Indian Mathematical Olympiad | 1992 Question 8
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- čas přidán 11. 09. 2024
- 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.
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This man used the entire alphabet to solve this problem. Great work
Indian though, it is like half African. Hardest math question in African, what would that be?
@@krishna-vb4ok India is the place where people literally worship cow.
@@AbirInsights Hindus do not worship cows, but a majority refuse to eat all meat, including beef. Cows play a central role in the history of Indian husbandry and culture (indeed, it is perhaps impossible to have a complex society without large domestic animals), and dairy products like milk and fresh cheese are staples of their diet. So the refusal to slaughter cows (which offer more alive in the long run) is seen as a first step toward Hindu vegetarianism, which is based on a principle of non-violence and belief in the soul of every living thing. So over time we have seen cows increasingly venerated in Hindu society, and they are important in certain Hindu celebrations. But no Hindu believes cows are divine in any sense or "better than people" or anything like that, and they do not worship them.
Other Indian religions like Jainism also promote vegetarianism and have varying levels of respect for cows, but none of them include worship of cows. The special place cows have in Indian society is perhaps in some ways analogous to the special place dogs have in Western society, where people typically find the slaughter of dogs for meat as far more repugnant than of other animals. The reasons are different (cows aren't pets in the same way as dogs), but the result is the same.
@@EebstertheGreat Thanks for that much needed clarification
@@AbirInsights I didn't know people has this kind of myth about India set in their mind.
I almost ran out of my stack memory trying to follow this.
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.☺
You just cracked the code to get more views.
Put a big flag on the thumbnail and watch the nationalists arrive.
@@Banzybanz no, he hid the square term. I just came to look at the question, but ended up watching the whole video. If I had seen the question, I wouldn't click on it because I knew the answer already, but didn't knew the proof.
@@stv3qbhxjnmmqbw835 he didn't hid the square term instead of writing "square" he symbolized square as ■
@Tangent of circle. Nationalist have no place in the modern world,All the Extremist party are nationalist,Free Kashmir
@Tangent of circle. Lol you have to be trolling?Nobody with alt right views would come open and tell that their Extrimist if you are really a nationalist then you can take the middle finger of mine,Stupid nationalist modick suppoter
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 crazyyyyy
Exactly this one is a little tougher than the general one
Usually 8th and 4th (last problems per day) are ether takes half of time or none depending on can you solve it or not
I think there just above 4 hours per day so like 1,5 hours on solving and 0,5 on writing or viseversa. Anderstending can you solve it or not is the hardest and most important part, and if you can solve it you should know if you can find a way to solve it in rational time. You don't have to solve everything you just need to solve most.
Also tbh it pretty easy problem, I think there must be 5 problems per day and it is 3rd per second day (basically last is coffin hard and 2nd from the end is just tough, so you should just take others as free ones) it also seems that 2^0+3^1=2^2 is a solution too, but I may missed the term in technical English that is for natural numbers and another for whole numbers starting from 0
In second case there probably family of roots where power of 2 is just 0 so we need to insure that there is fixed amount of this solutions (2^0 is 1 base 4, that may cause something in second part of solving to go wrong, basically can reduce amount of points for number by half) I think it easyly fixable with this solving, even if I am right, but most likely I just missed a word in problem (such missunderstanding cost me a lot..)
I actually wrong m, n are natural, so it is absolutely clear done number. I just focused on my solution a bit to much
Thanks Michael penn for making a video on a question from Indian mathematical Olympiad...NAMASTE
Thanks Michael for responding
I received a request for a problem from this very exam! I might put a card to your video as a shout out.
I got lots of requests for this problem. Everyone seems to really like it!
@@MichaelPennMath Yea, I really enjoyed how you get to do the same thing over and over to solve the problem :)
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
Try this one : Brazilian Mathematical Olympiad (OBM) 2019 Problem 3 Level 3
I'll put it on the list.
Bom ver um companheiro brasileiro aqui kkkk fiz a obm ano passado mas real essa tava difícil
Bom saber q n sou o único
Tive dificuldade até pra entender essa questão no dia
Boa sugestão parceiro, vou esperar ansioso pela solução dessa questão
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.
That's how I figured it!
Or he could stop at 16:30 by noticing no two powers of 3 are separated by 2, besides 3^0 and 3^1.
