What is the luckiest prime?
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- เผยแพร่เมื่อ 15 ก.ย. 2024
- We look at a nice number theory problem
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4:28 - 7:29 A much simpler way to get x = 4p-1 is to note that since p(p-1) = 2(y+x)(y-x) and y+x = p, that y-x = (p-1)/2. Then, x = (1/2)[(y+x)-(y-x)] = (1/2)[p-(p-1)/2] = (1/2)(p+1)/2 = (p+1)/4, and so 4x-1 = p
Man I have been watching too much news. I read the thumbnail as
"Find all prime ministers such that..."
You got p(p-1) = 2(y-x)(y+x), and then you determined that p = (y+x). That clearly means that p-1 = 2(y - x). Therefore 2p - (p-1) = 2(y+x)-2(y-x), which means that p + 1 = 4x, or p = 4x - 1. No quadratic necessary. I think that's a bit faster than going back to the original equation and subbing it in.
9:05
Have a great Sunday everyone, don’t forget to get enough sleep every night, your body needs it 🛌
I would but you’ve given me homework
@@jbthepianist You don’t have to solve it right now 😛
4:27
p(p-1) = 2(y-x)(y + x)
and p = x + y (1) . we can divide both parts by p =>
p - 1 = 2(y - x) (2)
let see on 2 * (1) - (2) =>
2p - p + 1 = 2(x + y) - 2(y - x)
p + 1 = 4x and from statement p + 1 = 2x^2 then
4x = 2x^2 => x = 0 or x = 2
Had to stop and rewind in a few places since my brain is rusty but i understood it in the end. Awesome.
I did it the hard (stupid) way. Checked all odd primes 3
The easy smart way, I'd say. Nothing wrong with lowballing a quiz answer first. Do the easy stuff first.
Pro tip to impress friends: Get your 7 some easy way, then, stare off into space, then, call out the answer. Richard Feynman would do this too, iirc.
How’d u prove there’s no sol for p>=11?
(Sry Im just starting to get into number theory)
@@mathlegendno12 Hope you'll love Number Theory, I'm a recent convert too.
John Conway, no longer living, wrote a book for it that helped me some. A character!
@@mathlegendno12 Man, I've been racking my brains all day but I just can't remember how I got that inequality. :/ And I can't find my notes. Try to prove it yourself using inequalities and/or modular arithmetics! Bon voyage.
Forgot to include the y, which if p is 7, then 2y^2 equals 50, so y equals 5
There seems to be possible to prove the same thing under the hypothesis that p is merely an integer: plugging p = 2x^2 -1 into the second equation, we obtain
(x^2)^2 + (x^2-1)^2 = y^2.
That is, x^2,x^2-1 and y are a pythagorean triple. In particular, they are part of the so-called twin pythagorean triples.
On the other hand, if one defines P_n to be the n-th Pell number (P_0 = 0, P_1 =1, P_{n+1} = 2P_n + P_{n-1}), it is known that all such triples are of the form
(2 P_{n+1} P_n, P_{n+1}^2 - P_n^2, P_n^2 + P_{n+1}^2).
Hence, we have two options:
1 - 2 P_n P_{n+1} = x^2. If q is a prime dividing x^2, then either q=2, or q divides P_n or P_{n+1}. Since P_n, P_{n+1} are coprime, then this shows that either P_n is a square and P_{n+1} is twice a square, or P_n is twice a square and P_{n+1} is a square. But this is easily characterised by a result of Ljunggren: the only square Pell number is 13^2, and the only one which is twice a square is 2. A direct verification shows that only the latter works.
2 - If P_{n+1}^2 - P_n^2 = x^2, then the same argument shows that either P_{n+1} - P_n or P_{n+1} + P_n is a square. But It is known that point of the form (P_{n+1}+P_n,P_n) are exactly the solutions to Pell's equation x^2 - 2y^2 = 1. In this case, a direct analysis shows that the only non-zero solutions to x^4 - 2y^2 = 1 are x= \pm 1, y=0.
