Using simpler cases to find a solution to an IMO problem

[@MichaelPennMath]

Hey KZbin: I started a podcast (kzbin.info/www/bejne/ap-Up3prdp2BfaM) with my niece Grace Pavloff and nephew Aleksei Pavloff! Alek is a small town sports journalist in Ohio while Grace is a student at Queen's University Belfast. Consider this comment the "soft launch" as we find our footing a bit more I will be publicizing it a bit more.

[@manucitomx]

You are working hard to make me like Number Theory. Thank you, professor.

[@Notthatkindofdr]

I worked on the four variable version and I believe the only solutions are (2,4,10,80) and (3,5,17,255). In each case notice that the largest number is the product of the smaller numbers (and this also happens in the 3-variable case), but I don't see a general reason why this has to be true.

[@d.l.7416]

theres an infinite family of solutions (k-1 , k+1, k^2+1, … , k^(2^(n-1))+1, k^(2^n)-1) for k = 3 or 4 the product of all but the last number is the same as the last number a=k^(2^n)-1 (a is the largest number) so the product of all of them is a^2 so the numerator is a^2-1 = (a-1)(a+1) so it is an integer if (b-1)(c-1)... divides a+1 = k^(2^n) (b-1)(c-1)... is (k-2)*k*k^2*k^4*…k^(2^(n-1)) = (k-2)*k^((2^n)-1) This divides k^(2^n) when (k-2) divides k, so when k = 3 or 4 Every solution for 3 and 4 variables happens to be one of these.

[@Notthatkindofdr]

@@d.l.7416 Excellent work! I think you might have the number of solutions and variables mixed up though. The number of variables in your solution is n+2, so it is 3 variables for n=1 and 4 variables for n=2. In each case you showed that there are only two possibilities (k=3, 4) that fit your formula. Your solution gives the two possible answers for the problem in the video (including the one Michael found) and the two possible answers for 4 variables that I found. Technically it also gives the only "solutions" for 2 variables (n=0) as well. And you have shown that for any greater number of variables there are two solutions. This raises the question of whether we can show that your formula gives the only solutions for any number of variables, as it does for 2, 3 or 4 variables.

[@goduck-x6u]

12:25 Once we get to the point of things like (2bc-1)/(b-1)(c-1) = 3, it is easier to rearrange and complete the product to (b-3)(c-3)=5 and get the answer b=4/c=8. Same goes for 3:52 for the 2 variable case. And for 16:38, once you get the bound of between (3,5), solve (3bc-1)/(b-1)(c-1) =4 is easy.

[@emanuellandeholm5657]

Using the substitutions A = a - 1, B = b - 1, C = c - 1, we get the analogous problem 0 < A < B < C ((A + 1)(B + 1)(C + 1) - 1 ) / ABC in N This gives ABC | (ABC + AB + BC + AC) => ABC | (AB + BC + AC) => A | BC By symmetry we also have that B | AC and C | AB. This gives us two nice inequalities since X | Y iff X < Y (We already knew A < BC)

[@emanuellandeholm5657]

@@petrie911 Yes! :( I totally missed that term

[@plislegalineu3005]

it's not iff because 2 < 3 but 2 !| 3

[@stevewolfe6096]

At 15 min + with the 3 subcases. The numerator is always odd and the the denominator parity is the opposite of “b” so “b” = 4 can be the only possible solution to check - similar to earlier arguments

[@yoav613]

When you watch and notice there is 1 min left and michael did not even start the next case.. the bell rings homework!!

[@jimmykitty]

*Wow! I was tryna solve this mathematical problem a few days ago! Thanking you for giving us the solution* ❤

[@robertveith6383]

"Tryna" is not a word. Use "trying to."

[@jimmykitty]

@@robertveith6383 Oh! Ok 😥

[@jadegrace1312]

@@robertveith6383 Tryna is a word. It's literally in several dictionaries.

[@Chill----]

@@jadegrace1312 lol

[@jimmykitty]

@@Chill---- 😊🌿

[@dneary]

You can get to the solutions a little faster with 2bc-1 = k(b-1)(c-1) and 3bc-1 = 2k(b-1)(c-1) which you can rearrange, respectively, to (k-2)^2bc-k(k-2)b-k(k-2)c+(k-2)(k+1) = 0 and (2k-3)^2bc-2k(2k-3)b-2k(2k-3)c+(2k-3)(2k+1) = 0 - and then use the factoring "trick" of completing the product: ((k-2)b-k)((k-2)c-k) = k+2 and ((2k-3)b - 2k)((2k-3)c - 2k) = 4k+3 to facilitate eliminating options after you limit the range of a. In the first case, for example, k=3 gives b-3=1, c-3=5 for the (2,4,8) solution, and in the second case, k=2 gives (b-4)(c-4) = 11, giving the solution (3,5,15). I believe these are the only 2 solutions.

[@goodplacetostop2973]

17:03 Bonus exercise 17:17 Good Place To Stop

[@threstytorres4306]

For Bonus Exercise, If 1≤A≤B≤C≤D the answers are: A = B = C = D = 2 and A = B = C = D = 3

[@alangivre2474]

@@threstytorres4306 thankj youu

[@jeremylengele]

@@threstytorres4306 (2, 4, 10, 80) also works

[@threstytorres4306]

@@jeremylengele If 4 is a Sol, The Output is approx. 3.14 If 10 is a Sol, The Output is approx. 1.52, and If 80 is a Sol, The Output is approx. 1.05

[@physicorum7107]

This was relatively easy afa IMO problems go in my opinion. Also would love to see a problem solving video for 1988 problem 6 , too great of a question to leave from the series

[@HAL-oj4jb]

That question really reminded me of problem 6 too! It's on my bucket list to solve it some day, and I was really afraid that this problem's solution would be very similar to it until I read your comment ^^

[@karman7203]

(a, b, c)=(3, 5, 15) is the only other solution.

