4486]

When solving the case k=l, it might be easier to just take everything mod3. If k

[@Goku_is_my_idol]

Thats the way i did it too! (I dont know wilson theorem yet. Still a student)

[@themathsgeek8528]

@@Goku_is_my_idol nice

[@andrijaadamovic5129]

Very elegant solution. Btw that problem is from Croatian state competition 2005, not from the Croatian Olympiad 2005.

[@kasraafshar255]

One of the best olympiad question and solutions channels but just one thing at the end of this video , considering k = 2 makes that k!=2 not k!=1 that you wrote on ths right side ; but not such a big mistake because it gives that 2l!= 2 + l! + m! , so we have l! = 2 + m! That for l>=3 there is no solution because u can take left and right side congerate to each other module 6 but RHS will never be congerate to 0 module 6 so l is less than 3 and in that case we also have not any solutions . From a high school student from iran who is in love with math 🙏🇮🇷

[@Reidemeistermoves]

Great video! I'd love to see a video where you talk about topics in your research area

[@ДенисЛогвинов-з6е]

Actually there is such a video. I guess it was done by redpenblackpen. That was a stream

[@mislav9951]

Im from Croatia! Its cool to know that youve been in my country lol

[@harish6787]

Oh he is from Croatia awesome

[@toljanginis2785]

I'm from Dubrovnik, it's amazing that Michael visits us from time to time!

[@sanelprtenjaca9776]

Cheers from Rijeka

[@matejcataric2259]

Meni je drago da ima Hrvata na ovom kanalu ❤️

[@toljanginis2785]

@@matejcataric2259 Treba širiti i naše znanje😄

[@jvranstify]

Writing this as K!L!=K!(1 + L!/K!) + M! implies that L >= K and similar argument for K >= L, so K = L

[@ThePharphis]

oh I like this one. Seems a lot easier and quite intuitive.

[@ThePharphis]

@@angelmendez-rivera351 On second thought I'm not so sure. In fact I don't think it does.

[@jvranstify]

@@angelmendez-rivera351 Because 1 + L! /K! then needs to be integer, etc...

[@jvranstify]

@@angelmendez-rivera351 must have been sleeping

[@Pika250]

You forgot to address a case of Wilson's theorem, where if n = p^2 where p is an odd prime, then n divides p * 2p and both p and 2p are strictly between 1 and n.

[@copperfield42]

15:05 why it can't be that k!/l! + m!/l! -> (k! + m!)/l! result in a integer?

[@mcwulf25]

How can you say m!/l!

[@HagenvonEitzen]

I think this is faster for the k 2, then m! =k! l! - k! - l! >= 6 l!-l!-l!=4 l!, so m > l, so k! = k! l! -l! -m! is a multiple of l!, which is absurd

[@digxx]

Maybe it is trivial, but I don't see it: What about the case m=k if l>k? For then we would have l!=2+l!/m! = 2 + l*(l-1)*...*(m+1). Why doesn't have this a solution?

[@hnnagarathna7286]

Just an hour ago this is the fastest time evr that I've got ur video as recommendation

[@donaldbiden7927]

My experience is that Number Theory proofs are the longest.

[@vigneshvenkataraman8]

I don't know a lot of Croatian , but does anyone else think that he looks like Rakitic

[@boubidebibou4547]

The most famous board in the world :D Mathematics are a way to dream... So thx Michael to keep us dreaming !

[@jkid1134]

I don't think your proof of wilson's theorem accounts for squares of primes, eg 9 cannot be decomposed into two distinct integers greater than 2.

[@adeolugboji3645]

I don’t understand the entire Wilson’s theorem part. He says (n-1)! is identical to -1 mod n, but I don’t see how that can ever be right.

[@adeolugboji3645]

Diego Marra What is a multiplicative inverse? Apologies but I am a complete novice at group theory

[@איתןגרינזייד]

I don't understand, what's the multiplicative inverse Modula n?

[@lejamtelpaul9944]

Good question, I don’t get it too

[@dariobarisic3502]

Multiplicative inverse of "x" (mod n) , let's denote it "y", is a number with property x*y = 1 (mod n). This is my guess based on his explanation of Wilson's theorem. For example, take mod 7. Inverse of zero is non existent, inverse of 1 is 1, inverse of 2 is 4 (because 2*4 =8 which is congruent to 1 mod 7), inverse of 3 is 5, and 6 is its own inverse. Same as in the video where he said for mod n, 1 and n-1 are their own inverses.