By checking directly, one obtains that x^2 = 4, x^2 -1 = 3, y =5 is the only solution to that problem. Notice we never used that p is prime...
HOMEWORK : Find all real solutions of the equation 17^x +2^x = 11^x +2^(3x)
SOURCE : The 28th Annual Vojtěch Jarník International Mathematical Competition
Solutions x=0 and x=1 are easy to find. We have to think if they are the only solutions tho...
Wow, no integer? That seems tough
@@eduard9929 those are the only 2 solutions
@1071608張鼎翊 This is a good idea ! This means x=0 and x=1 are the only solutions !
@@eduard9929 SOLUTION
*{0, 1}*
Let us rewrite the equation as 17^x − 11^x = 8^x − 2^x.
It’s easy to see that x = 0 is a solution. Fix x ∈ R \ {0} and suppose that it is a solution to our problem. Consider the function f(t) = t^x.
By the mean value theorem applied on the interval [2,8] there is u ∈ (2,8) such that
6f′(u) = f(8) − f(2) = 8^x − 2^x.
Again - by the mean value theorem on [11, 17] we get 6f′(v) = f(17) − f(11) = 17^x − 11^x.
Since x is a solution, we have 6f′(u) = 6f′(v).
Since x ≠ 0, we have 6xu^(x-1) = 6xv^(x-1) which implies u^(x-1) = v^(x-1) and (u/v)^(x-1) = 1. Therefore x = 1, since u < v, and it’s easy to check that it is also a solution.
Impressive problem, associated with an impressive explanation.
Just have a wonderful weekend. Congratulations Michael!
May be some one might like to comment on this approach.
Using the first equation in the second you get
(2x^2-1)^2+1=2y^2
Which simplifies to
2x^4-2x^2+1=y^2
So the left must be a perfect square. Re-write this equation as
x^4+(x^4-2x^2)+1=y^2
Now the left is only a perfect square if
X^4-2x^2=2x^2 or x^4=4x^2 which solves to give x=2
So we get x=2 and y=5. This then gives p=7
I started doing the problem as in the video but I thought this looks simpler and I think is ok. Also this is correct it shows p=7 is not just the only prime but in fact the only number. By the way ^ means power in my notation.
Regarding x^4+(x^4-2x^2)+1=y^2, the left side can also be a perfect square if
x^4-2x^2= - 2x^2, which would give x=0, that is the other solution for x found in the video.
Nice Place To Start at 0:01
Good teacher. Thank you
How do we know that y < p? He says that we can tell from the equation p^2 + 1 = 2y^2, which feels right, but how does one go to prove it's right?
2y^2 = p^2 + 1 < 2p^2 -> y^2 < p^2 so y < p
@@nurtasshyntas7745 thanks for the reply
Don't even need to check that x=0 solution because the problem states that x is not equal to 0.
You don't really need the restriction that x,y are positive.
First and easy to guess such prime is 7
is there any way to generalize the question, for instance changing nu mber (+1) from LHS of both equations to a constant like (a) and keep the same analysis... assuming a
If 1
Funny enough, p,x, and y are all prime in this case.
p=7
x=2
y=5
And y is 5 just to top that off.
Ther are an infinite number of prime numbers and 7 is the only prime that satifies those constraints. Huh. Cool.
By trial and error...
p = 7
x = 2
y = 5
Great. Btw, could someone explain the ‘y+x+mp’ joke?
Michael Penn... or Metroid Prime
@@goodplacetostop2973 I was wooshed so hard... mp = member of parliament for me :D And then I tried to craft a pun mp = empty. Which didn't work at all!
@@emanuellandeholm5657 - I thought about like math machine gun or sorts...
well done, understand your solution bro
Once at 7:56 you find P=4X-1 that means 4x=2x^2. Why complicate ?
That’s exactly what he got to 10 seconds later. Are you complaining he showed the substitution step? That’s not “complicating things”. That’s showing the steps to get there
I salute the brain of this guy, he can justify so many arguments
Loved this one!