The solution in the hint is the smallest solution.
@@erikb3799 and only solution
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.
Well, actually, if we look at the wording used on the official test paper, it says 'positive integers' only, not natural numbers(you can check on AOPS), really unsure why michael changed the wording
Can't agree with you Daniel. I studied always that the set of natural numbers start from 1 and go on and on.
Tech Toppers i'm with Daniel, set of natural numbers N starts at 0, and set N* starts at 1.
I think that we shouldn't use the terms natural numbers, because each country has their different definition for that.
The standard ISO definition has 0 as a natural number.
The set theoretic construction of the natural numbers requires 0.
Almost all modern number theorists regard 0 as natural.
The Indian examiners who set the paper said ‘positive integers’ .
Yet here we are again,
@@exacerbated no , I was taught zero is no natural numbers
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.
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.
That's how I solved it! Great approach
Yup. That is exactly how I solved it too.
Maybe if I watch enough of these I’ll actually be able to answer one in a contest
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.
Finally ! A question from my country's olympiad !!
Chabbi!!! Remember me?
Wooww!! Can't believe we met here. Chabbi, why did you disappear all of a sudden?
@@saminhaque13-52 what ??
@@steveleekyon1212 -_- I'm with u on plus lol
@@saminhaque13-52 what do u mean by disappear?
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.
Putnam 2002 Problem B5
A base b palindrome is an integer which is the same when read backwards in base b. For example, 200 is not a palindrome in base 10, but it is a palindrome in base 9 (242) and base 7 (404). Show that there is an integer which for at least 2002 values of b has three digits and is a palindrome.
Try this
I actually feel that this would have a better/elegant solution than just doing the same thing again and again
as an engineer, i think it is in fact more elegant to solve it using the same thing again and again
It's insane to solve it in an exam. Who would have the sanity levels to really believe that this approach is going somewhere and spend precious time of the exam? If there is no elegant solution, I guess no one would invest time to try it.
@@michaildokianakis5725 If it was asked in olympiad in INMO or International Maths Olympiad then they will have 3 hours for 6 questions
@@michaildokianakis5725 I think the solution presented is pretty similar to what most people sitting this who manage to solve it would come up with. If you have some experience at doing Olympiad questions, one of the main skills you pick up is the skill of being able to quickly tell if a particular approach is likely to work. In this case I think if you're about imo bronze level then it's pretty easy to tell this kind of method would work, without even having to get out some paper to make some notes. Personally if I was doing this question, I'd do it the following way, which is similar to the way in the video but just a little different (this proof is pretty close to what I'd write for an Olympiad, only thing I'd change is remove the comments in brackets and add a little detail about how to trace back the steps).
First let x^2 = 2^m + 3^n.
If m = 0 then by Mihailescu's theorem, this has no solutions if n > 1. (If you don't want to use Mihailescu's, then instead the factorisation (x - 1)(x + 1) = 3^n can be used). Then you just need to check whether n = 0 or 1 are solutions, and we find that the only solution for m = 0 is 2^0 + 3^1 = 4.
If m = 1 then we get x^2 = 2 + 3^n. There are no solutions for n > 0 because this would imply x^2 is 2 mod 3. Clearly n = 0 is also not a solution. Therefore there are no solutions for m = 1.
The last case is when m is greater than or equal to 2. In this case by using mod 4 we find that n is even, so let n = 2a. Then we can write the equation as x^2 = 2^m + 3^2a => (x - 3^a)(x + 3^a) = 2^m. Then let x - 3^a = 2^y and x + 3^a = 2^z. By subtracting these we then find that 2(3^a) = 2^z - 2^y and then dividing both sides by two gives 3^a = 2^(z-1) - 2^(y-1). Note that y = 1 otherwise the right hand side is even, whilst the left is clearly odd. Therefore this equation can be written as 3^a = 2^(z-1) - 1. If a = 0 then we get the solution 3^0 = 2^(2-1) - 1 and then tracing back the steps, this yields the solution 2^3 + 3^0 = 9. If a = 1 then this gives the solution 3 = 2^(3-1) - 1 and tracing back some steps gives the solution 2^4 + 3^2 = 25. If a > 1 then the equation has no solutions by Mihailescu's theorem (again you can avoid Mihailescu's if you want by using a factorisation).
