A mysterious factorial equation.

We solve a nice equation over the natural numbers involving exponentiation and factorials. Our approach employs Wilson's theorem and other modular reduction tricks.
Please Subscribe: kzbin.info...
Merch: teespring.com/stores/michael-...
Personal Website: www.michael-penn.net
Randolph College Math: www.randolphcollege.edu/mathem...
Randolph College Math and Science on Facebook: / randolph.science
Research Gate profile: www.researchgate.net/profile/...
Google Scholar profile: scholar.google.com/citations?...
If you are going to use an ad-blocker, considering using brave and tipping me BAT!
brave.com/sdp793
Buy textbooks here and help me out: amzn.to/31Bj9ye
Buy an amazon gift card and help me out: amzn.to/2PComAf
Books I like:
Abstract Algebra:
Judson(online): abstract.ups.edu/
Judson(print): amzn.to/2Xg92wD
Dummit and Foote: amzn.to/2zYOrok
Gallian: amzn.to/2zg4YEo
Artin: amzn.to/2LQ8l7C
Differential Forms:
Bachman: amzn.to/2z9wljH
Number Theory:
Crisman(online): math.gordon.edu/ntic/
Strayer: amzn.to/3bXwLah
Andrews: amzn.to/2zWlOZ0
Analysis:
Abbot: amzn.to/3cwYtuF
How to think about Analysis: amzn.to/2AIhwVm
Calculus:
OpenStax(online): openstax.org/subjects/math
OpenStax Vol 1: amzn.to/2zlreN8
OpenStax Vol 2: amzn.to/2TtwoxH
OpenStax Vol 3: amzn.to/3bPJ3Bn
My Filming Equipment:
Camera: amzn.to/3kx2JzE
Lense: amzn.to/2PFxPXA
Audio Recorder: amzn.to/2XLzkaZ
Microphones: amzn.to/3fJED0T
Lights: amzn.to/2XHxRT0
White Chalk: amzn.to/3ipu3Oh
Color Chalk: amzn.to/2XL6eIJ

[@prime_suntereh2241]

Wilson + inequality + bionomial + basic congruence = amazing problem

[@HarmonicEpsilonDelta]

10:47 I think we used the hypothesis when we multiplied 2*(n/2), because for that to be true, we need both 2 and (n/2) in the product, and for that, we need n/2 to be different than 2, since they come from the factorial, and therefore n>4, which implies n\geq 5.

[@goodplacetostop2973]

12:03 14:23

[@caesar_cipher]

Something very important happened at 12:03 - u must check it out

[@goodplacetostop2973]

Caesar Cipher Indeed! Thanks 👍

[@R0M4ur0]

In general, in this video we have a "that's a good place to stop", quantity squared

[@goodplacetostart9099]

Good place to start at 0:00

[@CM63_France]

@@R0M4ur0 along with "let's go head and do that","all the way {up|down} to","and so on and so forth"....

[@factsverse9957]

It appeared on the SMO (Singapore Mathematical Olympiad), Open Round 2 on the year 2008, as problem number 1. Hope it helps!

[@factsverse9957]

@Adam Romanov You can find the pdf of it online, I got a pdf from a website fadjarp3g, plug in SMO Round 2 2008 fadjarp3g in the Google search bar and I think you should find it, hope this helps

[@yusrizalakbar3983]

14:00 he said "that's true for n≥5" but 4^4 > 4!, why is it I'm sorry if it's a dumb questions

[@yusrizalakbar3983]

@qu-ack oh okay thanks

[@roboto12345]

I was trying that exact problem, it was in an LTE article. Thanks

[@chaitanyagaur7928]

Training for maths olympiad?

[@swayamprakashkar9664]

Very nice solution... Nailed it👏👏

[@Kettwiesel25]

I wasn't aware of Wilson's Theorem, but it's not too hard to see that n+1 has to be a prime in this problem, this is the easy part of Wilson's Theorem. If n+1 has a factor, which is smaller than it, then it appears in n! as well as (n+1)^k, so the two can not be just 1 apart (they are not coprime)

[@ffggddss]

First thing I thought when I saw the thumbnail: "Hey, that looks like Wilson's Theorem! . . Oh wait, not quite." Wilson's Theorem says that p | [(p-1)! + 1], iff p is prime. Here we would make n = p-1 and this becomes (n+1) | (n! + 1), iff (n+1) is prime; vs our problem: (n+1)ᵏ = n! + 1 which is a stronger condition on (n! + 1) than mere divisibility by (n+1). k & n are restricted to the +ve integers, so start with k=1: N+1 - 1 - n = n! Works for n = 1 and 2; no others. Solutions: (n,k) = (1,1), (2,1) k=2: n! + 1 will have to be a square. I think the only cases of that are n = 4, 5, and 7. n = 4 works for this equation; 5 and 7 do not: 4! + 1 = 25 = 5² 5! + 1 = 121 = 11² 7! + 1 = 5041 = 71² So: (4+1)² - 1 = 4! . . 24 = 24 Solution: (n,k) = (4,2) I suspect there are no others, but I don't see a way to prove that. Let's see what Michael has... ... Very nicely done! So my suspicion was borne out. Thanks to Michael for the proof. Fred

