Prime Powers | India National Mathematical Olympiad 2008 Problem 2

[@caoliva]

I got stuck with my solution using fermat's little theorem but got that p=5 will work but could'nt prove why it will be the only answer. Your solution was very much elegant. I found another gem on yt, thanks yt algorithm :>

[@letsthinkcritically]

Thank you!!

[@СВЭП-и4ф]

y^4 < p^x. for y > 4 p^x mod y = 4, p^x mod y^2 = 4. Not possible. => y

[@joaquinfernandez9232]

Sophie Germain likes this.

[@debayuchakraborti1963]

I can tell without watching the vdo how amazing its gonna be!! Thank You for doing this problem

[@siddharthabhattacharya3787]

Aap panditji faile hue ho

[@letsthinkcritically]

Thank you for your support!!

[@parthsingh3569]

Well congratulations for qualifying IOQM all the best for INMO. Good job debayu

[@debayuchakraborti1963]

@@parthsingh3569 wowo thenks

[@parthsingh3569]

@@debayuchakraborti1963 may I know your score in ioqm?

[@brinzanalexandru2150]

From the factorization we can write:y²-2y+2=p^a and y²+2y+2=p^b where obviously b>a and they are integers,if we divide both equations we get p^(b-a)=(y²+2y+2)/(y²-2y+2) now the fraction has to be an integer which is an easy exercise,we just do long division amd then get the inequality 4y≥y²-2y+2 which is satisfied only by y€[1,5] so after checking we conclude that only y=1 works but a trivial solution was y=0 which in my country is not considered as a natural number

[@sheldoncooper5494]

Wow u did this national level mathematics olympiad question with no hesistation and difficulty. I appriciate it and great work tho!! Could you pls tell me how to develope intuitive thinking to solve such problems.??

[@debayuchakraborti1963]

he is pro

[@sheldoncooper5494]

@@debayuchakraborti1963 ofc!!

[@h4z4rd28]

My tip is: start with basics and do A LOT of problems, then you can be also pro

[@letsthinkcritically]

Try to remember the ‘moral’ of the problem, but not the steps. What I mean is how did you got through the bit you were stuck while solving the problems. It’s usually some non-trivial or uncommon tricks. Those tricks might be useful when you see problems of similar kind.

[@pratikmaity4315]

The name of your channel suits your videos

[@shalvagang951]

I have found out one thing that when we factored out y^2+2y+2 and y^2+2-2y then their gcd is 1 relatively prime numbers so d is odd which divides them so d|4y d|y and so d|p^x but as d=1 so p^x, has a gcd of 1 so it means that 4y=p^b(p^a-b-1) then y cannot divide p^b and 4 cannot divide p^b as its odd it remains to one factor of 4y divide p^a-b-1

[@satyamsaurav7005]

Thanks for solving this problem.

[@brettaspivey]

x^4 + 4 = 0 mod 5 which simplifies the proof a little

[@jamesjames1549]

Only if x and 5 are co-prime

[@mariomestre7490]

Genial!!

[@tonyhaddad1394]

3:02 how ???

[@karthikkrishnaswami3164]

p|a and p|b => p|(a-b)

[@perseus_0741]

😍😍

[@Gaming_639_]

5:05 what is V2 ?

[@ittaloceara]

Nice

[@letsthinkcritically]

Thank you!!

[@Gaming_639_]

​@@letsthinkcritically 5:05 what is V2(y^4+4) ?

[@theandroidguy6032]

I just Randomly sleceted x=y=1 and prime number as 5 and got answer but this didn't work every time 😅

[@sumitsingh-bc1iw]

Can you please explain how we got that p can only be 2.? I understand that we had p^x=16(t) +4 but how do we ensure that p =2 is the only solution

[@srivatsansriram8328]

No no,it’s the greatest power of 2 is that 2 which means 4

[@moonlightcocktail]

P | 4 but the only prime that does is 2