So overall it's not too hard to come up with and not too hard to realise beforehand that the method will work. It involves a lot of very standard tricks which make it pretty easy to do if you know them. One trick is to deal with small cases first in order to use some modulo arithmetic and the other main trick is using difference of two squares factorisation a lot. Using Mihailescu's theorem is also handy to make the written proof a little bit shorter, but not knowing it probably wouldn't hinder you too much in finding a proof. Overall if you have a fair bit of experience then a question like this might take a few minutes to solve, and then a little while longer to write up a proof, maybe 15-20 minutes, but in most Olympiads this is a very reasonable amount of time, and in fact it probably wouldn't be a problem to spend a while longer than that, even an hour or two working on the question probably would be fine in an olympiad. In the international mathematical Olympiad, you get 4h30 to solve 3 questions, but a significant proportion of people only solve 1 question.
Nice solution. This was also a question in the secomd round of the British Mathematical Olympiad in 1996
Thanks! Greetings from Poland!
Polacy, złączmy się tutaj xD
no siema
Depending on how you define setN := 'the set of natural numbers' (is 0 element of setN?), you might want to add (m=3, n=0):
=> 2^m+3^n = 2^3+3^0 = 8+1 = 9 = 3^2
The set of natural numbers is {1, 2, 3, 4, ....}, the set of whole numbers is {0, 1, 2, 3, 4, ....}, so if this problem were using the set of whole numbers instead of natural numbers than there would be two solutions.
@@dclrk8331 I would prefer such a clear definition for natural numbers, but you shouldn't be so sure. There are different standards, see:
mathworld.wolfram.com/NaturalNumber.html
It says I am subscribed to you, but I swear this is the first time I'm seeing you. You're awesome.
Try this: proove that each set of n^2+1 people has a subset of n+1 people such that once ordered by height are also ordered by age. Thanks. Great contents as always!
What is 'n' here? Is it true for all natural numbers? And what do you mean by 'ordering'? Strictly ascending / Strictly descending or either?
Because if you consider trivial case n = 1, n^2 + 1 is 2 persons and n + 1 is also 2 persons.... So for any given 2 people, we can always have a subset of 2 people (the subset is same as the superset) and any ordering of age of the two people will also give ordering of the height, but it can be EITHER ascending OR descending....
Can you throw in some specificity?
@@050138 yep natural numbers and either ascending or descending.
They just need to be ordered in some way.
@Adam Romanov Wow! Didn't know such a theorem exists! I have exactly followed the steps backwards as you mentioned.... And arrived at the theorem statement on the lines of what you wrote.... Which I had been trying to prove since last two days....
The thing I am trying to prove goes like this.... Given a sequence of (n^2 + 1) numbers which are increasing.... Irrespective of whatever order you may arrange the numbers of this sequence, you will still have (n + 1) of the numbers either strictly increasing or decreasing.... It will be either increasing or decreasing, not both and it will depend totally on the way the terms of original sequence is jumbled
@Adam Romanov thanks for your reply. It's from Scuola Galileiana Unipd Admission Test. (Italian university math undergrad admission test)
Pleased try this one, it would be really helpful
Prove that each of the number √n+1-√n for positive integers n can be written in the form of
2cos(2kπ/m) for some integers k,m
I've put it on my list!
From PFTB? Seems like I have seen this somewhere before
@@shantanunene4389, yeah, i think i've seen it on Brilliant
Where is it from
Also, I think there's a typo. No number of the above form can be written as 2 cos(2 k π/m)
Once you have 2 = 3^i - 3^j it's immediately obvious that i = 1 and j = 0. How could any other powers of 3 (1, 3, 9, 27, 81, ...) be just two apart?
He just tell us the "mathematician's way" that can be used on the general problems
I think this is also a valid way, in fact its pretty elegant
I enjoy the way you can transform the ideas into math in very clever ways. Enjoy your content
I actually intuitioned that there would be only one when the pythagorean theorem came keeping in mind Fermat’s Last Theorem.
but you know that 's cheating isn't it
@@zouhairnouri6863 it's intuition as the person mentioned.. since we can't prove fermat's last theorem in the exam.. I think it's safe to assume that each component on the RHS can't be of power>2.. and from the video Pythageorean triplets visualised by 3blue1brown(awesome channel btw).. one can easily say one possible pair of m&n is possible..
