For the ecuation y! = 2^y - 2 , you can notice that y! is divisible by 4 , for any y >= 4 , but 2^y - 2 is not, so the ecuation hasn't got any solutions greater then 4. Anyway, amazing result !
If yy then x^ k+1 | x!, hence x^k | x!+y!. As k is the biggest number we have x!+y! = x^k, but x! > x^ k+1. Contradiction. No solution for x>y. For y>=x, we have that x-1 | (x!+y!) , --> x-1| x^y, also x-1 | x^y - 1, hence x-1 | 1 --> x=2. Therefore, y! = 2^y - 2, for y>4 there is no solution. Hence (2,2), (2,3) are the only solutions.
@@dailytopvideos1190 First x-1 | x!+y! and we know that x!+y!=x^y. Also x!=x * x-1 * x-2*...*1 and y!=y * y-1 * y-2 *...*1, as y>=x the factor x-1 will also be in y!. Example: y=8! and x=6!. So x-1 = 5! is in 8! and 6!. so x-1 divides both --> x-1 | x!+y!. We also know that x-1 is a factor x^y - 1 , as x^y and x^y - 1 are relatively prime then x-1 | 1 --> x=2.
In your first step, I am not sure how you get to the conclusion that x | (1+x!/y!). x can still divide y! even if x > y. For example, 6 > 5 but 6 divides 5!.
Hello there, Could you recommend any books to advance in competition maths. I'm getting stuck often and frustrating and it's clear I Dont know enough. If you could recommend some books for certain subjects, or beginner intermediate or advanced books, that would be great
@@kendricklenox5672 I've found some really good books since. Try the world scientific IMO books, try lecture notes part 1 and 2. They have a big bank of questions with many example questions and solutions
Can you suggest some books for being eligible to understand this level of questions , I am preparing for IMO but never tried these difficulties. Help me ......
wow very nice argument. I have heard of Bertrand's postulate before but never seen it applied. At first sight it would never have occurred to me that it could be used to solve this problem.
probably second time i see you using this postulate, and yet again i didn't manage to see this could be useful in this problem :( Disappointed i failed to do it myself, but your solution is obviously very nice
@@HB-cg4jv It is very easy to check, which I did but wasn’t going to bother to include. Take a range of values, say from 2 to 10. Now 10! Is on the order of 3 million. Use 2* 3^6 as a lower bound. The upper bound for the range is 10^10. The rhs grows must faster than the lhs bc it is exponential. So work the values, in a table if need be, until the rhs becomes ridiculously large. You will see that the answer is unique.
For the ecuation y! = 2^y - 2 , you can notice that y! is divisible by 4 , for any y >= 4 , but 2^y - 2 is not, so the ecuation hasn't got any solutions greater then 4. Anyway, amazing result !
Hello Eduard! consider look to my channel too for similar math olympiad problems. Thanks and regards.
Good
If yy then x^ k+1 | x!, hence x^k | x!+y!. As k is the biggest number we have x!+y! = x^k, but x! > x^ k+1. Contradiction. No solution for x>y.
For y>=x, we have that x-1 | (x!+y!) , --> x-1| x^y, also x-1 | x^y - 1, hence x-1 | 1 --> x=2.
Therefore, y! = 2^y - 2, for y>4 there is no solution. Hence (2,2), (2,3) are the only solutions.
I thought of factoring out but what made you think of the rest?
Can u explain second step, how x-1|x^y ?
@@dailytopvideos1190 First x-1 | x!+y! and we know that x!+y!=x^y. Also x!=x * x-1 * x-2*...*1 and y!=y * y-1 * y-2 *...*1, as y>=x the factor x-1 will also be in y!. Example: y=8! and x=6!. So x-1 = 5! is in 8! and 6!. so x-1 divides both --> x-1 | x!+y!. We also know that x-1 is a factor x^y - 1 , as x^y and x^y - 1 are relatively prime then x-1 | 1 --> x=2.
@@danilonascimentorj thanks
In your first step, I am not sure how you get to the conclusion that x | (1+x!/y!). x can still divide y! even if x > y. For example, 6 > 5 but 6 divides 5!.
Hello there,
Could you recommend any books to advance in competition maths. I'm getting stuck often and frustrating and it's clear I Dont know enough. If you could recommend some books for certain subjects, or beginner intermediate or advanced books, that would be great
I also need the assistance
@@kendricklenox5672 I've found some really good books since. Try the world scientific IMO books, try lecture notes part 1 and 2. They have a big bank of questions with many example questions and solutions
Just have a look through, they have so many questions from actual olympiad too.
@@kendricklenox5672 after that maybe try some more books from antitu andreescu for some higher level problems which hone in on the imo style questions
@@spazmoidectomorf6209 thanks man.
I highly appreciate this, want to attend next years IMO but my knowledge is lucking.
Can you suggest some books for being eligible to understand this level of questions ,
I am preparing for IMO but never tried these difficulties.
Help me ......
63rd IMO?
wow very nice argument. I have heard of Bertrand's postulate before but never seen it applied. At first sight it would never have occurred to me that it could be used to solve this problem.
Hello Richard! consider look to my channel too for similar math olympiad problems. Thanks and regards.
I'm nowhere near a genius, but how is using the postulate the intuitive way of solving this problem? Does it really come instantly to mind?
it's not the only method out there, he used it to assert the range where x exists, you can use other methods to do so :)
Hello there John, consider look to my channel too for similar math olympiad problems. Thanks and regards.
The problem asks only to find x, y satisfying the equation given. I did it by inspection. Was a uniqueness proof required in the competition?
Hi quark! consider look to my channel too for similar math olympiad problems. Thanks and regards.
This is so easy just notice that for x,y>2 x has to be even and analyse for 2 as factor
probably second time i see you using this postulate, and yet again i didn't manage to see this could be useful in this problem :( Disappointed i failed to do it myself, but your solution is obviously very nice
Pretty sure I would not have been able to do this. Although I may try and find another method!!! 👍
Turn on postifications
*Put x=2,y=2*
*Which is satisfying the equation.*
*So, x^y = 2²=4* 😎✌️
Well, x=2, y=3 ... 2!+3! = 2+6=8 = 2^3 is also a solution !! Now we have to prove these two solutions are unique ...
that's not how math works lol, you gotta find all possible solutions and prove they are unique.
@@HB-cg4jv 😜✌️
WHY WHY USE SUCH AN OBSCURE POSTULATE that NONONE WILL EVER KNOWNOR THINK OF..wouldn't Betrand himself even forget it? Seriously lol
it's not obscure?
Ho?
Can't you just say x = 2 y = 3 and be done with it
Prove it, because I check that x = 42069 and y = 314159 works.
asnwer=1
Solution by insight
2+6=8
2!+3!=2^3
x=2 y=3
"solution by insight"
"Solution by insight"
Ten seconds (2,3)
well done but that's a long way from a full solution
@@richardfredlund3802 The margin is too small.
@@padraiggluck2980 how do u know it's unique? the question is to solve for x,y which means give all possible solutions.
@@HB-cg4jv It is very easy to check, which I did but wasn’t going to bother to include. Take a range of values, say from 2 to 10. Now 10! Is on the order of 3 million. Use 2* 3^6 as a lower bound. The upper bound for the range is 10^10. The rhs grows must faster than the lhs bc it is exponential. So work the values, in a table if need be, until the rhs becomes ridiculously large. You will see that the answer is unique.
It annoying to see a video which is mostly see him writing
wtf, do u want him to dance?
its a math video
Why are you so temperamental today? Hormones?