a very classic "selection" problem

[@alex_schwartz]

also for the first case, "a" and "b" can be interchanged for another solution (p=3, a=1, b=2, c=2)

[@darkmask4767]

For the p=3 case, a=3^k, b=2*3^k, and c= 3k+2 for a nonnegative integer k is a solution (switching a and b also gives a valid solution).

[@zygoloid]

I think this was missed because the statement of Zsigmondy's theorem was incorrect -- it applies only when a and b are coprime.

[@minamagdy4126]

For a similar reason as the initial logic of the p=2 case, Zsigmondy's theorem proves that this is the only family of solutions for p >= 3

[@davidbrisbane7206]

Zsigmondy's theorem is doing a lot of heavy lifting in this problem. Its proof is not that trivial.

[@leif1075]

Isnt that a complete CHEAT and not fair as usual since no one is going to think of that theorem no matter how amart you are unless you've seen it before right?

[@davidbrisbane7206]

@leif1075 In this case, I think you have a point. Actually, I read that this theorem might not be allowed to be used in IMOs because high school kids would find it tough to understand how it could be proven.

[@goodplacetostop2973]

11:51

[@Rishabh00Shukla]

10 missed calls from Andrew Wiles! 😂😅

[@evreatic3438]

Zsigmondy Theorem was stated incorrectly. I believe the correct statement is: If a>b≥1 are *coprime integers*, then for all integers n≥2 there exists a prime divisor of a^n - b^n (called _primitive prime divisor_) that does not divide a^k - b^k for all integer k, 1≤k

[@Aman_iitbh]

a=6,b=3,c=5,p=3 is another soln ,every prime multple family is solution ,with p=3

[@bsmith6276]

I just had to check: 6^3 + 3^3 = 3^5 simplifies to 216 + 27 = 243 and that is indeed true. So it seems there is something wrong either with how Michael applied Zigmundy's theorem or some conditions on the problem and/or theorem are missing.

[@TheDerkus]

Babe wake up new Michael Penn video just dropped

[@devrimturker]

Resembles "Fermat's Christmas Theorem"

[@richardchapman1592]

Can you prove that for any added n primes divided by n it yields a prime when either n an integer and/or n a prime? This could be important on constructing an n dimensional space with derivatives and integrals based upon saw functions instead of circular ones.

[@supratimsantra5413]

Again a great boom 💥💥... congratulations sir from behalf of all Indian mathematics persuing students and mentors also❤

[@disoriented3971]

Unfortunately I wasn't able to follow this logic, got lost before 3:00

[@writerightmathnation9481]

Your statement of Zsigmondy’s theorem cannot be correct. If you read it, logically it implies that the part before the word except is always false. Moreover, the exception statement at the end does not imply by itself that p is equal to 3 because p is not mentioned in that portion of your statement of the theorem. In fact, this is part of the reason for the first part of my comment here: 3 always divides 2^3 +1^3 and 3 always divides 2+1 (in the integers). Finally, even if you make these corrections, specifying that a is equal to two and B is equal to one in the exception would be problematic because the order of a and B can be switched in a (possibly) correct statement of the theorem. In the above, I put the word “possibly” in parentheses, because when I looked up the statement of the theorem in Wikipedia, it doesn’t look the same, and I haven’t worked out why it means the same. en.wikipedia.org/wiki/Zsigmondy%27s_theorem

[@falquicao8331]

It seems to me that you can rewrite the theorem statement as "for all triples of integers (a, b, n) that are different from (2,1,3) and (1,2,3) there exists a prime p such that..."

[@writerightmathnation9481]

@falquicao8331 Trees but this isn’t the same as what’s on the board, and it’s not immediately clear that it’s equivalent to the Wikipedia version of the theorem statement.

[@evreatic3438]

Wikipedia isn't entirely accurate either. I posted the correct statement of the theorem in a separate comment.