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!

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

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

Many brain cells were lost trying to solve this problem.

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

Contradiction is such a powerful tool

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

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

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

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).

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

Now I know why exclamation mark is used for factorial!

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

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

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

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.

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...

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

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

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.

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

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

Wonderful!!!, it's a treat to see you go through the proof. Keep it up !!

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

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

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

Thank you so much for uploading this video. It is helping me to get through the pandemic!

Nice video Presh. I enjoyed it so much. Please share more videos like this about challenging problems.

And without using the Gougu theorem...

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

After you derive at a

I luv ur videos! Keep up the great work!

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

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

Your best problem in a long time. Keep it up, presh :)

This video is a gem! Thanks Presh

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

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

Wow, this was really awesome! Great job!

I have watched more than 3 times to follow the logic. Thanks for brushing my mind.

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

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

This channel reinvigorates my interest in maths. Generally I can follow the solution but this one lost me. But still, it's fascinating to watch. 👏😄

Love ur videos bro, keep up the good work.

how does anyone even come up with this

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

That was a treat to solve and watch! Indeed an Olympiad Number-Theory problem (though not as insane as the present ones by virtue of its age...)

Thank you for this video! Really enjoyed watching it :-)

I still don't get it :(

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

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.

Once you establish 3

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

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

Very nicely done! Thank you for that!

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.

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

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

One of your best proof ! Thank you

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

Halfway through i stopped listening and started reading comments.

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.)

Good content after a really long long time

Highly Impressive Solution!

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. 😁

I was so close to the answer

We also like to your video before watching, such a powerful voice and technology

i find it amazing that people can derive this prove out of thin air I have trouble following it even as I am watching the solution and I wonder how can someone take the necessary steps and making the correct deduction without knowing what the solution / direction to the solution is well done for having this problem, made my day

It’s not incredible. It’s *magnificent*

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

That's a great and incredible hard problem. I want more of this stuff… 😉

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

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

Could you add Turkish subtitles if I would like?

Hello Presh, can you please add in the description of your videos the requirements of what math concepts we need to be familiar with to solve each problem??

hallelluyah, first olympiad problem that i could resolve by myself.........keep making this kind of content, thanks, since any factorial greater or equal than 2! is even, the cuestion can resolve easily

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.

*Fresh tall walker*

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

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.

I'm done for the day!!!!

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 KZfaq is screwing with you (more than usual). I try to support creators, but... adblockers are calling.

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 KZfaq thing or a channel thing but just wanted to let you know!

This is one of the coolest problems posted on the channel.

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

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

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!

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

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

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

if Scalene triangle (all sides are different and angles are different ) has one side equal to 50cm, calculate the two side and the angles of the triangle

It's amazing how people can think of these solutions. And it's even more amazing that a solution like this was to exist.

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?

Got it in first look

Minor detail: You don't need to check the case a=0, since the description specifies positive integers.

incredibly elegant proof for this sound simple question

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”

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

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

This is literally incredible❗