Is this type of solution satisfying??

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  • Опубліковано 16 січ 2025

КОМЕНТАРІ • 134

  • @rennanchagas6174
    @rennanchagas6174 3 роки тому +62

    I have always felt more confident in proofs finding a horrendous irracional number as a solution than finding "no solution".

  • @Tsalnor
    @Tsalnor 3 роки тому +5

    m^m+1=n^2 => m^m=n^-1 => m^m=(n-1)(n+1)
    Since both n-1 and n+1 are factors of m^m, both must be powers of m. n+1 is a higher power of m than n-1. Thus, n-1 is a factor of n+1.
    (n+1)/(n-1)=k => 1+2/(n-1)=k, k is an integer
    This means n-1 is a factor of 2, so n must be 2 or 3. Neither case gives a solution in the original equation, and thus there is no solution.

  • @f5673-t1h
    @f5673-t1h 3 роки тому +99

    The result follows from Catalan's conjecture (i.e. Mihăilescu's theorem), stating that the only consecutive perfect powers are 8 and 9, since the question is asking if we can find m^m followed by n^2.

    • @picklesauce7983
      @picklesauce7983 3 роки тому +2

      true. Thx for the one-liner

    • @RexxSchneider
      @RexxSchneider 3 роки тому +1

      Only if you explicitly exclude 0 and 1

    • @zfnQRZJT
      @zfnQRZJT 3 роки тому +1

      @@RexxSchneider and 0 and -1, and -8 and -9, if you are OK with complex numbers

    • @Robert-jy9jm
      @Robert-jy9jm 3 роки тому +21

      Using the Catalan's conjecture for this problem is like using a nuclear weapon to kill a mosquito. :)

    • @zfnQRZJT
      @zfnQRZJT 3 роки тому

      @@Robert-jy9jm Well, no not really because a nuclear weapon could not actually kill a mosquito because they basically, so, so basically it wouldn't work at all in any case

  • @goodplacetostop2973
    @goodplacetostop2973 3 роки тому +15

    8:48 Anyone else watching the Olympics opening ceremony?

    • @jiyoungpark6233
      @jiyoungpark6233 3 роки тому

      no...😅😊😊

    • @mathlegendno12
      @mathlegendno12 3 роки тому +4

      Me, it’s very exciting, looking at ppl waving flags for an hour

    • @jiyoungpark6233
      @jiyoungpark6233 3 роки тому

      oh, really?...i just got home from work...i'll watch the replayed film of it, tomorrow🥰🥰🥰thank you💜💜💜💜💜

  • @jarikosonen4079
    @jarikosonen4079 3 роки тому +7

    Its nice to see that these actually are solvable...

  • @OscarCunningham
    @OscarCunningham 3 роки тому +60

    If we let ourselves go over the integers we get -1^-1 +1 = 0^2.

    • @mathlegendno12
      @mathlegendno12 3 роки тому +2

      Natural numbers* bc integers include -1 and 0

    • @OscarCunningham
      @OscarCunningham 3 роки тому +15

      @@mathlegendno12 My clumsy phrasing. By 'go over' I didn't mean 'go beyond'; I meant that m and n are allowed to vary across all the integers.

    • @ConManAU
      @ConManAU 3 роки тому +4

      And there can’t be any others because if m < -1 then m^m is not an integer.

    • @davidseed2939
      @davidseed2939 3 роки тому +1

      Not integers only Natural numbers ie positive integers.
      P

    • @omp199
      @omp199 3 роки тому +2

      @@davidseed2939 Sensible people define the natural numbers to include zero. If you want to talk about the positive integers, then we already have a straightforward term for that: "the positive integers". There is even a handy way of denoting that set using symbols: a blackboard-bold upper-case Z with a superscript plus sign. There is no need to waste the term "natural number" on something that already has a convenient name, while leaving the set of _actual_ natural numbers without a convenient name at all. ("Non-negative integers" is just too clumsy.)

  • @bprpfast
    @bprpfast 3 роки тому +6

    Based on the title and the thumbnail, I guessed there’s no solutions!

    • @abdullahyousef3596
      @abdullahyousef3596 3 роки тому

      hey, I did not expect you to be here for some reason :)

    • @d-rex7043
      @d-rex7043 3 роки тому +1

      @@abdullahyousef3596 hey, Abdullah, didn't expect to see you here! Hope you are good!

    • @abdullahyousef3596
      @abdullahyousef3596 3 роки тому

      @@d-rex7043 I am good alhamdolilah, thanks for asking.

  • @DeanCalhoun
    @DeanCalhoun 3 роки тому +5

    it’s just as important to prove things are impossible as it is to prove that they are

  • @CM63_France
    @CM63_France 3 роки тому +1

    Hi,
    For fun:
    0:15 : "or so on and so forth",
    4:29 : "ok, let's get rid of this and".