[@mihamihailovic4519]

If it isn't too much of a hastle can you please put the video link in the description for theorems you used and proved in another video (like wilsons theorem here)

[@a_llama]

10:47 im guessing n>5 because you assume that there exists an n/2 >= 3, meaning n>=6

[@UndeadFil]

Correct, and since n+1 has to be prime, the case n=5 can be ruled out because 6 obviously isn't prime.

[@mariomestre7490]

Una equació meravellosa. Genial

[@angelvalera1997]

I love those videos. Greetings from Spain :D

[@mathissupereasy]

I would love to see you working on inequality problems.

[@newkid9807]

Le Family Patriarch Bao Quoc Le

[@TwilightBrawl59]

3:25 Not a big mistake but the left hand side is even and the right hand side is odd.

[@kuldeepparashar7266]

Thanks sir

[@jamesfortune243]

Seeing your proofs is making me smarter. :)

[@mcwulf25]

Clever solution. Especially noticing n/2 and 2 are factors of n! when n even.

[@pendawarrior]

Fantastic

[@TheFlash-bw3hi]

Woow dope explanation.... Can u please try Putnam 2014 a5 , the official solution is unbearable 😭

[@math_infinity_]

Good question

[@TechToppers]

That was brain storming! Uff Will see again...

[@thayanithirk1784]

Please do Putnam 2006 B2

[@prunodagen]

You don"t need Wilson Theorem. (n+1)^k = n! + 1 n! + 1 is odd for n > 1, then n + 1 must be odd, then n must be even. (we don't need n + 1 prime, just the fact n is even to get k = 0[n])

[@Miju001]

Woo, that's slick

[@spiderjerusalem4009]

"just the fact that n is even to get k=0" how?

[@TechToppers]

Let s(n) denote the sum of the digits of a positive integer n in base 10. If s(m) = 20 and s(33m) = 120, what is the value of s(3m)? Can someone please help me with this? It came in India. PRMO 2019 25th August Paper

[@WriteRightMathNation]

Nice problem, but at 3:28 (someone else may have already pointed this out...), you mixed up "left-hand" and "right-hand"... 😊

[@abhishekbhattacharjee4039]

This problem is beautiful

[@joaopedrobmenezes2977]

Vp(x!) >= 2*Vp(x) for all non prime x such that x> 4, then you can see that if n>4 and it is not prime( which is the case for n >4) k would have to be at least n, but (n + 1)^n - 1 > n! for n > 4. This is just an alternative solution:)

[@gregsavitt7176]

Just got out of Calc 3 thinking that I had gotten pretty good at math. Then I watched this and my brain melted.

[@MysticMagik]

This is mere child's play compared with calc 3

[@spiderjerusalem4009]

​@@MysticMagikquite the opposite. Lots of calc stuffs are straight-up chug-in-plug-in dumb play, whilst number theory and combinatorics are whole different unpredictable beast

[@chhabisarkar9057]

A nice little problem from INMO 2014 , problem 2 Prove that [n] + [n/2] + [n/3] ....+ [n/n] + [√n] is even , where [n] is the floor of n Plz see it . Thank you

[@Kettwiesel25]

Proof sketch: Notice that f(n) - f(n-1), the increase from n-1 to n, is the number of factors of n, except if n is a square number, then it is 1 more. Now use that a number has an odd number of factors if and only if it is a square.

[@TechToppers]

Hello 😂

[@chhabisarkar9057]

@@TechToppers hmmmm

[@spiderjerusalem4009]