There are other solutions. You need to look at the edge cases m=0 and n=0. (0 is in ℕ, so it is a solution to consider).
2^0 + 3^1 = 4 = 2*2
2^3 + 3^0 = 9 = 3*3
These are solutions of you consider 0 a natural number. I take the more standard definition that N is made up of positive integers.
@@MichaelPennMath Standard in the US ;o)
In Europe, we have ℕ*={1,2,3... and ℕ={0,1,2...
That is very interesting and explains a lot when I see talks by Europeans. Anyway, the problem originally read "all positive integers". I shortened it to \in\N to take up less room on the board. Also, how do you get the \mathbb{N} ie the blackboard bold N in the comment?
@@MichaelPennMath This is a unicode symbol. I have a list of them on my phone ready to be copied/pasted. Here are a few:
Fractions ⅟ ½ ⅓ ⅕ ⅙ ⅛ ⅔ ⅖ ⅚ ⅜ ¾ ⅗ ⅝ ⅞ ⅘
Die ⚀ ⚁ ⚂ ⚃ ⚄ ⚅
Chess ♔ ♕ ♖ ♗ ♘ ♙ ♚ ♛ ♜ ♝ ♞ ♟
Parallelograms ▱ ▰ ⏥ ⏢
Mathbb 𝕒 𝕓 𝕔 𝕕 𝕖 𝕗 𝕘 𝕙 𝕚 𝕛 𝕜 𝕝 𝕞 𝕟 𝕠 𝕡 𝕢 𝕣 𝕤 𝕥 𝕦 𝕧 𝕨 𝕩 𝕪 𝕫 𝔸 𝔹 ℂ 𝔻 𝔼 𝔽 𝔾 ℍ 𝕀 𝕁 𝕂 𝕃 𝕄 ℕ 𝕆 ℙ ℚ ℝ 𝕊 𝕋 𝕌 𝕍 𝕎 𝕏 𝕐 ℤ 𝟙 𝟚 𝟛 𝟜 𝟝 𝟞 𝟟 𝟠 𝟡 𝟘
You can easily find them (and many, many more) online, the thing is that websites and apps do not necessarily support all of them. I never used some of the ones I just copied/pasted on the YT comments section, so some funny characters might be displayed instead.
And for math formulas, I also have a list of things I wrote in the past:
∂f/∂t = ∂²f/∂x²
∀ x ∈ ℝ
∫ e⁻ᵗ dt
The more I'm exploring your videos on modulo arithmetic the more I'm falling in love with this again...😄
The teacher's teaching is very good, it is very good to talk about how to solve these problems. Students need to follow his teaching. If this is not followed, it may not be possible to understand No. Mathematics is difficult, but even harder, you have to learn
7:30 by saying that 4^a is congruent to 0 modulo 4 you don't take into account the case where a=0 (4^0=1 is congruent to 1 mod 4). This case gives another solution m=0 and n=1 (2^0+3^1=1+3=4=2^2)
and the same goes for when n=0, another possible solution is m=3,n=0 (2^3+3^0=8+1=9=3^2)
This is an elegant solution to a beautiful problem.
However, I believe it can shortened somewhat.
At 13:01, we have 4^a=2^c.2^d, so 2a=c+d (this relation was not noted).
On the other hand, at 14:17, we have d=1.
So once we find c=3 at 18:52 we get 2a=3+1=4, a=2.
So the working to find a from 19:34 to 21:30 is not really needed.
Hello! If you want you can try the following problem: Let an be a series, where a1=p (p is a prime number) and an=2*a(n-1)+1. Prove that exists a natural number i, such that ai is not a prime number. (Romanian Mathematical Olympiad)
I like this problem. I will give it a go. btw: I have a collaborator who is originally from Romania!
I think the general term would be 2^n*p+2^{n}-1, now settings n=k(p-1) implies the whole term is divisible by p.
Very interesting question! Was able to solve it using Fermat's Little theorem.... More over, that 'i' for which it is true is the prime 'p' itself, for all odd primes p....
