Equation on Factorials

[@isaacchan8159]

Sir you are such a legend. This video just appeared in my recommended list. I don't like math before but after I watched your video, I started enjoying math for the first time. Your tutorials are so much better than teachers in my high school classes. Good job and keep going, can't wait to see you get 100k subs!

[@yoav613]

i have started watching your videos few days ago and they are really great. very enjoyable

[@jarikosonen4079]

Good... Usually these factorial related diophante cases have only small numbers included, like a,b

[@gastoncastillo9946]

How did you come up with the idea of doing that claim? It seems a little bit out of nowhere. Apart from that, the explanation is superb. Keep on!

[@mcwulf25]

It's a simple claim anyway, as those prime terms on the LHS must both be odd (3 is not a factorial) so c^2 must be even. So C must be even and c^2 must be a multiple of 4.

[@HagenvonEitzen]

9:30 Or: If a=b then c²+1 = (a!-1)²m which would match only 0²+1=1² (but here we consider 0 not to be in N)

[@shreyamjha3058]

Amazing problem with a nice motivation

[@HagenvonEitzen]

4:40 or just note that (Z/pZ)* is cyclic of order p-1, so cannot have an element of order 4.

[@gervasiociampin2062]

Can you explain please?

[@joaquingutierrez3072]

Why not? (Z/5Z)* has an element of order 4, namely, 2. I would say that (p-1)/2 must be even because if it is odd we have that - 1 = 1 (mod p) 2 = 0 (mod p) So p|2, so p = 2, which is a contradiction because we assumed p odd.

[@Deathranger999]

@@joaquingutierrez3072 I think this is if p is congruent to 3 mod 4, since then we would have to have 4 | p - 1, which is clearly impossible.

[@242math]

understand it now, great job

[@dionisis1917]

I think that the claim is all the exercise . Also with the same method find all natural number x,y,z such that: 4xy=z^2+x+y

[@alainrogez8485]

10:08 0 is a natural number. You asked for solutions for any n in N, not N*.

[@mryip06]

seems you were using an iPad and having a bug keeping flashing, right? any method to solve this issue?

[@subhrayanbarman1654]

Very easy

[@law1409]

How did you come up with those bounds?

[@-basicmaths862]

(a,b)=(2,3),c=2

[@mcwulf25]

Is there a gap in your proof? The LHS will always == 1 (mod 4) for a,b >= 4 as will the RHS. Which is what we want, no?

[@typha]

When you say that it remains to show only 6 cases, you still haven't yet considered cases where at least one of (a!-1) or (b!-1) are not odd (of which there are infinitely many). You cover those cases about 2 seconds later, but I think if I were to ask a student "where did he rule out (a,b)=(1,7) as a possible solution?" they might become confused. Fun problem though. And besides that small thing, well done :)

[@adiaphoros6842]

“When did he rule out (a,b) = (1,7)?” When he ruled out all the cases a ≥ 4. Without loss of generality, he assumed a > b. So (a,b) = (1,7) is considered to be the same as (a,b) = (7,1).

[@typha]

@@adiaphoros6842 In that portion of the argument he only ruled out the cases a ≥ 4 where (a!-1) and (b!-1) are both odd. (6:23) In the narration he says that these two conditions "must" happen, but in actual fact (b!-1) being odd does not follow from a ≥ 4. (b!-1) being odd is an additional assumption that is required to make that part of the proof work (as presented, the assumption is actually not required and yet by invoking it he has restricted himself). That is exactly the mistake I was expecting people to make ;) He actually covers those cases where (b!-1) is not odd at 8:46 (or really at 8:59), as I alluded to in my original comment.

[@gervasiociampin2062]

Why mod 4 and not mod 2?

[@joaquingutierrez3072]

(p - 1)/2 is even So (p-1)/2 = 2n (p - 1) = 4n So p = 1 (mod 4)

[@omsinghvashishtha6398]

"MIND your decision" , you replaced c -> c^2.

[@ImaginaryMdA]

This one was really difficult.

[@wojciechwisniewski6180]

I really like your videos, but your diction is something I can't handle. I think it's very important to speak clearly when you are teaching and things like "zee-vo" ("zero") can destroy even the best video...