(p-x)^2 is an integer
I see what you did there with that *"luckiest"* prime thing
Nice and sneaky. Respect
I think you misspoke at 2:11
Parabolic
By direct proof
Whats a good book to learn number theory from? for a beginner/non-mathematician. Thanks !!!
Try Elementary Number theory by David M Burton first then after getting used to this book you can do modern Olympiad number theory by Aditya Khurmi(This one has nice explainations to the topics of Olympiad number theory and contains past problems from different countries). Hope you will enjoy
What about the case if p-1 divides y-x or y+x
Based on this we know that y = 5
This was fun to watch
you know, should've guessed the solution with that title...
Sir can you please share the link to the video where you had solved a question wherein we had to find all n belongs to N such that n^[1/(n-2)] is an integer
If anyone else can, please do share
Hi Michael, I love your videos in general. But, there seems to be too much editing in your videos lately. I don't know the reason for it, but the video appears very choppy. Also, I imagine that it makes editing very cumbersome. I noticed edits in the video every 5-10seconds! Is there a reason for these edits? Your early videos didn't have them.
P is odd imply that p^2 is odd
And y=5
Nice video again :D
Nah lame place to stop I wanna know X and Y
(Fine I guess x=2, y=5, but eh wanted to watch it)
If the integers must be positive or zero, why didn't you just write that they are natural? x,y belong to N
Hi. Im starting at college this year, and I am not quite sure what to study yet. The chose is between sivil engineer in applied mathematics where they fokus on differential equations and stats, or just to study pute mathematics. What do you guys recommend?
what is "pute mathematics"?
Those 2 things are wildly different. My only advice is if you only like mathematics don't go into engineering thinking it pays good so you'll get through.
Wizardry.
P is odd
I have big problems
How I can send you a copy to solve it
I didn't understand the joke mp.
mp=Michael Penn
Idol in math
4:20 because x+y = p, we get that x has to be 2 because difference between the 2 larger primes is even.
30 seconds were enough to figure out the answer, empirically 😁
How did you "empirically" prove there are no other answers
@@AntoshaPushkin I realized that the only answer is 7 after the video 😅
What lesson do all students want in a math video?
You are the best
Interessante que essa questão tem no Poti kk. Eu tinha feito ela essa semana.
The equation p=2X^2-1 seems to be a good generator of primes. Only x=5 failed for the first several natural numbers. Is this a thing?
Long time no 👍🏼, Sir.
But this one was fun 🤗
p=7 just hit and trial
But how do you know that it isn't also true for, say, p = 101. There's an infinite number of primes, and the question asked for ALL primes that meet that condition.
Nice
1 is a prime number as well because 1 is devided by itself. Therefore p=1 is another solution
primes are defined to be positive intergers >1
THERE IS HAVE A NEGATIVE 1 ALSO A PRIME.
Indeed, -1 acts like a prime, as it has just 2 factors: itself and 1. -1 is a solution to the 2 given equations, x=0 and y=1.
@@davidwalker3804 GOOD JOB DAVID WALKER!!! 😁😁😁😁😁😁
Primes are specifically defined as natural numbers greater than 1 that have only 1 and itself as its factors. Negative numbers would not count as they are not natural numbers.
@@xevira True. Besides all other negative numbers have at least 4 whole number factors. So the “natural number” part of the definition of “prime” eliminates this single negative integer.
@@davidwalker3804 TRUE, THAT ''NEGATIVE 1'' IS A PRIME/NATURAL NUMBER.
I thought of 7 at first look 😁😁 #msd_fan
What confuses me is that besides p=-1 and p=7, there's also solution p=1 which did not pop out as solution to your equations. Of course 1 does not count as prime but it feels like you missed something.
Also, the role of p in the factorization 2(x-y)(x+y) depends on the fact that p is prime, so it excludes any non-prime solutions.
@@TJStellmach I don't see that as valid reason, -1 is not a prime either and with regards to divisibility of those factors it's not any better than 1.
@@TJStellmach Also, it seems to me 1 and 7 are the only solutions even if we drop the requirement on p being prime. At least there's no other solution for p
Muito bom