Let's represent nth term as a[n]
Since a[1] = p and a[n] = 2*a[n - 1] + 1
we have a[2] = 2*a[1] + 1 = 2p + 1
a[3] = 2*a[2] + 1 = 2*(2p + 1) + 1 = 4p + 3
So we can see that in the nth term, the coefficient of p will be 2 multiplied (n - 1) times and the constant term will be sum of all the powers of 2 from 0 to (n - 2), or
a[n] = 2^(n - 1)*p + (2^(n - 2) + ...... + 2 + 1)
or a[n] = 2^(n - 1)*p + [2^(n - 1) - 1]
or a[n] = 2^(n - 1)*(p + 1) - 1
or generally or a[i] = 2^(i - 1)*(p + 1) - 1
Now, Fermat's Little theorem states that if 'p' is a prime and 'a' is a natural number relatively prime to p (gcd of (a, p) = 0), then
a^(p - 1) = 1(mod p)
So for every odd prime (all primes except 2), as 2 is relatively prime to p, we have
2^(p - 1) = 1(mod p)
We also have (p + 1) = 1(mod p) obviously
Multiplying the two congruences,
2^(p - 1)*(p + 1) = 1(mod p)
which means 2^(p - 1)*(p + 1) - 1 is divisible by p
But 2^(p - 1)*(p + 1) - 1 is nothing but a[p]
So we proved that for all odd primes p, a[p] is divisible by p and hence not a prime
If p is 2, then we have a[1] = 2 and
a[2] = 2*2 + 1 = 5 which is an odd prime, for which we already have proven that there will be an i such that a[i] is not prime
QED
@@angelmendez-rivera351 Thank you! ☺️
I really loved your approach towards the question. Can you suggest how do I develop such a thought process akin to your's ??
Beautifully explained 🙂🙂
16:22 from here the fact that one of the factors is 1 immediately follows, since numbers that differ by 2 cant both be divisible by 3
Good observation.
Lior orz
@@II_xD_II How do you keep finding me smh
Great video! Maybe you can do some problems involving quadratic residues next? eg: This problem- Find all positive integers x,y such that x^3+7=y^2
Quadratic residues and quadratic reiprocity are way beyond IMO maths.
Wow, thats a real masters level problem and that solution reminds me of my Number Theory class taken in 1969. Great piece of logic
Try this one:
S_n={k\leq n | k^2 has a zero in its decimal representation }
Find, limit n to infinity ( |S_n| / n)
Where |A|= cardinality of set A
Damn he took three hours out his life to solve this for us. It would be awesome to solve it with shapes and not all the maths
How can I be this patient in problem solving? I really think I could do this one but I give up quickly. Every time I run out of ideas, you come up with a keen observation.
Salute you!
Regards
India
Same here.
Hello, a lot of doiphantine equations have similar ideas and by learning those idea, you will be able to solve even hard question
some of them are:
checking modulo
delta method (ax^2+bx+c=0, then b^2-4ac should be a perfect square)
checking the highest power of p (vp(n))
factorization
and so on, good luck
@@hooceiorgo1122 what's p(vp(n))
Hello again ! I love the way you explain these, it makes the audience feel smart thanks to your unconventional teaching. Keep it up !
Michael: names powers as “a”, ”b” ,”c” ,”d” ,”I”,J”. “e”, “f”, “g”, “h”: am I a joke to you?
I think I would've pulled a Fermat. M=4, N=2. There is not enough space in the CZcams comments for the proof, and hope for the best.
Just wondering how someone could think up something that long.
People spend years to learn topics to slove this problem
I think it's that you build an intuition for it over time, and once you do one step, there's only a finite amount of things you can do and you keep going and going until you're there!
You drink and you think up things.
Let ABC be an acute triangle with circumcircle omega and let H be the intersection of the altitudes of the triangle ABC. Suppose that the tangent to the circumcircle of triangle HBC intersects omega at points X and Y with HA = 3, HX = 2 and HY = 6. The area of triangle ABC can be written as m sqrt ( n ) where m and n are positive integers and n is not divisible by the square of any prime. Find m + n.
This question is from the AIME 2020.
..... "tangent to circumcircle of triangle HBC interesects omega....."
But tangent at which point?
Worked your way through a third of the alphabet on this one -- reusing "a", "b", "i", and "j".
