Solving the hardest question of a British Mathematical Olympiad

Thanks to Nahian for the suggestion! This is a difficult factorial problem.
British Mathematical Olympiad 2002-2003 problem 5
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[@MindYourDecisions]

I am always excited and nervous when I post a video like this! One the one hand, I am thrilled to share a challenging mathematical proof. On the flip side, proofs require perfection and it is very challenging to make a video with no errors. You guys have great attention to detail, so if you see any mistakes, let me know! For major mistakes I will repost a corrected video; for minor mistakes/typos I will leave a note in a comment. Hope you enjoyed this problem!

[@hardestmaths6975]

Hy sir i wana talk to you

[@tentic]

ily! and ur videos!!, they help me learn so much! (I have completed pre calc course and I'm in middle school lol)

[@ilias-4252]

You wanted solutions on the positive integers so you didnt have to check a=0 at the start of the video. Very minor mistake.

[@replicaacliper]

More challenging stuff like this please!

[@jameswan2963]

13a=13+a please help me for the solution

[@sounak5853]

"Pause the video if you would like to give this problem a try" Thanks for your concern, I'll just skip 10 seconds instead.

[@Azimsajid1729]

kzbin.info/www/bejne/omm2YaNjbpabbaM

[@tfdtfdtfd]

No need to pause.....the short interval provided was enough to put together the proof

[@RosidinAli]

😅😅😅

[@wushu1984]

He should have said turn off your device for a week as you attempt to give this problem a try...

[@mydkarthikmecharena9010]

😂

[@samuelking4723]

I understand the steps. What I don’t understand is how *anyone* was able to figure them out in the first place.

[@user-jn4sw3iw4h]

This was indeed more of a "keep watching the video for the answer" than a "keep watching the video to figure out how to solve this" video

[@josmin5384]

Exactly, and imagine if someone was able to solve this type of question in the exam. These are the sort of questions that are said to be "easy" for Math Olympians which just doesn't make any sense.

[@hydrochloricacid2146]

I'd agree that this feels more like randomly stumbling into the solution and less like reasoned mathematical logic...

[@daominhdung5347]

I did participate in some of the same competition, and actually, you can use some kinds of ideas for such problems; but as this is a high-level competition, you must be very creative. With the kind of problem like the one above, the idea is you will try to EXPERIMENT with properties related to integers (for example, divisibility), and with each experiment, you will try to NARROW DOWN the range of value for each number. It is not easy; some experiments will only narrow down very little, others can help a lot. As a result, the competition often takes a lot of time, and the students should be very creative

[@josmin5384]

@@daominhdung5347 Interesting. I will take note of that.

[@maanasagarwal7518]

Many brain cells were lost trying to solve this problem.

[@derekhudgin6660]

Many were lost watching this video

[@jongyon7192p]

yeah people become dumber after trying to solve math problems

[@dey9293]

Brain cells get activated if one is interested in the subject!😀

[@joekerr5418]

F

[@MagruderSpoots]

The carnage was awful, brain cells laying everywhere. Some where still alive and trying to crawl away, others were gasping their last breaths. The lucky ones died quickly.

[@JSSTyger]

After 10 minutes of fuming, I proved that my desk is really, really strong.

[@RocketRaju2212]

@xd hello he was banging his desk all the time

[@jokerjackass8461]

😮mine too

[@Zavstar]

Contradiction is such a powerful tool

[@fburton8]

No it isn't! (Sorry. :P)

[@prithujsarkar2010]

@@fburton8 xd

[@Trockenshampooleopard]

@@fburton8 Yes, it is!

[@ramone.chacon5084]

@@fburton8 yes, it is x2

[@jongyon7192p]

Assume contradiction is a weak tool. We have a counterexample showing contradiction isn't weak. Thus by contradiction, contradiction is a strong tool.

[@Chadicle7]

This. THIS is what I’m talking about! Keep this content up Presh!

[@carterqualls5266]

He ripped off this exact problem from Michael Penn, posted two weeks ago. kzbin.info/www/bejne/gp7HpoaYfbKBjdk

[@hybmnzz2658]

@@carterqualls5266 this is an olympiad problem lmfao there is no trademark. That being said y'all should check out Michael Penn's channel if you want to see many math problems like these!

[@alexeyrb1807]

Nice problem! Some alternative ways that I found when solved it first: Rewrite the equation as (a! - 1)(b! - 1) = c! + 1, check cases like 0 and 1 and then look at (a! - 1) = (c! + 1) / (b! - 1). This allows to establish that a >= 3 and c > b. For case a = b: solve a quadratic equation in terms of a!. It gives a! = 1 + sqrt(1 + c!). This then gives a = 3, c = 4, but also an upper bound on c, since if c can't be cubic or higher expression in terms of a (eliminating some low number cases first). But it's sloppy, your proof on c

[@Bry10022]

If you need the not equal to, less than or equal to, or greater than or equal to symbols, Here they are: ≠ ≤ ≥

[@json6573]

this is what i thought too.

