This problem writer is clever.

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Опубликовано:

19 ноя 2021

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Комментарии : 69

@goodplacetostop2973 2 года назад

0:03 The problem writer is Pierre Haas from Luxembourg 9:18 Homework 17:39 Good Place To Stop

@sharpnova2 2 года назад

assume twin prime conjecture is false, then this wouldn't be a good place to stop but it was a good place to stop, therefore via contradiction, TPC = true QED

@keinKlarname 2 года назад

So RU-vid is a good place to solve great unsolved problems in Mathematics!

@mcwulf25 2 года назад

Haha but only if we were asked if there was an infinite number of solutions. Which is a good place for me to stop.

@mcwulf25 2 года назад

What I like about this problem is that the question hints that it's not possible to find all p and q. What I like about the solution is that it's pure algebra. No messy inequalities or guess-and-check.

@emuhast 2 года назад

It would be great if you made some videos explaining the thought process of creating some of these problems

@GreenMeansGOF 2 года назад

These problems are from Math Olympiads. Michael doesn’t create them. A lot of the videos are viewer suggested problems.

@emuhast 2 года назад

@@GreenMeansGOF I know, but someone had to come up with these. Michael probably has some tips for creating such problems

@mrpenguin815 2 года назад

And probably comes up with his own questions for his students, which will often have nice answers at the end.

@random03245 2 года назад

Equation can be written as (q-(p+1))×(n+2) = 4q(p+1), where p and q are odd primes. From this equation q-p=2 and n=4(p+2)(p+1)-2.

@RexxSchneider 2 года назад

For those who want to check their working on the HW. For the first part of the HW, we need to show that gcd(p+1, q-p-1) is 1: gcd(p+1, q-p-1) = gcd(p+1, q-p-1 + p+1) because adding p+1 to the second number will leave the gcd unchanged. gcd(p+1, q-p-1) = gcd(p+1, q), but that must be either 1 or q since q is prime. Suppose gcd(p+1, q) = q, then q | (p+1), but that is impossible, given that we already have q > p+1. So the gcd(p+1, q-p-1) = gcd(p+1, q) = 1. For the second part of the HW, it easier to work with the relationship 1/(p+1) - 1/q = 4/(n+2) - which was derived by 3:45 - and then solve for n: (q-p-1) / q(p+1) = 4/n+2 (n+2)/4 = q(p+1)/(q-p-1) n = 4q(p+1)/(q-p-1) - 2 Check p = 2, q = 5: n = 4*5*3/(5-2-1) - 2 = 60/2 - 2 = 28. Check p = 2, q = 7: n = 4*7*3/(7-2-1) - 2 = 4*21/4 - 2 = 19. Check q = p+2: n = 4(p+2)(p+1)/(p+2 - p - 1) - 2 = 4(p² + 3p + 2)/1 - 2 = 4p² +12p + 6.

@makotoniijima862 2 года назад

When p = 2 & q = 7, n = 19

@diniaadil6154 2 года назад

someone does their homework !!

@potawatomi100 2 года назад

You’re amazing my friend.

@lucastellez2558 2 года назад

Loved this video

@assassin01620 2 года назад

When I used Desmos and put in the equation, but specified that q=p+2, and n=4p^2+12p+6, it seemed to work for all real numbers. Interesting.

@copperspike 2 года назад

I'd assume that's because the restrictions is p & q being prime and n being a natural number isn't given. This would make it just a polynomial function n(p) = p^2 + 14p + 6, a fine quadratic, but not what we're looking for

@gamingbutnotreally6077 2 года назад

As it should... nowhere are you restricting p and q to being primes, otherwise I mean you just have a quadratic...

@assassin01620 2 года назад

@@copperspike I forgot that n had to be natural. Lmao

@assassin01620 2 года назад

Actually there's a lot I didn't consider xD

@Kriophoros 2 года назад

Is that a typo, cuz the solution is n=4p^2+12p+6. However, as long as p and q are two consecutive odd numbers, this initial equality is always satisfied, even if any or both are composite.

@stmmniko7836 2 года назад

Nice problem!

@JanxakaJX 2 года назад

A clever maths problem writer? Outrageous

@renerpho 2 года назад

A really clever problem writer would have proven the Twin Prime Conjecture first, then boiled down the proof into something that students could find in a competition, and design a problem around this that would be equivalent to the TPC. Any good student, upon realizing they had proven the TPC, would immediately doubt their work. You'd want to make sure to invite some psychologists to observe the students' reactions.

@brabhamfreaman166 4 месяца назад

8:04 Why on Earth would you launch into an unnecessarily messy, technical proof by contradiction when the statement follows directly from the fact that $q otequal p+1$ as noted just moments earlier (a trivial consequence of $q>p+1$ ?