  • @LightPhoenix7000
    @LightPhoenix7000 3 роки тому +49

    You say that you're not doing modular arithmatic, but isn't even/odd just modulo 2? :P

  • @GaryFerrao
    @GaryFerrao 3 роки тому +33

    By looking at your solution for the even case, i was thinking why can't b = 0 be a a solution. And then i saw m as the exponent and thought it best not to open that can of worms 😂

    • @joshmyer9
      @joshmyer9 3 роки тому +3

      It'd probably be great for YT comment "engagement" numbers, though. 🙃

    • @OscarCunningham
      @OscarCunningham 3 роки тому +2

      There are two debates here, 'Is 0 natural?' and 'Is 0^0 = 0?'. I think opinions about these questions might be anticorrelated. Thinking in terms of sets leads to wanting 0 as a natural number, because you can define the naturals as the cardinalities of the finite sets. But then you'll probably define n^m as the number of functions between sets of sizes m and n. So there probably aren't too many people out there thinking that 0^0 + 1 = 1^2 is a solution.

    • @nottaexpert2690
      @nottaexpert2690 3 роки тому +1

      @@OscarCunningham 0^0=1. I never understood why until watching a video recently, I think by redpenblackpen.

    • @AlfredJacobMohan
      @AlfredJacobMohan Рік тому

      @@nottaexpert2690 As a limit, it is true. We must be careful to state that x tends towards 0+.

    • @lox7182
      @lox7182 Рік тому +1

      ​@@AlfredJacobMohan e^(-x^2) tends to 0 and 1/x tends to 0 as x tends to infinity
      but (e^(-x^2))^(1/x) = e^-x tends to 0.
      You can even make the limit tend to numbers other than 0 or 1.

  • @PunmasterSTP
    @PunmasterSTP 3 роки тому +1

    The ending of this video was two-tally awesome! Thanks so much for making all of these great videos and sharing.

  • @petersievert6830
    @petersievert6830 3 роки тому +7

    It is. I dare say, it is even the most satisfying to prove, there are no possible solutions :-)

    • @Kuratius
      @Kuratius 3 роки тому

      I think there is one if you define 0^0 as 0 and take n=1

    • @omp199
      @omp199 3 роки тому

      @@Kuratius But what would be the justification for defining 0^0 as 0? Defining it as 1 seems much more sensible. You always start with 1 as the base from which you start multiplying.

  • @francescfloresgamez657
    @francescfloresgamez657 3 роки тому

    i think it's easier to see that mᵐ = (n+1)(n-1), then with n+1 = mᵃ and n-1 = mᵇ show that mᵃ - mᵇ = 2 has only solutions for m=2 and m=3, and then show that these values do not work in the original equation.
    P.D. : finding values for mᵃ - mᵇ = 2 might be more obvious if you notice that mᵇ(mª⁻ᵇ - 1) = 2 and 2 is prime, so mᵇ is either 1 or 2.

  • @mcwulf25
    @mcwulf25 2 роки тому

    Nicely explained as always.
    We could use the second part in a similar way for even m. The GCD of the RHS factors would be 2 for consecutive even numbers. One factor will be a^m and the other 2b^m (or vice versa with the 2).

  • @RandyKing314
    @RandyKing314 3 роки тому

    Satisfying? Yes! Because that is all that is needed.

  • @alexismiller2349
    @alexismiller2349 3 роки тому +1

    This is also just a special case of Mihăilescu's theorem: the only two consecutive perfect powers are 8 and 9 :D

  • @manucitomx
    @manucitomx 3 роки тому +1

    A non-number theory number problem!
    Thank you, professor!

  • @alainrogez8485
    @alainrogez8485 3 роки тому +1

    6:15 I thought Michael would fall on stage.

  • @wannabeactuary01
    @wannabeactuary01 3 роки тому

    The even case was fun but the odd case was hard work...

  • @iooooooo1
    @iooooooo1 3 роки тому +1

    I'm not sure I understand the particular proof that b^2 + 1 = n^2 is impossible - what reasoning is used to derive b^2 < (n+1)^2?
    (I do see a number of ways to do it. (edit: that's what I get for commenting before continuing! I see a similar way to one of those below is also shown.)
    One way is to show that differences of squares are at least 3 if the smaller square is not 0: (k+1)^2 - k^2 = 2k + 1, which is at least 3 if k >=1 and is 1 only if the smaller square is 0. This contradicts 1 = n^2-b^2 since b is a natural number.
    Another might be that if b^2 + 1 = n^2, then 1 can be expressed as a difference of squares and then factored, 1 = n^2-b^2 = (n+b)(n-b). If n < b, we have a product of one negative term and one positive multiplying to a positive, contradiction; if n = b, this results in 1=0; if n > b, we have two numbers, one of which must be at least 1 and one of which must be strictly greater, whose product is 1.)