[@romanshevtsiv3526]

Yeah did it more or less the same way. Looks faster and cleaner for me than the original one.

[@heysqualito]

@@romanshevtsiv3526 I also thought of the same

[@anshumanpanigrahi7323]

Correct eqn will be (a! - 1)(b! - 1)=c! + 2

[@GravitationalRealism]

Great solution! One small enhancement I noticed - for the section starting at 4:41 to prove a = b, it is quicker to divide the equation by b! (rather than a!) and notice a! / b! is an integer iff a = b (and everything else are integers).

[@schungx]

Yes, symmetry arguments are so powerful. a!/b! is an integer iff a >= b. However, since the equation is symmetric wrt swapping of a and b, the only solution is a=b. Once you get that, it is a no-brainer. Take n = a!. n^2 = 2n + c!. n^2 - 2n - c! = 0. Use the quadratic formula to find that n = a! = sqrt(c! + 1) + 1.

[@thexoxob9448]

Your way still works even if you divide by a!.

[@tarunrathitra1158]

It's not easy to even understand this proof I can't even start to imagine solving this on a timed test These Olympians are probably superhumans

[@advaykumar9726]

IMO aur inmo Ka level dekhe ho?

[@marvinmartinez3818]

Not so much, it's matter of practice and interest in the subject

[@tosca1883]

not really, after solving a few problems you start to learn new methods and ways to understand math, this problem was really fun, and it wasn't difficult at least for me, but the solution presh used is actually more complicated that the one i had, so its just practice

[@advaykumar9726]

@@akshaj7011 what?

[@sweetpotato_35]

Yeah if you had in mind divisibility this would have been an easy problem. It is just different kind of maths that you aren't used to.

[@bhaswatibhattacharya4201]

Imagine someone writing "easy" in the comments for this one.

[@rustemtehmezov9494]

Easy!

[@daddahamza6831]

@@rustemtehmezov9494 HAHA

[@justingolden21]

Easy. Just find the upper limit because factorials grow exponentially, and multiplying beyond a certain point will always be greater than adding, and then experiment and realize it's impossible with a or b as 1 or 2, and that 4 is an upper bound, and then that a and b must be 3, and then solve for c. (Except this doesn't work because you don't know if there could be a c that's exactly large enough to compensate for the difference between multiplying and adding a! and b!)

[@tusharkumar3998]

We're simpletons it is not easy for us 😂

[@giuseppebassi7406]

@@justingolden21 i used a different method but with the same trick, the product of two factorials had to be greater than the sum of three at some point

[@dyosorkan]

As a french viewer,your videos make me practice both english and maths... And how to think outside the box. Genius

[@Microplastics2]

Oui, moi aussi

[@mohdhassan2677]

Now I know why exclamation mark is used for factorial!

[@vontabi]

The last time I was this early, the Gougu theorem was still called the Pythagoras theorem

[@ericzhu6620]

you mean Gougu? It's Chinese way to call it, but I believe the Europeans still call it Pythagoras theorem, btw I live in Portugal.

[@EXOPLANETnews]

Hey i have an interesting channel about space science and mysteries if ur curious about it do visit my channel once pls 🙏 🙏🙏.

[@Grizzly01]

@@ericzhu6620 It's a joke, Eric. Varun was making a joke. Watch a fair selection of Presh's previous geometry-based videos, read the comments sections of those videos, and hopefully it should become clear.

[@yosuakrisnando]

@@Grizzly01 , he might know it before you "advise"...

[@chanlokhim3182]

I am from Hong Kong, we call that Pythagoras theorem too tho, but we also know the name of GouGu Theorem

[@VinaySingh-jm8iw]

Great proof, but how would one know they needed to prove that b

[@typingcat]

Probably with lots of previous experiences with similar "proof" questions that used this kind of boundary checking/contradiction tests.

[@bigboboluager5003]

You don't know that. The way he presents this is not the way you start thinking about the problem. I started out, trying a few easy values like 1 and 2 for a, b and c. You immediatly get a contradiction so you can keep that in the back of your mind. Sometimes this will come in helpfull later on and sometimes it wont. If you have some practice you will learn more tricks, that you can just kind of throw at the problem and see what happens.

[@thunderstorm8653]

Experience

[@frankie391]

Wow, it's super duper rare to encounter a Factorial problem like this. Thank you Presh Tall Walker, what an amazing solution. And also, can someone tells me how can we think of the ideas at the beginning of the solution? Like how can Presh know we need to prove b=a or c

[@GeorgeFoot]

Olympiad problems tend to be open like this, you have to creatively think of ways to narrow the field. Rather than trying to find the different numbers, maybe we can find three numbers all the same? Or prove that it's impossible. What about a b or c being really low, like 0, 1, 2? You just have to explore things like this and watch out for any good leads. Also know that where factorials are concerned it's almost always something to do with divisibility or primeness, as the fact that all factorials divide each other is very powerful.