@kevinmartin7760 2 года назад

7:00-9:00 Wouldn't it be slightly easier to immediately observe that gcd(q, q-p-1) must be either 1 or q because q is prime, then note that clearly q > q-p-1 > 0 so q cannot divide q-p-1 so the GCD must be 1. Note that we still must use the observation that q>p+1; if you don't do that and p>>q q could potentially be a divisor of (a very negative) q-p-1 (example: q=3, p=11, q-p-1=-12 which has q i.e. 3 as a factor).

@Kriophoros 2 года назад

This is a weird problem. It looks complicated at first, but once you find the solution you'll discover it to be quite redundant. I actually solved it without using q>p+1 or p is prime (note his solution didn't use that either)

@Kriophoros 2 года назад

Also n can always be found as long as p and q are two consecutive ofd numbers, no matter if they are prime or not. So why do we need p and q to be prime again??

@jkid1134 2 года назад

HW: 2/3+8/7=14/21+24/21=38/21=2*19/(19+2)

@trdi 2 года назад

Wouldn't the first 5:20 be easier to do just by saying q=p and seeing that right side is always smaller than 2, while left side is 2*(1+positive number), so always >2?

@arnavdas3139 2 года назад

HW : add p+1 ✌️😁 noob instincts kick in

@26-dimesional_Cube 2 года назад

Can I have my own problem. Just 8th grade geometry. No problem Given any quadrilateral ABCD, let E, F, G, H are the midpoint of AB, BC, CD and AD respectively. Proof that EFGH is a parallelogram (Hint: There are 5 ways to proof a shape is a parallelogram: - A quadrilateral with opposite sides parallel to each other is a parallelogram (note the "s" at the word "sides". It means that if EFGH wants to be a parallelogram then it must have EF//GH AND EH//FG) - A quadrilateral with opposite sides equal to each other is a parallelogram (Same thing as the above. It means that if EFGH wants to be a parallelogram then it must have EF=GH AND EH=FG) - A quadrilateral with only one pair opposite sides and equal to each other is a parallelogram (It means that if EFGH wants to be a parallelogram then it must have either two conditions: + EF=GH AND EF//GH OR + EH=FG AND EH//FG) - A quadrilateral with all pair of opposite angles equal to each other is a parallelogram (It means: + angle E = angle G AND angle F = angle H) - A quadrilateral that has 2 diagonal line intersect at midpoint of each line is a parallelogram (Called the intersection point as O then we have EO = OG = EG/2 AND FO = OH = FH/2) Good luck!

@9WEAVER9 2 года назад

GET A 'ROOM'

@goodplacetostop2973 2 года назад

I like how you give the problem AND a thousand of hints to solve it. It’s not a problem at this point, it’s a tutorial 😂

@resilientcerebrum 2 года назад

It's a really trivial question

@kylecow1930 Год назад

homework, prove that there are infinitely many prime solutions

@xyz.ijk. 2 года назад

It would have made a lot more sense to simply solve the Twin Prime Conjecture, which would also have proven the last case. I don't know ... I'm not used to you taking shortcuts. Otherwise I thought the video was pretty good.

@BobbyKenny 2 года назад

Weird Qs

@TJStellmach 2 года назад

Finding the gcd of q with q-p-1, you've got a prime (q) and a number that's less than that prime, so their gcd is obviously 1. 1 and q are the only possible divisors of q, but q is too large to divide q-p-1. The video basically says this with needless extra steps.

@9WEAVER9 2 года назад

so do you though, technically with even more steps than the video because you in essence require all the video steps and then you use all of your steps, which have an essentially 0 probability of being performed in this precise way in absence of his video, UNLESS YOU COLLABORATED...

@cobalius 2 года назад

I'm just thinking of all these cool mathy girls

@xulq 2 года назад

down bad

@CrossMax122 2 года назад

European GIRLS math olympiad? sad...

@captainsnake8515 2 года назад

What

@nicepajuju3900 2 года назад

Why

@klausolekristiansen2960 2 года назад

@@nicepajuju3900 Because it sends a signal that girls are incable of competing with boys.

@nicepajuju3900 2 года назад

@@klausolekristiansen2960 no it doesn’t

@NoNameAtAll2 2 года назад

@@nicepajuju3900 are there boys-only olympiads or is it sexist?

@villagegreen7765 2 года назад

Good video, and Twin Primes has been proven btw.

@comexk 2 года назад

No, it has not.

@captainsnake8515 2 года назад

This is incorrect.

@cletushumphrey9163 2 года назад

source?

@stewartzayat7526 2 года назад

Source: trust me bro

@goodplacetostop2973 2 года назад

As of today, we only know there are infinitely many pairs of primes (p, p+246). If you assume Elliott-Halberstam conjecture is true, it falls to (p, p+12) pairs.