I immediately thought to 2^3+3^0=9=3^2 which is a solution you didn't tell. You were working on N and not N* ^^
Of course there is also 3^1+2^0
with wolfram alpha with this as formula (2^m+3^n=o^2) it gave these answers
m = 0, n = 1, o = ± 2
m = 3, n = 0, o = ± 3
m = 4, n = 2, o = ± 5
16:34 When he wrote 3^i , I thought it is becoming complex.
Nice Problem! This ones a great example of the common technique of using differences of squares in solving exponential diophantine problems.
Please try Indian math olympiad 2020 problem 2.would be excited to know that you have put it one your list
If we allow zero as a value, m = 3 and n= 0 also works: you get 8 + 1, which is 9, a perfect square.
The ISO 80000-2 definition of "natural number" includes 0, which leads to two more solutions m=0, n=1 (1+3=4) and m=3, n=0 (8+1=9). (I'm not saying this video was wrong; this video just used a non-ISO definition of natural numbers.)
For those interested, please look for the step in this video that excluded these solutions.
For those more interested, you might have noted that 4 and 9 are squares of 2 and 3; please show that the resultant squares being powers of the two LHS numbers is more than a coincidence.
For those even more interested, please further prove or disprove that these are the only two additional solutions allowing m or n to be 0.
No need to do the mod 3 on 3^b = 2^(c-1) -1 because c-1 = 2a-2 and we can see that it's even. And then no need to introduce k. Also, once you have the equation in terms of a and b, it turns out to be a bit cleaner to go to 3^(2b) = (X+2^a)(X-2^a) rather than the other way round.
The second half (after 11:00, where m=2a and n=2b have been established) goes much faster when you realize we are in fact looking at an nontrivial primitive pythagorean triple (short: nppt) (2^a, 3^b, x) with (2^a)^2+(3^b)^2=x^2. It is primitive as obviously 2^a and 3^b are coprime, and nontrivial as both 2^a and 3^b are not equal to 0. Each nppt is of the form (u²-v², 2uv, u²+v²) or (2uv, u²-v², u²+v²) for natural numbers u and v, with 0 < v < u, u and v are coprime, and not both odd (Euklids formula or use the rational parameterization of the unit circle).
3^b cannot be of the form 2uv, as 2 divides rhs but not lhs of the equation.
Hence 2^a =2uv, and both u and v must be powers of 2, say u=2^i and v=2^j with 0 j=0
Using this, we know 3^b = (2^i-1)*(2^i+1) is a power of 3, so both factors (2^i+1) and (2^i-1) are also powers of 3;
if the smaller factor is not equal to 3^0 = 1, it must be congruent 0 (mod3),
the larger factor is bigger by 2, hence congruent 2 (mod3) => contradiction.
It follows that the smaller factor (2^i-1) is equal to 1 and hence i=1.
Put all together we end up with u=2, v=1 => 2^a = 4 and 3^b = 3 => a=2, b=1 => m=4, n=2 as the only solution.
around minute 18, it seems like if (2^k-1)(2^k+1) is a power of 3, then both factors are a power of 3, but the only powers of 3 that differ by 2 are 1 and 3. Same for 4^a+9=x^2 implies 4^a = (x-3)(x+3) and the only powers of 2 that differ by 6 are 2 and 8, so x=5. Seems a bit easier to me.
Amazing video! The explanation flows so naturally!
2^3 + 3^0 = 3^2, or 8+1=9, is also a solution.
For the part around 6:00, we can just reduce 2^m mod 3 as (-1)^m, which is even iff m is even
You missed a solution n=1 m=0.!! At around 8:24 in, you reduce 2^(2a) + 3^n = x^2 in modulo 4 to 3^n = x^2 mod 4. But this is only true whenever a is not 0. Otherwise set a = 0 then you get 1 + 3^n = x^2 modulo 4, and some modulo arithmetic reasoning will result in n having to be 1. And indeed n = 1 and m = 0 is a solution!
Haha... You're right
With the same reasoning m=3 n=0 is also a solution.
didn't know neil patrick Harris is such a good mathematician
This is absolutely amazing the modulo is the mvp here
You forgot to check the n=0 cases and the m=0 cases: some other solutions for (m,n,x) are (3,0,3) and (0,1,2).