[@leonais1]

I'd consider that to be a summary rather than insight. Factorials can be hard to deal with, however you can work with values such as n!/m! so cases where n>m, n=m are immediately interesting.

[@kevinm1317]

A lot of it boils down to experience and intuition, which only comes from practice - its often difficult to explain why exactly you try something. But I'll try my best to explain here. It's very natural to assume a=3, you might realize that 2 and 3 must divide b!, so 2 and 3 must divide the RHS as well. If b and c are much larger than a, its instantly clear this is impossible, as the RHS would be odd+even+even. As such, exactly one of b or c must be equal to a or a+1, from which we get c>=a+2 (as c>b). It immediately follows that b=a or a+1, and from there its quite natural to try to see which one it is/can be. If you try b=a, then of course you'll get the solution mentioned, and if you try b=a+1 then you'll get the same equation as in the video.

[@loglnlg]

He knew it because he solved it before making a video When I was solving this problem at that moment of progression (getting frormula at 8:06) I was stuck on this general problem, so I divided them into subproblems where c=a, c=a+1, c=a+2, etc. and then I firstly thought about divisibility by 4 and proved that if c>=a+4 then there is no solutions, so I got only 4 little subproblems (instead of infinity of them) and each of them could be solved more easily than the whole problem. For me it is helpful to divide problems into cases (if we could do it, sometimes it is just could't be done), not even to solve whole problem by them, but to see particular examples and work with something little at the beginning and maybe see something more general in them

[@PrashantSharma-nw6gc]

In such types of problem you must have to make some assumptions and be dare to do so because if they are wrong and your method is right, the great genius mathematics itself told you their errors. May be there some contradictions occurs by which you are able to say that the assumption is wrong.

[@WOLLEY104]

I'm a GCSE student and these kinds of videos are always fun to watch. I love maths but never understood how people even know where to start with these questions. I lost track of what was going on like a minute into the video.

[@matthewcarlyon5133]

Wow, great video Presh. I especially loved the section of proof at 5:00. Brilliant maths.

[@Macieks300]

Nice puzzle. I solved it kind of similarly but when I got to a

[@rorschachkovacs7867]

I did the same way that you do, and eventually you can get the conclusion that n=0 and m=1, just aplying that n! is even if n is equal or greater than 2

[@latenter1428571]

There is an easier way to show a contradiction. Once a < b < c, then b! divides a!*b!, b! divides b! and b! divides c!. But b! cannot divide a! Then c < b (that's not our case) or a = b.

[@dey9293]

Once you eliminate the impossible, whatever remains must be the truth.- Sherlock Holmes😀 This is the only theorem that solves the above problem.

[@ihaveamathquestion7377]

After you derive at a

[@ibrahimmalki150]

I never would’ve thought combining elementary concepts such as divisibility with basic middle school stuff like factorial and algebra would be so complicated

[@marvinmartinez3818]

That's math, going from the simple to the complex

[@ivica974]

You are rare channel , who don't use false advertising of high school homework math problems like they are some math olympiad problems. And this is indeed beautiful problem and solution.

[@kalpanarms9597]

Some people like these videos before even watching them, that is how much we love your videos! :) nice video btw....

[@gaganupadhyay4978]

Nope that's just how KZbin algorithm works.

[@ahasabamirraiyan8031]

Yeah

[@laavanayaggarwal6671]

‌‌

[@gordoncharles741]

No, that's just blind stupidity.

[@EXOPLANETnews]

Hey i have an interesting channel about space science and mysteries if ur curious about it do visit my channel once pls 🙏 🙏🙏.

[@AlbertEinstein49607]

Hello, British Mathematical Olympiad here, thanks for reviewing this question. I did actually get this question right in the olympiad but there is a much more simple way!

[@gfasoulas]

Whenever I watch a video from this KZbin channel my brain is twisted.

[@laavanayaggarwal6671]

small brain small pp

[@EXOPLANETnews]

Hey i have an interesting channel about space science and mysteries if ur curious about it do visit my channel once pls 🙏 🙏🙏.

[@Mrpopperer]

nah smol brain usually results in big pp xD

[@ABHISHEK-xm5mt]

@@Mrpopperer Power of belief, Dare to believe 👇 kzbin.info/www/bejne/r5jGXnWVhK2tb9k

[@piman9280]

Whenever *I* watch, my brain wants to Go to goo!

[@charlesaudet9035]

The part showing a=b may be simplified (Time 4.44 to 7.44). Instead of dividing by a!, divide by b!. The left-hand side will be an integer a! and the right-hand-side of the equality will be a fraction a!/b! + 1 + c!/b! (since a

[@shaiavraham2910]

"Find all solutions..." There is only one solution. lol 🤣

[@piman9280]

.... and *the* solution can be found after a few seconds of "guess and check!" 0! = 1, 1! = 1, 2! = 2, 3! = 6, 4! = 24. Then 6 x 6 = 36 = 6 + 6 + 24.