  • @drygulch1000
    @drygulch1000 3 роки тому

    Mass to the mass power plus one equals moles squared

  • @vh73sy
    @vh73sy 3 роки тому +1

    A trivial solution is
    m=-1 & n=0

  • @DaneBrooke
    @DaneBrooke 3 роки тому +1

    Perhaps more direct to move 1 to the rhs, ln [both sides], differential operator [both sides], move the 1 to the rhs, massage the rhs and then the proposition is that ln(m) = (1 + 2n - n²)/(n² - 1), which fails cuz rhs is rational but ln(m) is never rational for rational m, QED.

    • @DaneBrooke
      @DaneBrooke 3 роки тому

      @@petrie911 My thinking was wrong in "differential operator" (it has not been established that dm = dn). I have been musing whether or not it can be fixed up... I think so, but then the proof is quite longer and less approachable than MP presented.

  • @emanuellandeholm5657
    @emanuellandeholm5657 3 роки тому +1

    m must be odd, otherwise we have a square + 1 = a square which has the only solution 0^2 + 1 = 1^2, but m^m is 1, not 0 for m = 0. m also cannot be a square, eg m=9 gives a square + 1 on the RHS. m^m + 1 is 1 mod m, so n must be +/- 1 mod m.
    Trying to solve this without using difference of squares and gcd.

  • @jamescollis7650
    @jamescollis7650 3 роки тому +2

    I don't understand how you get from gcd(n+1,n-1)=1 and (n+1)(n-1)=m^m to n+1=a^m, n-1=b^m for some a,b with gcd(a,b)=1

    • @jamescollis7650
      @jamescollis7650 3 роки тому +3

      @UCW5g8lu6PcvgfUSZEkw50gA I did understand that bit. I think I get it now though, for two factors of an mth power to be coprime the individual factors themselves must be a power of m otherwise there will be prime factors in common

    • @franksaved3893
      @franksaved3893 3 роки тому

      @@jamescollis7650 yep

  • @General12th
    @General12th 2 роки тому

    Hi Dr. Penn!

  • @jhonnyrock
    @jhonnyrock 3 роки тому +1

    Let's start a war: m^m=0, n=1

  • @mrl9418
    @mrl9418 3 роки тому

    No more Vertex Algebras?

  • @ImaginaryMdA
    @ImaginaryMdA 3 роки тому

    ooooh, I did the n odd case first, and then tried to find the solution for n even in a similar way... missing the most obvious part of the solution!

  • @nah496
    @nah496 3 роки тому +5

    the ending made me cry 😭

  • @m4riel
    @m4riel 3 роки тому

    it certainly is

  • @parameshwarhazra2725
    @parameshwarhazra2725 3 роки тому +8

    You are really a great teacher who spends so much valuable for the students and provide important problem solving method totally free on UA-cam. How on earth a man can be so kind and helpful? Michael Penn sir I really learn a lot from your videos. Thanks once again on behalf of the entire Indians.❤️❤️❤️❤️❤️❤️❤️

  • @VSN1001
    @VSN1001 3 роки тому

    Hey Michael, I used a different method and was wondering if my method is valid. I first did mapulations S.t. m^m=(n+1)(n-1). I let m = (m-k)+ k where k > m-k. Then, through grouping :
    n+1 = m^k -(1)
    n-1 = m^(m-k) -(2)
    (1)-(2) :
    [m^(m-k)][m^(2k-m)+1] = 2
    Using a bit of mathematical logic and the fact that k & m are bounded to natural numbers, i found that (1)-(2) cannot hold true since
    Minimum of [m^(m-k)][m^(2k-m)+1] = 6 > 2. So I conclude that m^m+1=n^2 has no solution for n&m are natural. Does this work or did I made a mistake

    • @gmncnr
      @gmncnr Рік тому

      I think its true

  • @Andreyy98
    @Andreyy98 3 роки тому

    Just another solution about the second part(m is odd). I find it interesting, bcs it goes from the logic of m even, and is in the end reduces to it,
    even tho with a bit harder calculation.
    Lets suppose m = 2a+1 then (2a+1)[(2a+1)^a]^2=(n-1)(n+1). As n-1 and n+1 are relatively prime then 2a+1 | n-1, or 2a+1 | n+1.
    Let's look at the first case(second is similar) : 2a+1 | n-1 => n = k(2a+1)+1 => n+1 = k(2a+1)+2
    Then [(2a+1)^a]^2 = k^2(2a+1)^2+2k(2a+1) => (2a+1)k^2 + 2k - (2a+1)^(2a-1) = 0.
    D= 4 +4[(2a+1)^a]^2 = 4([(2a+1)^a]^2+1)
    As k is whole we know that the discriminant must be exact square => [(2a+1)^a]^2+1 is exact square, which contradicts as explained with m - even

    • @abdullahyousef3596
      @abdullahyousef3596 3 роки тому

      Man, you really wrote or that on a computer keyboard, I salute you.

