Indian Mathematical Olympiad | 1992 Question 8

We present a solution to question 8 form the 1992 Indian Mathematical Olympiad. The goal of this question is to classify all perfect squares of the form 2^m+3^n.

[setupchess]

This man used the entire alphabet to solve this problem. Great work

[usptact]

I almost ran out of my stack memory trying to follow this.

[manthing]

Each time, i see the problem of your video, i sit back and think to mayself: I wonder, how is he gonna tackle this one. And at the end of the every video, i see that i have much to learn. Well done.☺

[jogeshgupta]

Thanks Michael penn for making a video on a question from Indian mathematical Olympiad...NAMASTE

[BharathReddy]

You just cracked the code to get more views.

[tracyh]

17:21 Another way to see this is that 3^i and 3^j are two powers of 3 with a difference of only 2. This only occurs when i=1 and j=0.

[jerry]

Maybe if I watch enough of these I’ll actually be able to answer one in a contest

[samb]

A slightly quicker way to show that m is even is to note that 2 = -1 (mod 3), so 2^m = (-1)^m = 1 (mod 3) => m is even

[ProfOmarMath]

I received a request for a problem from this very exam! I might put a card to your video as a shout out.

[garari]

Great problem as always! There's a little simplification: at 12:46 we can notice c+d=2a. With this fact we can immediately conclude from d=1 that c is odd and so c-1 is even (without referring modulo 3), and also when we find at 18:52 that c=3 we can immediately say that a = (c+d)/2 = 4/2 = 2. Since at that moment we also determine that b=1, the rest of the solution becomes trivial.

[chhabisarkar]

Finally ! A question from my country's olympiad !!

[ഡിങ്കനേഷ്]

You took 22 minutes, when probably you already knew the answer. How much time these Olympiad guys would have took when they don't know the answers already. I find this one

[onlinecircuitsimulation]

First step is to prove both m, n are even. Which makes that is a Pythagorean triplet. So let m=2k, n=2p. also we have 2st=2^k, where s is even and t is odd, but the last equation forces t=1 and s=2^y. Also 3^p=2^{2y}-1. Note that the RHS is factorizable, so just have to consider few cases.

[matthewstevens]

Nice solution. This was also a question in the secomd round of the British Mathematical Olympiad in 1996

[tranbachnguyen]

Hi guys,
First of all, thanks Micheal for this video which makes me kinda fall in love again with Math.
Then I have a suggestion to simplify the solution (sorry if this was mentioned before) :
Let's go from 10:54.
We have 3^(2b) = x^2 - 2^(2a) = (x-2^a)(x+2^a). Using the same strategy in the video we get x+2^a = 3^p (1) and x-2^a = 3^q (2), with p>q and p+q = 2b. Substract (1) by (2) gives us: 3^p - 3^q = 2^(a+1). Here we easily see that q = 0 (otherwise 2^(a+1) divisible by 3), so we go directly to the point that 3^(2b) = 2^(a+1) +1.
Similarly as above we got something like 2^u - 2^v =2
Or 2^(u-1) - 2^(v-1) =1 with u+v = a+1.
It goes easy from here!!!
We got v-1=0 so v=1 and u=2 then a= 2 and b=1.

[joaosousapinto]

Super fun problem, and great presentation! Once you show that the perfect square needs to be of the form 4^a + 9^b, there's another approach that's perhaps less elementary but simpler. Namely, (2^a, 3^b, sqrt(4^a + 3^b)) is a primitive (i.e. all co-prime) Pythagorean triplet, so it must hold that 2^a = 2mn and 3^b = m^2 - n^2 for some m > n (thanks Euclid!). This implies both m and n are powers of 2, and that n = 1 (since 3^b is odd); from that, m=2^(a - 1) follows. So we must have 3^b = 2^(2*(a - 1)) - 1 = (2^(a - 1) + 1)(2^(a - 1) - 1), so both factors are powers of 3 and differ by 2, so they must be 3 and 1 respectively, implying a = 2. Finally, 3^b = 3 implies b = 1.

[Thiago-kbbu]

Try this one : Brazilian Mathematical Olympiad (OBM) 2019 Problem 3 Level 3

[Daniel-nlug]

When you did the step at 3:37, this is only true if n > 0. There is one solution when n = 0, which is (m, n) = (3, 0). Later there was a similar thing reducing mod 4 which relies on m > 0, and when m = 0 there is the solution (m, n) = (0, 1). "Natural numbers" is a slightly ambiguous term. Some people consider 0 to be natural and others don't. If the olympiad paper said natural numbers on it then I would suggest including these solutions as well and mentioning that they are only valid if 0 is considered a natural number. However, in the olympiads I've done, I can't think of a time when the phrase natural numbers was use. Instead they normally write "positive integers" when they don't want 0 included, and write "non negative integers" when 0 is included, and this way it is not ambiguous.

[mridulsachdeva]

I actually feel that this would have a better/elegant solution than just doing the same thing again and again

[amircanto]

I enjoy the way you can transform the ideas into math in very clever ways. Enjoy your content