[@RGP_Maths]

So? You don't know there's only 1 solution until you prove this. There was no need for the question to tip you off as to how many solutions you are looking for.

[@pdpgrgn]

@piman that doesn't prove that there aren't other solutions

[@piman9280]

@@pdpgrgn - who said it did? I merely commented on the fact that the only solution is very simple to find. The very fact that the "problem" is so shamelessly contrived speaks volumes about the "contriver."

[@GRBtutorials]

But you also need to prove there's only one.

[@lchen1144]

At time, 4:5, an alternative is to divide both sides of the equation by b! (instead of a!), this leads to a! = a! / b! + 1 + c! / b!. Because 3

[@n.rv..n]

As an Olympiad Student, this is quite trivial for a Diophantine Equation Try this harder problem: Find all triples of positive integers (a, b, p) such that p is prime and a! = b^p + p

[@danyd1583]

You are just amazing. You must have spent days on this video... Thanks for this awesome explanation. Wondering how people solved the exercise in minutes during the olympiade lol

[@eytansuchard8640]

The method of negation is seen over and over again in integer equations. Thank you for this enlightening video. While performing other tasks I watched your video and it reminded me that sometimes in order to think we need to stop doing what we are used to do.

[@helo3827]

You are my favorite youtuber, I never miss any of your videos

[@niejocelyn1635]

Cool

[@advaykumar9726]

Me too as I have turned on the bell

[@janathpeiris3431]

Due to symmetry, you can assume that a = b . Then the equation reduces to (b!)^2-2*(b!)-c! =0 . solving the quadratic equation --- > b! = (1 +sqrt(1+c!)). Ignored the negative solution. To have an integer after taking the square root, sqrt(1+c!) need to be a square. c can be 4 and 5. 4 is a feasible solution . b! = 1 + sqrt(1+4!) = 6. Hence, b = 3. We assume a=b . Hence a =b=3 and c = 4.

[@yurenchu]

We can't just assume that a = b . There may be solutions where a doesn't equal b. (It wasn't stated that the problem has only one unique solution.) Furthermore, we don't even know in advance if there exist solutions where a = b.

[@billy.7113]

It took me longer than 11 mins to understand the whole proof.

[@itsjustanapple5452]

For all or us.

[@harris6498]

😂😂😂

[@arianrolex6456]

Bill Y. Dude I got no idea why I’m watching this I’m 13 what the heck is he talkin about . Why am I watching this I have math paper 2 tomorrow...

[@sadkritx6200]

@@arianrolex6456 don't worry bro. Come back to this video in a few years, maybe it will become easier as you will probably know more about mathematics then than you do now. I noticed this many times. When I was 13/14 many math videos seemed too hard to even understand. But now when I'm 17 the impossible to understand videos have gone down but not become 0 lol😅

[@DmitDmit1]

Nice problem. Solved following the same steps, but prooving some of them slightly differently. When you proove that a==b, I think it's easier to consider divisibility by (a+1) when you have b!=1+b!/a!+c!/a!, it's clear that the left side is integral and the right side is not.

[@honyi1995]

When you said "as always" at the end, I immediately said in my head: "Stay awesome, bros!"

[@JuanMataCFC]

actually amazing proof, i think u covered every possibility. i don't think i've ever seen contradiction used better than this before!

[@pokmaster4475]

This is one of the best math proofs, I have seen

[@sasharichter]

a=b can be established almost immediately by realizing that if ab then due to equation being symmetric in a and b, we can choose a>b. dividing both sides by a! we find that b! = 1 + b!/a! + c!/a!. but because a>b then the second term is a fraction less that 1 and so the third term must be a fraction (also less than 1) otherwise the right hand side is not an integer. thus b! = 1 + 1 = 2 which is a contradiction! so we have 3

[@Timotheeee1]

how does anyone even come up with this

[@typingcat]

Drugs. Lots of drugs.

[@nelser1160]

For real, how would someone know what to prove at the start?

[@user-qs3ih3ll5f]

miracle, I guess.

[@rustemtehmezov9494]

@@nelser1160 So, when you solve so much of questions like these, it's easy to check these metods and get bounds.

[@nelser1160]

@@rustemtehmezov9494 I wonder how many I have to solve to get that lol

[@logannasty3240]

Constructing these problems, that’s the truly magnificent feat! I imagine it takes weeks or months to come up with such a beauty. Then this is only one of many problems in an Olympiad. So much intellect invested into competitive maths...