Let f(n) be the above expression f(n-1) =Σⁿ⁻¹ₖ₌₁ ⌊ⁿ⁻¹/ₖ⌋ + ⌊√(n-1)⌋ = Σⁿ⁻¹ₖ₌₁ (⌈ⁿ/ₖ⌉-1) + ⌊√(n-1)⌋ (Here, we used the fact that ⌈ⁿ/ₖ⌉=⌊ⁿ⁻¹/ₖ⌋+1 for any n,k∈ℕ) f(n)-f(n-1) = n-Σⁿₖ₌₁ (⌈ⁿ/ₖ⌉-⌊ⁿ/ₖ⌋) +⌊√n⌋-⌊√(n-1)⌋ Let Δ(k,n)=1 if k|n, 0 otherwise hence ⌈ⁿ/ₖ⌉-⌊ⁿ/ₖ⌋=1-Δ(k,n) summing from k=1 to n, it equals Σⁿₖ₌₁ (1-Δ(k,n)) = n-Σₖₗₙ 1 = n-τ(n) Thus f(n)-f(n-1)=τ(n)+⌊√n⌋-⌊√(n-1)⌋ •》Case 1: n is perfect square implying ⌊√n⌋=1+⌊√(n-1)⌋ hence f(n)-f(n-1)=τ(n)+1, which is even since τ(n) is odd if and only if n is a perfect square •》n is not perfect square implying ⌊√n⌋=⌊√(n-1)⌋ hence f(n)-f(n-1)=τ(n), which ia even in either case, f(n)-f(n-1)=k for some even k summing from 2 to n f(n)=(n-1)k+f(1) = (n-1)k+2 = 2((n-1)(k/2)+1) which is even for any n∈ℕ, as desired.

[@stevenwilson5556]

6:53 why is n + 1 a prime? Does this cover cases where n + 1 is not prime?

[@chhabisarkar9057]

Reverse wilson

[@subversively6680]

By applying Wilson 's theorem, can we find the rest of prime numbers that are >100 (theoretically speaking)

[@ruairigarrett6834]

Good luck calculating those factorials lol

[@iabervon]

@@ruairigarrett6834 You only need them mod p+1, so at least you're not dealing with numbers more than about p², but you're still doing p multiplications and might as well just try dividing p by everything less than p.

[@hans-juergenbrasch3683]

n=1,2 imply k=1. Now for n>2 we must have n+1=p is a prime, otherwise n+1 would have a proper divisor 1 < d =< (n+1)/2 < n which yields a contradiction modulo d: 0=n!=(n+1)^k-1=-1 and hence we can rewrite p^k-1=(p-1)! with p>=5 or 1+p+...+p^(k-1)=(p-2)! Note that gcd(1+p+...+p^(k-1),p-1)=gcd(k,p-1) Since (p-1)/2 < p-2 we must have k = t (p-1)/2 for some t. Since p^(p-1)-1= = (1+(p-1))^(p-1)-1>(p-1)^(p-1) we must have t=1, i.e. p^((p-1)/2)-1=(p-1)! For p=5 we get a solution corresponding to n=4, k=2 and for p>5 we get p^((p-1)/2)-1= =prod_(i=1)^((p-1)/2) [i(p-i)]> >[p^2+(p-3)^2-5]p^[(p-1)/2)-2]> >p^((p-1)/2), a contradiction. Note, that we have split the product into two factors, one for i=1,2 given by (p-1)2 (p-2) and the other for i=3,..., (p-1)/2

[@user-A168]

Good

[@sylviabayard6117]

A way to skip Wilson's Theorem: Add 1 to both sides, to get (n+1)^k = n! + 1 Well, we know that n! + 1 is coprime to any number less than or equal to n, so any primes dividing (n+1)^k must be greater than or equal to n+1. However, by the very form of it, any primes dividing (n+1)^k must be less than or equal to n+1. Therefore the ONLY prime dividing (n+1)^k must be n+1.

[@factsverse9957]

Then? How can we proceed, it's unclear for me

[@blackmephistopheles2273]

@@factsverse9957 Continue the argument in the video, after the point where Wilson's Theorem was used; this argument above shows that WT is unnecessary. So, continue from 6:56 (edited to find time reference).

[@Pedro-fg4sw]

LTE Lemma

[@spiderjerusalem4009]

but... that is just a rephrase of the converse of wilson's theorem which was applied in the vid....

[@rustemtehmezov9494]

Where is guy says everytime "answer=1"?

[@masonholcombe3327]

Video idea: Prove the Hardy-Ramanajan Series for partitions of n :)

[@geometrydashmega238]

He has a playlist of partitions and I think he was going to do that, or did it already. Check that out

[@jideofororamah4399]

Because K is the power and n=2 (n+1)^k = n!. 1 is a constant so how did you get k=1 where n=2

[@RealLifeKyurem]

3^k = 3 What k satisfies this equation?

[@minh9545]

@@RealLifeKyurem what do the 3 stripes and the -1(mod p) mean?

[@RealLifeKyurem]

TNT ≡ is read “is congruent to.” -1 mod p means if you divide a number n by a prime p, the remainder is -1. In other words, it satisfies the expression: apm - 1; for any integer a. The reason why ≡ is used instead of =, is because any multiple of p satisfies the expression.

[@LiangLao2]

just a suggestion, can you open a new channel? one channel focuses on math question especially the Olympic question, one channel focus math theory such as analysis algebra number theory, etc.

[@Polpaccio]

1 year later , 200k more subs

[@donaldbiden7927]

Beautiful solution ! Kindly tell the motivation for the inequality.