0 probably isn't a seen as a natural number in the Indian math Olympiad
Yes, factorization is a good way but I just wonder if Pythagoraen Triples are working. Write m^2-n^2=2^a and 2mn=3^b or m^2-n^2=3^b and 2mn=2^a. While the first case is not obviously appropriate then just solve the second one!
Yes, once you know that both exponents are even it is clear that you have a primitive pythagorean triple and can use the parameterization you mentioned. This is a very nice and potentially shorter solution. I didn't approach it this way to keep the argument more self contained.
@@MichaelPennMath if I were you I would have used the same way😂. In our junior summer and winter MOP camps, we generally use mod and factorization while solving diophantine equations just like this. Also in competitions and olympiads, the official solutions are based on this method!
It would be really nice to see a visualization of this, with the level sets of 2^m + 3^n (where m & n can be any non-negative *real* number, not just integers) at all perfect squares. These curves should only intersect exactly one lattice point, at (m, n) = (4, 2).
Great work. Just discovered this CZcams channel. I'm subscribing right away.
Back to 1992, 0 May not be a natral number. I am not sure about this. But for now, 0 is. It means that 1+3=4 is another solution. So there are two solutions here. I think this would be a remark here.
And it is the same way you can show they are the only solutions.
@@yanlonghao5012 There is also the solution 8+1=9. I think the three solutions (16+9=25, 1+3=4, 8+1=9) are the only ones.
@@philippenaturel7337 👍. I did not do the math. Thank you for point out this. I think using the difference of square could prove this.
I must admit I edited the statement of the problem in order to take up less room on the board. The original statement required that m and n be positive integers (which is the most common definition of N). You are right about additional solutions if you let m and n be non-negative integers (the other common definition of N).
Sir mathematician, which book should I buy so that I get into all the higher algebra u use in ur awesome videos, I like solving them all.
I'm a 12th pass student from India, n love mathematics!
4:35 in this case it would've been more elegant to note that 2 is congruent to -1 mod 3 and then argue that m cannot be odd
what about m=3 and n=0?
Instead of the decomposing proof in the 3^a + 1= 2^(b), you can also show a < 2 then work on the remaining a cases. If a >= 2, 3^a+ 1 > 8 is a power of 2 > 8, so 8| 3^a + 1. But 3^a + 1 is always 2 or 4 mod 8.
Please try this one
A is a positive real number. n is a positive integer number. Find the set of possible values of the infintie sum
x₀ⁿ+x₁ⁿ+x₂ⁿ+... where x₀,x₁,x₂,... are all positive real numbers so that the infinite series x₀+x₁+x₂+... has sum A.
Source: Bangladesh Mathematical Olympiad (BDMO) 2019, Higher Secondary, Problem 4
Taking n=p^1008 with p prime gives infinitely many examples of when this is happening (answering your tangential wondering )
Really enjoying your videos. I've never seen equations reduced by modular arithmetic before, is there a good source for more information on this? Can you make a video about it?
I reduced this mod 3 then showed m must be even. So put m=2k.
Then difference of squares leads to
3^n = 2*2^k + 1.
From here we can show there is no solution for n>2 by reducing mod 27, because the RHS can never be zero mod 27.
So n=2 is the only solution unless we count m=0, n=1.
Hey Michael this is me from India again .....this is a question from Pre Regional Mathematics Olympiad.....”How many triangles ABC are there upto similarity such that the magnitude of A ,B and C are positive integers and satisfy “cosAcosB + sinAsinBkC = 1” for some positive integers k where kC doesn’t exceed 360 degrees? Please make a video on this.....I have great hopes from you god 🤗🤗🤗🤗
Is it (CosA)*(CosB) + (SinA)*(SinB)*k*C? Or is BkC one term?
What if we allow m = 0 or n = 0? I can think of two other solutions in this case: m = 3, n = 0 and m = 0, n = 1. Are there any others?
natural numbers doesn't include 0.
@@MegaGuiaGamer That's a matter of opinion. -- @mushroomsteve That's all: 2^0 + 3^n = x^2 means 3^n =(x+1)(x-1) and so on the right we have two powers of three with difference only 2; this works only if they are 3 and 1, so x=2.- Similarly 2^m + 3^0 = x^2 means 2^m = (x+1)(x-1) and on the irght we have two powers of two with difference 2; this works only if they are 4 and 2, so x=3.