[@abysslight2490]

You could’ve proven the fact that a=b much easier if you divided both sides of your equation by b! instead of a! at 4:53. This is because we already knew that a is less than or equal to b, so the case where they are not equal would mean aa (as previously shown), c!/a! is an integer, and 1 is obviously an integer, but since b>a, a!/b! is not an integer, thus a! isn’t an integer. And that’s a contradiction. It’s a lot simpler and quicker way to get to the same conclusion.

[@QuantumHistorian]

There's a much more elegant solution using modulo arithmetic. Honestly, when dealing with integer problems, modulo arithmetic should be the first port of call. Some of it overlaps with the video a lot so I'll leave the details out, but overall its a much tidier presentation and doesn't require so much guesswork about what to prove next. Really, there are only three steps here. As a prelude, consider x! mod y!. If x >= y, then y! | x! and so x! mod y! = 0. If x < y, then the modulo base is larger than the number, and so x! mod y! = x!. First, assume as before that b >= a. Taking mod b! of both sides, we get: a! b! mod b! = a! mod b! + b! mod b! + c! mod b! 0 mod b! = a! mod b! + 0 + c! mod b! (*) There are two cases to consider. Firstly, a < b. That implies 0 = a! + c! mod b!, and thus that the last term is non-zero, hence that c < b, which in turn implies that c! = b! - a!. Substituting that back into the starting equation we have a! b! = 2 b! which clearly has no solution. Thus, we move on to the second case, where a = b, leading to a! mod b! = 0 and so (*) tells us that c! mod b! = 0 and therefore that c >= b. The case c = b can be discarded by noting that it would lead to the equation a! a! = 3 a!. Thus, we have now restricted ourselves to the cases a = b < c. Going back to the original equation, we have a! a! = 2 a! + c! a! (a! - 2) = c! a! - 2 = (a+1)(a+2) ... (c-1) c (**) Where we have divided out a! in the last line. Taking mod 3 of both sides (and assuming a >= 3, the smaller cases can be checked by hand) we have a! mod 3 + (-2) mod 3 = (a+1)(a+2) ... (c-1) c mod 3 0 + 1 = (a+1)(a+2) ... (c-1) c mod 3 If the RHS contains 3 or more consecutive terms, then one of them will be a multiple of 3 and so the modulus will be 0. If it has 2 consecutive terms, none of them can be a multiple of 3 so we have (3n+1)(3n+2) mod 3 = 2, so this is also impossible. Thus, the RHS can only contain one term which is exactly 1 above a multiple of 3. This tells us that c = a + 1 and a = 3n. Substituting back into (**) we have (3n)! - 2 = 3n + 1 (3n)! - 3n = 3 (3n)! mod (3n) - 3n mod (3n) = 3 mod (3n) 0 = 3 mod (3n) n = 1 Thus (a, b, c) = (3, 3, 4) is the only possible solution. Evaluating it shows that it does indeed work. Note that we don't have to deal with the complexities of n=0 because we ruled out a=0, 1, 2 earlier. I know that this is similar in many details to the video, but a much more streamlined chain of thoughts IMO. Remember, when dealing with integer problems, modulo is your friend. _Especially_ with factorials flying around. As an aside, there's also a proof of this result using Wilson's Theorem, but that's unnecessarily complicated here.

[@Authen194]

Bro what you do? A teacher😀

[@QuantumHistorian]

@@Authen194 haha, no, but I know how to write up proofs in a detailed and understandable way

[@Authen194]

@@QuantumHistorian can you help me

[@HollywoodF1]

4:10 Proposing that we have not yet shown that b < c because we’ve yet to prove that a solution exists. I suppose, when including later steps that proceed similarly, the proof proceeds on the assumption that a solution exists.

[@meenashetty9782]

Halfway through i stopped listening and started reading comments.

[@targetiitbcse1761]

same😂

[@mathboy8188]

The step where already have 3 < = a 0 for this part). Instead of dividing a! b! = a! + b! + c! by a! as in the video, divide both sides of that equation by b! to get a! = a!/b! + 1 + c!/b!. Then a!/b! = a! - 1 - c!/b!, which is an integer because b < c. But if 0 < a < b, then a! / b! is not an integer, so that's impossible. Therefore, since a

[@williamleinonen4888]

yeah, right. Glad I gave up after a couple minutes.

[@indrashisbhattacharyya7854]

This made me cry

[@zapzordx5440]

3rd time I solved any of your questions... I'm a 6th grader(your education system might treat 6th grade differently) and solved it and it made my day as I got it right :) Well I just used trial and error method...I know they asked for a real explanation but I'm still happy to get the answer 3,3,4. 😁

[@vanditseksaria5897]

the main crisp of the problem is to show that there are no other solutions

[@isaacdeutsch2538]