[@physicorum7107]

Did you bash to get the solution ? Or a programming script?

[@bsmith6276]

What about (2,4,8)?

[@alexey_burkov]

@@bsmith6276 what about thinking before writing? The one you commented gained in the video.

[@mithutamang3888]

So, the equality holds for the 2 variable version is a and b is a=b when the problem, ab-1/(a-1)(b-1) is a natural number where a and b are also natural number. The only two solutions are a=b=2 and a=b=3, the equation is satisfied! 😁👍

[@mithutamang3888]

The condition is a=b for a and b are the equality holds!

[@yuanwang8136]

14'10" since numerator is odd, then b-1 must be odd so b must be even, therefore be can only be 4, you don't need to check b=3 and b=5

[@DeletedUser410]

14’ does not mean 14 minutes in time. That only works for longitude and latitude (and is stupid anyways). Same for “ as seconds. Just use a colon between minutes and seconds

[@billbill1235]

5:50 it is not strictly less than 2???!!!

[@rimidalal7020]

video on cauchy-schwartz inequality please

[@mithutamang3888]

Assumed that, a=b but a and b are bigger than or equal to 4 that this is object there is no natural numbers!

[@mattl-dlgx3994]

I don’t really understand, how is a less than 2 (and b less than 3) in the version with two variables ? I thought it was 2

[@thejelambar82]

Time stamp, please?

[@mattl-dlgx3994]

@@thejelambar82 around 3:06

[@hirshx7188]

We have that a/(a-1)=1+1/(a-1) which is monotonically decreasing in a, so if a gets bigger, a/(a-1) gets smaller, thus the biggest value is attained at a=2.

[@GreenMeansGOF]

@@hirshx7188 agreed. This bothered me too but a quick look at the graph shows everything is fine.

[@mattl-dlgx3994]

@@hirshx7188 thanks I’m convinced now 😁

[@ImaginaryMdA]

I can prove the case for 4 variables is solvable in finite time, using this trick, but I can't be bothered. XD

[@chaosredefined3834]

Is there not enough space in this youtube comment?

[@ximoto123]

Michael, please show that 1/1-a^2 + 1/1-b^2 +1/1-c^2 > 189/62 ,where a+b+c=1

[@alibaranseloral]

Only two solution. (2,4,8) and (3,5,15)

[@conorquinn607]

(3, 5, 15)

[@Generalist18]

How did you solve case bash?

[@JM-us3fr]

Doesn’t this problem have to do with Carmichael numbers? Carmichael numbers n have the property that p-1 divides n-1 for all prime factors p of n.

[@Notthatkindofdr]

It looks a bit similar, but it seems like the solutions don't seem to be related to Carmichael numbers.

[@gusmichel7035]

Checking the factors of the first few Carmichael numbers, the problem is that while each term in the denominator divides the numerator, you don't have enough 2s in the numerator to cover all those in the denominator.

[@ayoubabid714]

I will just study the case when the inégality not hold

[@yannobzh]

Should be more careful at what is done at 3:13 :/ You get the good result but by a completely false way. Anyway good video :p (One should consider a/(a-1) = ((a-1)+1)/(a-1) = 1 + 1/(a-1) and there u can bound things)

[@timpani112]

What Mike is saying is not wrong. The only thing you can really argue here is that his argument is not fully rigorous, but this is not at all the same as saying that he's getting to his results in a "false" way. Generally, it is impossible to answer the question of how much rigour is required for a solution to be "correct", but I think what Mike did is enough. This is because I think the fact that the function (x+1)/x is a decreasing function for x>0 is well known enough that it doesn't have to be proven every time it's being used.

[@yannobzh]

Well I completely agree with you. I still think that he should mention what he is using (after all he mentionned the first case a-1 / a-1 = 1 ). Imagine kids watching this thinking that in order to majorise a fraction u majorise both numerator and denominator ? Maths teachers would feel bad (im french btw, sorry for my english). Thank you anyway 3)

[@robertveith6383]

@@yannobzh That is (a - 1)/(a - 1) = 1, a > 1. Make sure you use grouping symbols.

[@timpani112]

@@yannobzh Yeah, it's hard to decide where to draw the line between rigour and brevity/clarity/simplicity in one's presentation. Ultimately it will come down to the intended target group I guess. I find the explanation adequate, but I agree that there are many people (especially younger viewers) who might get some strange ideas if these kinds of things are glossed over.

[@giulioverzeletti513]

12:20 Who is it always odd?

[@physicorum7107]

Because 2bc is a multiple of 2 and a multiple of 2 - one would be odd

[@ralphmb980]

2*(integer) is always even, 2*(integer) -1 is always odd

[@giulioverzeletti513]

@@ralphmb980 oh right

[@stupidtalks8011]

911st like which y'all know what's up with it

[@cadaver123]

😯

[@jesusthroughmary]

If a,b,c are all greater than 1, why bother saying they are in Z when they are all in N?

[@m.walther6434]

But the 'smales Version' a-1/a-1 is no solution at all. To be a solution b*c=1 and (b-1)*(c-1)=1 must hold, but has only complex solutions.

[@luisaleman9512]

There is no b and/or c in the smallest version, only a.