Perfect numbers and Mersenne primes
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- Опубліковано 29 кві 2024
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michael never fails to do somthin that I have no freaking idea about
Well, he fails to mention that it is called Euclid-Euler theorem, who proved sufficiency and necessity respectively.
open any elementary number theory book, you'll find this section (David Burton one would be foremost recommend)
@@wesleydeng71 he searches for content not sufficiency
You must be new :) This is a very famous result. There's even a Veritasium video about it from a month ago.
15:50 It was many years ago but I’m pretty sure I had this claim in a final exam
This is the Euclid-Euler theorem. Is that a record for longest gap between the birth years of two people a theorem is named for?
to be much more precise as to why a=1
if a>1,
σ((2ᵐ⁺¹-1)a)≥1+a+(2ᵐ⁺¹-1)a=2ᵐ⁺¹a+1
which would contradict the given equality, therefore 1≥a, but a is a natural number, thus a=1
σ(q)=σ(2ᵐ⁺¹-1)=2ᵐ⁺¹=q+1
⇔ q prime (since a number is prime if and only if its divisors are only 1 and itself, that is, if and only if the sum of its divisors is 1+itself)
⇔ (2ᵐ⁺¹-1) prime
⇒ m+1=p is prime (otherwise, 2ᵐ⁺¹-1 could be factored into two numbers, each of which is greater than 1, which would then imply that q is not prime)
⇒ q=2ᵖ-1 Mersenne prime
⇒ n=2ᵖ⁻¹(2ᵖ-1)
It's also called the Euclid-Euler's theorem since Euclid proved the sufficiency (the forward implication), and Euler proved the necessity (the converse, that is the form of all even perfect numbers)
2^n - 1 being prime requires n being prime. Written in binary (2^n - 1) is "1" written n times. If n were composite and divisible by k, the number (2^k - 1) ["1" written k times] would divide the original number easily.
@15:15 when did we show m+1 is prime. All we know is that 2^(m+1)-1 is prime q, where did we get p from?
He glossed over the details, but early in the video he mentions that if 2^k-1 has k composite then it can be factored and is therefore not prime. Since we know 2^(m+1)-1 is prime we can conclude m+1 is prime.
@@BridgeBum ah right, I forgot that part by that point in the video. I guess he could’ve mentioned that again quickly as a reminder, but my bad. Thanks!
Very nice proof at the end! For some reason, I thought this proof was more complicated, perhaps because it was first proven by Euler!
Had a hw question using this result and it eventually came up in the final exam lol. The professor really loves this question
Hell. I'd be stoked if he just explained how the Mersenne primes worked and why
Thanks!
3:27 so we pretty much Don't know anything about them lol.
13:15 onwards: there was no reason to complicate things with inequalities; we have established that σ(q) = 2^(m+1) * a, and q = (2^(m+1) - 1) * a, so clearly
q+a = 2^(m+1) * a = σ(q)
Hey, i had one doubt can you help me. I want to know how we are claiming m+1 to be prime 😅
@@puneetbajaj786 we know it has to be prime because at that point we know q is prime. He mentions near the start of the video that to be prime a number of the form 2^p - 1 must have p prime.
It took me a moment to realise why the second half of the proof only applied to even perfect numbers, since it didn't seem to explicitly use that fact anywhere.
But, I believe the key part is when writing "sigma(q) = a + q + other stuff", since this only works if a is a proper divisor of q. But if q is odd (so m=0) then it shakes out that a=q, and this step doesn't work.
It's used in the fact that σ(n) = 2n. This is only true for perfect numbers since the definition of a perfect number is that it is the sum of its proper divisors.
@@allanjmcpherson Right, yeah, I just meant the "even" part. Like, it wasn't immediately obvious why you couldn't just say "let n be a perfect number" instead of "let n be an even perfect number".
@@mrphlip oh I see what you mean. I suppose just because the first half only applied to even perfect numbers? I agree that restriction isn't strictly necessary, but I suppose it preserves the symmetry of the argument.
@@mrphlip I think? I've honestly already forgotten the details of the forward direction. I think it only applied to even perfect numbers, but I'd have to go back and check to be sure.
It is trivial to verify that if a and b are distinct primes, then sigma(a*b) = sigma(a)*sigma(b). But the more general statement (at 10:03) that this holds if a and b are coprime is not obvious to me. Is this proved in another video?
If a and b are coprimes σ(ab) = 1 + Sum(p_1q_1)^j(p_2q_2)^k... where every p^j and q^j don't share any common factor and their exponents are determined biunivocally (there is no overlapping) so σ(a)σ(b) gives exactly the same result as σ(ab). If they were not coprime some factor q in b would be of the form q=kp^n and exponents would overlap., that is, there would be a divisor repeated in the sum of σ(a)σ(b) which is not repeated in σ(ab).
This conventional math is what's holding most back and teachers luke this who are afraid to learn more than what they teach...I could teach you sir new mathematic freshly created and it can run in accordance with some current mathematics and applications
Relevant: ua-cam.com/video/Zrv1EDIqHkY/v-deo.html
all perfect numbers are even - they’re isn’t a formal proof that all perfect numbers are even but it looks that way.
I have assuredly discovered an odd perfect number but the UA-cam comment field is too small to contain it ;-)
@@zh84 fermat last theorem moment
They are only perfect when a grouping of them makes them equally perfect, otherwise, a candidate just might be a flesh wound of the others, and they accept that candidate to hide the guilt.
sorry Michael, I'll never stop disliking pure number theory
All perfect Numbers are indeed even, right?
He said it is unknown that there are any odd ones and it hasn't been proven that there are any odd ones.