We immediately notice that the equation is invariant under a swap of a and b, so we may assume, without loss of generality, that b is at least a. Now, we don't know that c is at least b. We know that a! divides c! (reducing everything modulo a!). We would like to know that b! divides c! as well, which is equivalent to c is at least b. So consider. Note that since a is at most b, a! is at most b!, so a! + b! is at most 2b!. Suppose c < b. Then c! < b!, so a! + b! + c! is at most 3b! < a!b! once a > 2, so the only solutions with c < b have a = 1, 2. Suppose a = 1. Then b! = 1 + b! + c!, so c! = -1, a contradiction. So a is not 1. Suppose a = 2. Then 2b! = 2 + b! + c!, so b! = 2 + c!. Since we're assuming c < b, reducing modulo c shows that c divides 2, so c = 1, 2. But 2 + 1! = 3 and 2 + 2! = 4, neither of which are factorials. So there are no solutions with c < b, thus c is at least b. Now at this point, we could do a more convoluted argument that I did first, but I realized a simplification to my argument as I type this up, so in the interest of my thumbs I will present this more streamlined argument*. Since c! is divisible by b!, reduce the original equation modulo b!. Then we get that a! = 0 (mod b!), so b! | a!. But then since a! | b!, a! and b! are associate, and since they're both positive integers, this means a! = b!. But then a = b. So we can reduce the original equation to a!² = 2a! + c!, and dividing everything by a!, we get a! = 2 + c!/a!. Suppose for the moment that a > 2. Then 3 | a!, so reducing modulo 3, we see that c!/a! - 1 is divisible by 3. Thus c!/a! is not divisible by 3, and thus c = a, c = a + 1, or c = a + 2. c = a means that a! = 3, impossible. c = a + 1 means that a! = a + 3, leave that for now. c = a + 2 means that a! = a² + 3a + 4. Reducing modulo a, we see that a | 4, so a = 1, 2, or 4. Since we are supposing that a > 2, a = 4, but that. polynomial at 4 yields the value 32, which is not a factorial. So the only option is c = a + 1, which entails solving a! = a + 3. Now, for x > 3, we have the following inequalities of functions: x! > 2x > x + 3, and thus since we've assumed a > 2, our equation can only hold if a = 3. Indeed, 3! = 6 = 3 + 3, so (3, 3, 4) is a possible solution--indeed it is, 3!3! = 3! + 3! + 4!. This is thus the only solution for a > 2. Finally, recall that if a = 1, we get a contradiction--we did not use the assumption c < b for that. So assume a = 2. Then we wish to solve b! = 2 + c! with no restrictions on b or c. Since we know there are no solutions when c < b, suppose not, and reduce modulo b. Then we see that b | 2, so b = 1, 2. If b = 1, we get c! = -1 again, impossible, and if b = 2, we get c! = 0, still impossible. Thus there are no solutions if a = 2 either, and the only solution in the positive integers is (3, 3, 4). *In my original argument, I reduced modulo 2 instead of modulo b!, which shows quickly that b = a or b = a + 1. I then treated these cases separately (obviously, from the argument I have presented, the case b = a + 1 ended in contradiction). Some interesting stuff came up, but it's not as elegant as reducing modulo b! and showing directly that a = b.

[@NaHBrO733]

nice!! its good to have some really challenging problema once in a while👍🏻

[@daigakunobaku273]

At that step when you have to proof that a = b there is an easier way. Just divide both sides by b! instead of a! and you ll get an integer (a!) equal to a non-integer (1 + c!/b! + a!/b!, since c > b, c!/b! is an integer, and since a! < b!, a!/b! is not, hence the whole sum is not an integer)

[@benheideveld4617]

And without using the Gougu theorem...

[@lazymello6778]

good one, i had another way using some modular arithmetic: by some neat factorization we get c!+1 = (a!+1)(b!+1) I took 3 cases, a

[@afbanales]

Your definition of wonderful is WAY different than my definition of wonderful!

[@ABHISHEK-xm5mt]

Power of belief, Dare to believe 👇 kzbin.info/www/bejne/r5jGXnWVhK2tb9k

[@CraigNull]

Once you establish 3

[@kritika1413]

Thumbnail: solve Me : NO Because I can't 😂😂😂

[@JohnDlugosz]

The numbers are small, so trial and error will take less than ten minutes.

[@kaisarsihaloho975]

Thumbnail : solve Me : Yes Brain : Can't Me : No Presh : HAH! I KNEW IT!

[@neg2sode]

This proof is incredible. Love the way you presented it.

[@hamster8706]

Fun fact: A prime number in russian is "простое число", which can mean simple number too,therefore, a prime number is the opposite of a complex number

[@yurenchu]

_Prime number_ is the "opposite" of _composite number_ . A composite number is a (positive) integer that can be written as the product of two (positive) integers that are both not equal to 1 ; in other words, a composite number is _composed_ as the product of two (non-unit) integers. (For example, 35 is a composite number because 35 = 5 * 7 . Another example: 9 is a composite number because 9 = 3 * 3 .) The word _complex_ also means "composite"/"composed", but in a different way: a complex number is composed as the sum of a real number and an imaginary number. (Since the real number may be 0 and/or the imaginary number may be 0i, a complex number may also have just a real value, or just an imaginary value.)