@@HagenvonEitzen I know it works with 0, but it's sth obvious, even if some people add 0 to the natural numbers.
@@HagenvonEitzen Nice!
@@mushroomsteve so I guess that the problem is the sentence, 'cause it's open to multiple interpretations. (I'm sorry if my english is bad, isn't my mother language). You're right, bro :p
Your videos are the best at math. I’ve learned a lot of new stuff in your videos. Thanks again for such great videos. Btw. Hi from Thailand :)
It seems like there is a connection from this problem to Fermat’s Last Theorem.
10:28
It seems a simpler way to check that is that 3^(2b+1) cannot be divided into equal factors 3^x times 3^y with x+y=2b+1 , since one of these factors will have at least one factor of three more than the other: 2b+1 is odd and thus not divisible by two.
Edit: then again, this would interrupt the flow of the video, so it might not be as good.
Just a quick question: if m and n are natural numbers, shouldn't n=0, m=1 ; n=3 ; m=0 also be solutions?
Not necessarily, 0 isn't always seen as a natural number. Often the natural numbers are defined N = {1,2,3,...}, it depends on the convention. I suppose in the Indian Mathematical Olympiad this was the convention used.
Hii I'm fan of yours.
I did the middle completely differently.
I also started by showing in the same way that both powers had to be even. So 2^n = 2^(2*i), 3^m = 3^(2*j)
Let a be the smallest of 2^i or 3^j, and let b be the other term and c^2 is the square sum.
I noticed that when a^2+b^2=c^2, for a (c-a).
In our case (c+a)*(c-a) is a power of either 2 or 3, so (c+a)/(c-a) has to be either 2, 3, or 4.
But also b^2 = (c+a)/(c-a) * (c-a)^2, showing that (c+a)/(c-a) has to be a square, therefore this quotient can only be 4.
This proves that b^2 = 2^n and a^2 = 3^m and (c+3^j) = 4*(c-3^j) This can be rewritten to c = 5*3^(j-1).
So 2^n = 4*(5*3^(j-1) - 3^j)^2 = 4*(2*3^(j-1))^2, but because 2^n has no factor 3, this means 3^(j-1) has to be 1 and j=1.
Therefore 3^m = 9, c=5*3^(j-1)=5, and 2^n = 16. Q.E.D.
Wow, I thought this would be an easy proof at the start. I was wrong! While it maybe wasn't too difficult, you had better be fluent in modulo math! Great job!
But m, n are N not N* so e.g. we could say 2^0+ 3^1=4=2^2, idk if there are also other such solutions but here is one he overlooked
How come there's only one solution? Wouldnt you be able to use any Pythagorean triple?
Why didn't you consider the solutions m=0 n even and n=0 m even
0 isn't always considered a natural number.
Thanks!! I'm so happy I was able to follow and understand the resolution to the end :) , I think I'm beginning to get to grips with modular arithmetic.
2^m+3^n=np mersenne prime number. modular square seven is perfect number two term equal multi term computation time theorem.
ΙΜΟ 2011 P2(the windmill problem) PLEASE,that would be my DREAM
I'll put it on the list.
Awesome Video as always....
☺☺
With a lot of effort, I could follow your resolution of the problem. But the real question is, how do one come up with that solution to begin with?
Another proof that P is not equal to NP (most likely anyway).
Hello!
I would suggest o problem which I think it is by far the hardest ever problem to a competion. I am not exagerating , not one bit!
It is the problem 6 of RMM 2016 (Romanian Masters of Mathematics). When I saw it for the first time, I've said that no student will be able to do it! But , one did it ! one in 113. A few took 1 or 2 points and that's it ! Please consider watching it! Thanks
Is this channel is yours? Or is it a group work? I’m fascinated C;
The channel is all me! I have done a few collaborations with other channels in the past and have some others upcoming. Also, I think I will try and and have some "special guests" also.
m=0, n=1
I've tried with 3 and 5 and found no natural number solution....
Just out of curiousity if both of them are distinct primes(other than 2) what can we say about that?
We found „a” ealier. When you were searching „b” you noticed, that a=c+d, and You found that! d=1 and c=3 (even b4 knowing „b”!)
14:10- we know d
18:35- we know c
The explanation was so detail ! Thanks and good job