[@3manthing]

When examining whole numbers modulo some number n, the set colapses only to 1,2,3,...,(n-2),(n-1). Every number bigger than n is being represented modulo n by one of these integers 1,....,(n-1). There is a special rule when adding and multiplying these numbers so when adding/multiplying two numbers modulo n, you still remain in the same set modulo n. Knowing this, multiplying different numbers modulo n might result in these two numbers resulting 1 when multipliyed. That means that the first number is multiplicative inverse modulo n of the other and vice versa.

[@איתןגרינזייד]

@@dariobarisic3502 thanks

[@dariobarisic3502]

@@3manthing I'd also add that if we deal with (mod p) where "p" is prime, than inverses are unique, at least for all n0), so there is "x" which gives z=1.

[@bhanusri3732]

17:45 l>m doesn't always mean m divides l??

[@bhanusri3732]

@@angelmendez-rivera351 He said that m divides l because l>m but its not always the case for example 5>2

[@bhanusri3732]

@@angelmendez-rivera351 Ohh thanks a lot. I misunderstood it

[@kozokosa9289]

Didn't mindyourdecisions already cover this?

[@rubyjha4298]

Love your videos ❤

[@pokoknyaakuimut001]

So much thanks from Indonesia 😇😇😇

[@leif1075]

I'm curious Michael why you chose to specialize in that area or branch or field of math..representation theory and vertex operator algebra and not any other?

[@MonzennCarloMallari]

For a second I thought you just had to shout the letters

[@protectionsecularism6090]

Please topology and more advanced differential geometry

[@aldues00]

what if a=b?

[@arshadkhan6562]

exact same question came up in a British Mathematics Olympiad Paper!

[@hassanalihusseini1717]

Thank you for this number theory proof!

[@suryanshsharma2451]

Great teacher

[@mathmatrix66]

What the hell is mod aha ?

[@Tome281]

Drazen Adamovic is one of my professors, the world really is a tiny place

[@user-A168]

Good

[@throwawayuser9931]

Very nice prob solving in this channel actually....... Tho a lot of bashing involved

[@jomama3465]

This channel derseves more subscribers.

[@anggalol]

You should make patreon

[@bsuperbrain]

13:37 "we know that one is a whole number" - do we? Prove it! :D

[@skrill500]

lol just use peano axioms

[@thephysicistcuber175]

You technically forgot to consider the case k=l=0, but that trivially yields no solutions. EDIT: Misleading thumbnail.

[@vaxjoaberg9452]

It's been a long time since I took a math class but I'm pretty sure 0 isn't a positive integer. :P

[@thephysicistcuber175]

@@vaxjoaberg9452 Oh sh!t. Misleading thumbnail.

[@nathanisbored]

@@thephysicistcuber175 unless he since changed the thumbnail, i dont see what's misleading about it. the set N conventionally does not include 0

[@thephysicistcuber175]

@@nathanisbored uhm no? 0 is conventionally natural.

[@nathanisbored]

@@thephysicistcuber175 ok so i looked it up and apparently both conventions are commonly used. the way i learned it was N = {1,2,3,...} W = {0,1,2,3,...} Z = {...,-3,-2,-1,0,1,2,3,...} and if you want negatives only, i guess you would do: Z- = {-1,-2,-3,...} and Z+ would just be equal to N it seems in some conventions, Z = W, which is even more confusing. But anyway, i think its safe to assume in the thumbnail (and most of his videos), he was using the convention that N = {1,2,3,...}

[@PachinkoTendo]

Didn't expect this. Wilson's theorem is overkill, you can solve this with basic arithmetic and using an even/odd parity argument.

[@leif1075]

Right did you just kind of plug and chug guessing starting with small integers?

[@eldattackkrossa9886]

let's go

[@Macisordi]

Ragusa Is beautiful

[@shikhapandey7423]

First

[@suryanshsharma2451]

I was first

[@guntherklausen6891]

@@suryanshsharma2451 no, he actually WAS first, sadly

[@hamiltonianpathondodecahed5236]

@@guntherklausen6891 idk why but your reply made me giggle