[@bobjordan5231]

Well done Presh. Not a verbal pause or miscue, which shows you know this material ice cold. Impressive... most impressive...

[@em_the_bee]

Or that he has prepared a script?

[@bobjordan5231]

@@em_the_bee Perhaps, even reading something with no pause or miscue is impressive!

[@justinmay2295]

I was so close to the answer

[@harshitvarma7867]

I bow down to you my lord.

[@harish6787]

Good

[@TechToppers]

How close?

[@carlosmspk]

What did you do?

[@justinyoung6342]

The next-to-last step where you showed ca, we know c=a+1. Further, we can see that c itself is congruent to 1 mod 3.

[@johnjordan3552]

The funny thing is the question it self isn't really hard, just need to keep track of your logic string yet that seems to be the hard part

[@orionspur]

(a!-1)(b!-1)=c!+1. Noting factors shows c>a,b. From there 3,3,4 easily follows.

[@williamadams137]

I have another number theory problem that I’ve been stuck on. Find all integers u such that u³ + 2u + 1 is a perfect square. (You must have a complete solution.)

[@ananthchopra106]

Only one integer that is 1

[@williamadams137]

Ananth Chopra nope.

[@williamadams137]

Actually it’s much harder than it looks.

[@yurenchu]

u = 0, u = 1 and u = 8 are solutions, but I don't know if those are all solutions. There are no negative integer values of u that are solutions (because it can be easily shown that for any u ≤-1, f(u) = u³+2u+1 ≤ -2 which is negative and hence cannot be a perfect square). There are also no other positive integer values of u ≤ 32 that are solutions (confirmed simply by hand). However, I haven't been able to prove that there are no solutions where u > 32 . I can prove though that (u+1) cannot be a multiple of 4, nor a multiple of any of the prime numbers 5, 7, 13, 23, 29, 31, 37, 47, 53, 61, 71, 79, ... (the rest of the list of prime numbers probably goes on infinitely but is apparently unpredictable though).

[@trnfncb11]

It is somewhat easier to prove that a=b, for if b>a, then a!=a!/b!+1+c!/b!. Here the 1st term on the rhs is 1/n < 1 and the last term is either integer (which wouldn't work) or 1/m. But then 1/n+1/m must be 1 so a=2 and 2b!=2+b!+c! or b!=c!+2 with no solution.

[@ninad7694]

dude you make such good videos, you inspired me to make a youtube channel! keep up the good work!

[@user-pl7tr9dv6l]

Hi, after factoring the original equation into (a! - 1)(b! - 1) = c! + 1, you'll notice that both product terms on LHS have to (obviously) be less or equal than the RHS. This implies that a≤c and b≤c (there's one special case of c=1 for which this does not hold but it can be checked manually that this does not give any solutions). Next, dividing the original equation with a! and observing that both sides of the equation have to be integers, we find that b!/a! has to be an integer. Doing the same with b! we find that also a!/b! has to be an integer. The two above conditions, taken together, can only be satisfied if a=b. This turns the whole original equation into a quadratic, and working out its discriminant details (and requiring that it be perfect square) you'll notice that c! has to be of the form '4k(k+1)' where k is a natural number. The only factorial of this form is 4!, hence c=4, and consecutively a=b=3

[@facilvenir]

It was hard enough to follow your explanation! Thank God I didn't even try to solve it myself!

[@viktoriavadon2222]

There is a faster and more elegant proof for the a=b part. Just consider that b! divides the left hand side, so it must divide the right hand side too. We know it divides itself, and divides c! since c>b has been shown, necessarily b! divides a!, and so a>=b. But we supposed a

[@PSquared-oo7vq]

Great video, but I have never, ever gotten more ads in an 11 minute video than this one. At least seven, two of which were > 3 minutes long, and one less than 15s into the video (and that's after a double pre-roll). If the creator set the video up for this many ads, then boo you. If not, then KZbin is screwing with you (more than usual). I try to support creators, but... adblockers are calling.

[@manswind3417]

Well, I fortunately didn't encounter any ads at all, and in general I don't get as many ads as you say, despite using the regular KZbin version, so evidently the creator hasn't set the vid up for these many ads - it's youtube's doings Btw, are you on aops, coz your the guy with your username there is a 13 yo IMO medallist...

[@earthbind83]

This was a beautiful proof. I had fun seeing all the pieces fall into place.

[@armacham]

1:48 "suppose A is either zero or one" -- why would I suppose that A could be equal to zero, when the problem clearly states that the values of A, B, and C are limited to positive integers?

[@theqwertyman9309]

For the sake of argument, I guess?

[@yurenchu]

In some languages/cultures/education systems, 0 is considered positive (as well as negative). Presh was just being rigourous.