[@ericslavich4297]

Once we know k is a multiple of n, you look at the original equation again. You realize the exponent on the left hand side is at least n. Also note n^n > n!, so the left hand side is growing more rapidly than the right. The rest is details and being rigorous.

[@accountname1047]

The form of the numbers is similar to the brown numbers

[@xinfuchen5355]

No Wilsons Theorem is needed. When n>4, i) (n+1)^k= 1+n!= 1 mod (2). This imples that n is even ii) n even implies that (n-1)! = 0 mod (n) iii) k = ((n+1)^k-1)/n mod(n) = (n-1)! mod (n) = 0 mod(n) (iva) k=0, no (ivb) k>=n, Then (n+1)^k > n^n -1> n!-1. No solution, Hence, when n>4, there is no solution.

[@Teodyk]

Obviously.

[@rustemtehmezov9494]

You get (n+1)^mn-1>nⁿ>n! but isn't it was (n+1)^mn-1=n! and this means n!>nⁿ? İ think it was typo

[@oussemabouaneni992]

That's the contradiction. He supposed that such an n exists and proved that its existence is impossible.

[@rustemtehmezov9494]

@@oussemabouaneni992 but if we get n!>nⁿ then this will be contradiction, because this never holds for these rules. But he got nⁿ>n! which is clearly true. He should be write n!>nⁿ (he got it at the end but wrote it wrongly i think, but i understood his proof method clear)

[@oussemabouaneni992]

@@rustemtehmezov9494 No no he wrote it perfectly correctly. He proved that (n+1)^k - 1 > n^n > n! which means that (n+1)^k - 1 is never equal to n! for n >= 5 which is what he set out to prove.

[@rustemtehmezov9494]

@@oussemabouaneni992 Oh, now i understood what you mean, yeah this is accetable. (Almost same as i supposed) Thanks for clarification👍

[@oussemabouaneni992]

@@rustemtehmezov9494 Cheers mate.

[@tomholroyd7519]

Count Olaf

[@user-vk6nl6mi1d]

0 not a natural number?

[@mcbeaulieu]

What is Mod? 😭😭😭😭

[@kruksog]

Mod gives remainder. Some examples: 5 mod 3 = 2 because 5/3 has remainder 2. 4 mod 2 = 0 because 4 is divisible by 2 (i.e. no remainder). Once again, 7 mod 2 = 1 because 7/2 has a remainder of 1. It's important to remember when doing elementary number theory, we work in the integers and naturals. Hence all the whole number concerns.

[@devrajdas6544]

Is there any problem you can't solve?

[@vaxjoaberg9452]

Global climate change?

[@xinfuchen5355]

No Wilson's Theorem is needed. When n>4, i) (n+1)^k= 1+n!= 1 mod (2). This implies that n is even ii) n even and n>4 imply that (n-1)! = 0 mod (n) iii) k = ((n+1)^k-1)/n mod(n) = (n-1)! mod (n) = 0 mod(n) (iva) k=0, no (ivb) k>=n, Then (n+1)^k-1 >=( n+1)^n -1>=n^n>n!. No solution, Hence, when n>4, there is no solution.

[@seroujghazarian6343]

For integers, n^n=>n! Is true. But beyond that, it's not always true. Case in point:(1/2)^(1/2)=sqrt(2)/2 < (1/2)! = sqrt(pi)/2

[@HagenvonEitzen]

13:33 Or just (n+1)^n -1 >= (n+1)n(n-1)... 2 - =(n+1)! -1 > n!

[@mandelbrotsugee]

じっと見ていると、n=4と分かりますな。 K=2

[@tyjensen8366]

God he’s such a chad

[@guibaroleo]

hahahahahaha

[@sdspivey]

You say K must be a multiple of N, but the solutions given have N as a multiple of K.

[@axelperezmachado3500]

all solutions with n >= 5 must have k as a multiple of n. That does not necessarily apply to the given solutions.

[@jideofororamah4399]

I got confused at how you got k=1 at n=2

[@ddp8511]

Just put n=2 U get 3^k=3

[@jideofororamah4399]

Angel Mendez-Rivera That solves it. Thanks 🙏❤️ @michael penn

[@keithmasumoto9698]

Immediately I don't get why n+1 is prime. If n=5 then n+1=6 is not prime.

[@RandomGamer9]

I had to rewatch to understand it but if you reduce (n+1)^k-1=n! (mod n+1) you get 0-1=n! (mod n+1) then you rearrange to get the form from Wilson's theorem and that n+1 is prime. The zero in the above equation come from the fact that (n+1)^k is divisible by (n+1)

[@CM63_France]

Hi, This was before hair cutting, so I understand my request about the camera is not taken into account.