quite a nice couple of problems

[@xinpingdonohoe3978]

That was a cool quantum calculus reference. I enjoyed those videos; didn't expect them to be utilised here, or really anywhere.

[@supratimsantra5413]

Very very cool problem sir..... thanks you so much for sharing with us.....we are here with motivation for your valuable wonderful lectures.

[@joelklein3501]

4:04 I am not very familiar with discrete mathematics and number theory, but I believe that the argument here is incomplete. We didn't show that a≠b, and therefore we can't say for sure that both a and b show up in (p-2)! , when including multiplicity. Unless there is some theorem I'm not aware of that says that no factorial is a square of a natural number (besides 1, I guess) then I fail to see how this necessarily works

[@joelklein3501]

Actually, since p-1 is always even, we can always choose a=2. Then, b = (p-1)/2 > 6/2 = 3 >2 = a So we found two different factors of (p-1)😁

[@pavlopanasiuk7297]

I guess it is cooler not to reference q integral here, but to instead demonstrate this to be a Riemann sum. If you select x_n = q^n, you get Δx_n = q^n (q - 1), which tidies the sum a lot and leads to the same integral

[@ahzong3544]

it's also quite cool from the content creating perspective to engage viewers into watching his other videos on quantum integrals, id guess that this is the main motivation of him presenting the solution this way

[@rupeshkapoor5380]

5:21 air quotes "big enough"

[@Your_choise]

If a=/= b then p-1| (p-2)! If a=b, then p-1=a^2, with p-1 ≥6, so a ≥3, so 2a=3a-a ≤a^2-a

[@WhatIKnowAboutThat]

@@Your_choise Thank you

[@Xeroxias]

I did the first one by myself. Where Michael went for a contradiction, I inferred highly restrictive properties of p that make 2, 3, and 5 the only candidate solutions. Otherwise, the problem is simple enough that there aren't many differences in the approach. We can check that this holds for p = 2, 3, 5, but not for p = 7. Let's assume p is at least 11 and see if there are any more. (p - 1)! = pⁿ - 1 = (p - 1)(1 + p + p² + ...) (p - 2)! = 1 + p + p² + ... reduce modulo p - 1 (p - 2)! = n So the product of all nonzero elements of the integers modulo p - 1 must be congruent to n (that is, if p = 7, the product is 1*2*3*4*5). But that means there can't be a pair of divisors a, b such that 1 < a < b and ab = p - 1, because then the product would collapse to 0 mod p - 1. Since p - 1 is composite for p > 3, so the product will include at least one non-unit divisor of p - 1, the only way out is if p - 1 = a², because a will only appear once in the product. (The fact that 2 - 1 and 3 - 1 are not composite is the loophole that permits them as solutions!) But again, if a is composite, a = cd, then you can write a² = (c)(ad) = p - 1, expressing p - 1 as the product of two distinct non-unit factors. We conclude that there must be some prime number q such that p = q² + 1. That means that p and q must have opposite parity, so one of them is 2, and that must be q, meaning p = 5 is the only prime 1 larger than the square of a different prime. Thus, the three primes found in the beginning are the only such primes.

[@andrea-mj9ce]

Why does one have the equality that starts at 4:53?

[@Charliethephysicist]

The same results for the first problem follows even if p is not prescribed to be prime.

[@alexchan4226]

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