[@SambhunathMaitra]

I have solved the problem by applying Wilson's theorem with the conditions a=b and c

[@yashwantsingh6703]

before this video: presh's problem are easier nowadays after this video:🤐

[@nasrullahhusnan2289]

We can directly get c=a+1. From b!=1+(b!/a!)+(c!/a!) with b=a it becomes a!=1+1+(c!/a!) LHS: 3 | a! as a>=3. As c >b --> c>=b+1 Thus c>=a+1 as a=b. RHS>a+3. In order RHS divisible by 3, a must be equal to 3. Hence a=b=3 and c=4.

[@laavanayaggarwal6671]

*Fresh tall walker*

[@HagenvonEitzen]

Wlog a>=b and right from the start use b! - 1 = b!/a! + c!/a!, note that on the right are only integers or numbers >0 and = b = a - or a!=2, b!=c!=1 (which doesn't work). So a!^2 = 2 a! + c!, or: a! - 2 = c!/a!. In particular, the right hand side is positive, so a > 2. As (2a)!/a! > a!, we see c

[@timnauwelaers6876]

I still don't get it :(

[@justingolden21]

Out of all the problems on this channel, I believe this one is the most "how genius do you have to be to figure that out" while still having each individual step be so simple. I wonder if we'll ever get AIs to "think" like this. It's really hard to imagine.

[@bangchui4465]

8:2(1+3) = 16 or 1

[@afuu794]

8:2 (1+3) = 4 (4) = 4 × 4 = 16

[@perishedgalaxy1732]

It will be 16:1 as it is given in ratio format.

[@mikasa3427]

Simpler, but much less elegantly, once you've got the upper bound on b and c in terms of a you can just exhaustively list out all the possibilities because the left hand side grows faster than the right hand side.

[@gojosan1376]

Finally Presh bro is back with his god level stuff.. And yes I missed gogu 😂😂

[@piman9280]

*Nobody* misses "gogu!"

[@Tiqerboy]

It's Gougu and I knew this would be the longest Presh video where that mighty word wasn't spoken without even watching it!

[@warrior1265]

There were a lot of ads that kept playing during the video so I had trouble keeping my train of thought while trying to follow along with your explanation. I don't know if this is a KZbin thing or a channel thing but just wanted to let you know!

[@rubyjha4298]

I'm done for the day!!!!

[@typingcat]

That's what she said.

[@robertotessera]

Interesting fact: assuming that a=b, it leads to (a!)^2 = 2a! + c!, solving the second-degree equation for a! one gets a! = 1+ sqrt(1+c!). So if a solution exists with a=b, 1+c! must be a perfect square. But this is exactly Brocard's Problem, a still open problem in maths (en.wikipedia.org/wiki/Brocard%27s_problem kzbin.info/www/bejne/Y3XNm2mmm7dsgbc): 1+c!=m^2. The only known solutions are (c=4,m=5),(c=5,m=11),(c=7,m=71). Only the first one works for our equation: c = 4 => a! = 1+ sqrt(1+4!) => a! = 6 => a = 3. So a = b = 3, c=4. I know this is not the right demonstration, but if another solution with a=b exists (but it doesn't as your demonstration shows), it will had solved Brocard's Problem :).

[@vikaspandey3537]

Got it in first look

[@jonasdaverio9369]

Obviously 😂

[@adityamukhopadhyay6803]

This problem is one of the most interesting ones I have ever seen.

[@jasonzheng5839]

This problems looks like something along the lines of “if Paul bought 10 cookies and gave his friend 2, how many kilos of meth do I have in my pocket”

[@kukulroukul4698]

:)))

[@justanonverifiedyoutubechannel]

Sugoi!! That was awesome 👌 I had a lot of fun watching this one too .....keep up the good work

[@abdulkadiralat5197]

Could you add Turkish subtitles if I would like?

[@kienthanhle6230]

You can practice English listening bruh

[@kienthanhle6230]

@Emma Beckman Vietnamese and English

[@leung77]

incredibly elegant proof for this sound simple question

[@pavelavietor1]

hello I give it a try and fail 🙂 🙃. thanks for sharing 👍 saludos

[@Denis_Ovchinnikov]

But there's an easier way to show that a=b. After you've proven c>b divide the equation by b! and consider wholeness of each member.

[@sragav4559]

If u think u r good mathematician then solve his questions to know where u r

[@ashishpradhan9606]

Which question

[@sragav4559]

@@ashishpradhan9606 all questions are amazing. For example take this only question which u wrote the comment.

[@cyberinfos7877]

import math for i in range(1,100): for j in range(1,100): for k in range(1,100): if math.factorial(i) + math.factorial(j) + math.factorial(k) == math.factorial(i) * math.factorial(j): print(i,j,k) Ans : 3 3 4

[@aayushtripathi3799]

eh while loop would be easier

[@inikristian]

what the-

[@mj9765]

This video is a gem! Thanks Presh

[@srinivasadireddi]

First