a beautiful prime number equation
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- Опубликовано: 14 окт 2024
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At 9:51, we rule out the cases p = 19 or q = 19, which do *not* correspond to pq | 19.
However, if p ≠ 19 and q ≠ 19, then pq | p-q+19, as he writes later on.
tldr: Logic is sound; just ignore the pq | 19
I mean this problem is way easier than you made it seem. Once p=2 and q=2 are dealt with, just use a size argument; clearly we need p=p+2, left hand side will be bigger than RHS by expansion of binomials for instance.
Came here to say this. That said, it's nice to see a completely different argument.
Thank you Michael. The difficult non-standard problems that you tackle on this channel are very stimulating ! ( But I missed the back flip this time ).
At 10:00 you don't have to check cases, you already have pq|19 which can't happen because 19 is prime.
He had pq|19*a and then checked, if p or q can be 19 ("pq|19" should be 19|pq). Both is not possible and therefore, pq|a.
8:55 missed minus sign
And that should be the place to stop
Alternative way to see that if (p-1)(q-1)=3, q>=3:
From the equation we know q is congruent to 19 mod p, which is equivalent to q=mp+19 for some integer m. Since p>=3, if m>=1, then q>=22, but q is prime, so q>=23 and then q-1>=22>20. Since p-1>=2, (p-1)(q-1)=3, but q is prime, so q=3, which means p is either 3 or 7 and eliminates q=3 and q=11. Now, p is congruent to -19 mod q, so if p=3, then q=11, which is already eliminated. If p=7, then q=13, which contradicts the correspondence above (if q=13, then p=6).
Case p >= 3, q >= 3:
p^q-q^p = p*q^2 - 19 mod q => p = m*q-19 (FLT), for some integer m (I)
p^q-q^p = p*q^2 - 19 mod q => q = n*p+19 (FLT), for some integer n (II)
Now multiply (I) by n and (II) by m and replace the values of n*p and m*q to obtain the following equations:
q = 19*(1-n)/(1-m*n) (III)
p = 19*(m-1)*(1-m*n) (IV)
Bounds for m and n can be obtained by observing that p = 19*(m-1)*(1-m*n) >= 3, producing:
m*(19+3n)>= 22 => m>=1 (since m is not negative because p = m*q-19 >= 3 and q>=3) and also n>=-6 (since 19+3n >= 0 => 3n >= -19 > -21 => n > -7)
By replacing the possible values in (III) and (IV) it is easy to check that no solution is possible:
For n>=2, q=19*(1-n)/(1-mn) = 19*(1-2)/(1-2*m), which forces m = 1. But m cannot be 1, otherwise p = 0 (contradiction)
For n=1, q = 0 (contradiction)
The other cases can be checked as homework by the reader.
@ 8:51 Should be p is congruent to -19 mod q.
Since p divides q-19, then p < q. Since q divides p-19, then q
8:55 It should be q | p-19. That changes |p-q+19| at 11:55 to |-p-q+19| which luckily doesn't change the logic that follows.
15:54
Hold on, something goes wrong at 8:54
They should have used this problem at the Balkan M.O. 2019 instead of 2004 🙂
Why?
@@khoitruongbao5118 Tradition. A lot of the M.O. number theory problems contain the year's digits in the pb statement.
So how do M.O.'s work? Is it team or individual? Is there a time limit? Is there partial credit? Do contestents work at desks or on the board? How many contestants?
I would be totally flumaxed at one of these. I think I'm fairly good at applied math (when I need the answer as a physicist/engineer) but I'm not fast at it, and never do it the first time in public!
@@TomFarrell-p9zIn my country Belgium, the MO is individual with a 90min time limit for 30 questions, working on a desk. Since there are 3 stages in the BMO (qualifiers, semi-finals and finals), the starting number of contestants can be as many as possible, since only a fixed number of them can pass through to the higher stage. The questions are either MC or any number between 0 and 999 included. At the end, you simply submit your paper to be reviewed on the spot, so you can immediately have your score, and then get the results sent to the organisator. You get 5 points per good answer, 2 points per abstention, and 0 per wrong answer.
@@TomFarrell-p9zIn the International M.O. there are always 6 questions. The first three questions are asked in the first day in a 4-hour exam, the second three on the second day in a second 4-hour exam. The answers are written on paper using pen or pencil. No notes or electronics allowed. The answers are graded for correctness, completeness, and creativity, and there are partial credits. The format (2 sessions of 3 questions in two consecutive days, 4 hours each) is fixed. By tradition, the first three questions are easier than the second, and normally as you go forward, the questions become more difficult. I believe that questions 2 and 4 are always about geometry, but I have not looked at many IMOs, so, it's just a guess. You have to be fast, but there is no social anxiety of presenting your solutions.
And that's a good place to STOMP 🦶😜
Is this my lunch?
Yes, Double U, it is your lunch
@@titan1235813 I almost cracked my teeth trying to eat this video.
@Double_U_tau_Phi then try going to the math dentist to get those teeth fixed
@@titan1235813 Go to Michael for this
I don't understand how there's anything left to check after 10:53. If pq | 19 (p-q+19), this implies either p or q must have 19 as a factor, and since they are prime, one has to be equal to 19. And since at that point we've already shown that having either one of them equal to 19 leads to a contradiction, why do we need to check the case where pq | p-q+19?
I believe that your logic here is incorrect. If pq | 19 (p-q+19), it is possible that pq simply divides into (p-q+19) and the 19 factor in front is irrelevant. For example, consider the similar case of : pq | 19 (p-q+17). p=3 and q=5 is a solution of this equation, and neither one has 19 as a factor.
You are completelly correct.
We still know that p|(q - 19) and q|(p + 19). I believe Penn could have used this to rule out the small cases faster, but his work was solid enough.
It was something wrong in 8:54
14:30 what about p=5 and q=5? 4x4 is also less than 20.
He already discussed that p cannot equal q
Yeah, but 5^5-5^5 is still zero, which is definitely *not* equal to 5^3-19.
11:20 pq divides 19 times a bracket. You show pq can't divide 19 and say that means pq must divide the bracket. Can't one of p and q divide 19 and the other divide the bracket?
@@theartisticactuaryif any one of them divides 19, the other must be 19. And he checked this case (both become 19)
@@TheEternalVortex42 right... I missed that 👍
Nice
Endless ...............................................................................
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pas de raison de dire pq÷19 car impossible
ruclips.net/video/EGHnkT8WoSA/видео.html
15:22 you saw it here first! 15 x 15 = 75! 🤣
huh? that's 3x5^2 = 15 x 5 = 75. He made two mistakes, this is not among them. One when he checked pq | 19 (not necessary, already impossible, 19 is prime), and the second when he mixed a +- sign. None of them breaks the solution.
Booooooooring .....
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