    • @Andreyy98
      @Andreyy98 3 роки тому +1

      @@abdullahyousef3596 Ah as a software engineer I'm used to it... :D

  • @alainrogez8485
    @alainrogez8485 3 роки тому

    I know it was impossible because of the Catalan conjecture (which was proven by the way).

  • @tokajileo5928
    @tokajileo5928 3 роки тому

    I wonder if the 2 in n^2 is any number x (n^x) where x is natural, at what values of x is there a solution?

  • @idontknowwhatido3972
    @idontknowwhatido3972 3 роки тому

    I love your videos.
    Please I have a small suggestion :) You always says you stop. After most videos I would love if you would more elaborate about the problem. Talk about it in more depth. You stop, but I remain to wonder in emptyness, wanting to know more (in philosophical, mathematical, aplication or whatever sense). It would make conclusion of your great videos much more smooth then overly rapid.
    I understand you wont do it always, but please maybe sometimes you would more talk about it and elaborate your thoughts :)

  • @syedmdabid7191
    @syedmdabid7191 3 роки тому

    Sir ! What's the 20th term in the expansion of the multinomial (2x--3y+4z--5z) ^20 in the dictionary order ?

  • @RexxSchneider
    @RexxSchneider 3 роки тому

    I think you may have to also explicitly deal with or explicitly reject the case where m=0, since the proof that m can't be even doesn't work when m=0. It's always a nuisance that "natural numbers" are variously defined as non-negative integers or as positive integers.

    • @wyboo2019
      @wyboo2019 3 роки тому

      m=0 gives you 0^0 on the LHS which is undefined, so m=0 is not a solution

    • @RexxSchneider
      @RexxSchneider 3 роки тому

      @@wyboo2019 First of all, having an undefined value does not disqualify a potential solution, the best you can say is that it might or might not be a solution, but you don't know.
      It is also notable that the limit as x tends toward 0+ of x^x is demonstrably 1, although that value for m^m would not provide a solution, since 1+1 is not a square.
      Secondly if Michael Penn wishes to exclude m=0 as a solution, he needs to explicitly say so, because as I said above, I don't believe the "even case" logic applies when m=0.

    • @wyboo2019
      @wyboo2019 3 роки тому +1

      @@RexxSchneider the fact that 0^0 is undefined means the LHS cannot be evaluated; if we don't have a working definition for a term, we can't evaluate expressions involving that term, which disqualifies it as a solution
      the limit of x^x as x -> 0 is 1, but the limit of 0^x as x -> 0+ is 0, so this doesn't mean 0^0 = 1; the limit of sin(x)/x as x -> 0 = 1, but that doesn't mean division by zero is now defined and sin(0)/0 = 1
      he shouldn't have to disqualify m=0 is a solution; one, because not everyone agrees 0 is a natural number, and as said 0^0 is generally agreed to be an undefined value, especially in algebra

    • @trelligan42
      @trelligan42 3 роки тому +1

      @@wyboo2019 Michael has consistently used the non-zero definition of natural number. Though he is not consistent in always stating this...

    • @RexxSchneider
      @RexxSchneider 3 роки тому

      @@wyboo2019 The point is that because 0^0 is undefined means that we can't evaluate expressions including it. If we were trying to use a value for it in a proof, we couldn't. But the point is that we can't use a value for it to _disprove_ a proposition. You can't say "0^0 is not a square ... etc." precisely because it's undefined. You have to explicitly exclude m=0, not rely a disproof that doesn't work.
      There is a real difference between sin(x)/x when x=0 and x^x when x=0. The former has the value of 1 by any definition of sin, and the function sin(x)/x is continuous and differentiable everywhere. Whereas we agree that 0^0 is undefined, and x^x is discontinuous for non-positive x.
      He should have to disqualify m=0 as a solution because: (1) not everyone agrees 0 is _not_ a natural number (and that includes ISO 80000-2); and (2) you can use a property of an undefined value to _disprove_ a proposition. I hope that's clearer for you.

  • @sphenisc3
    @sphenisc3 3 роки тому

    Can someone explain the n-1=a^m step because gcd=1. Thanks

  • @titassamanta6885
    @titassamanta6885 3 роки тому +1

    I did like this: let n-1=m^k. Therefore n+1=m^(m-k) => m^(m-k) - m^k=2. Now, we can show that for any odd or even m minimum non-zero difference between m^(m-k) and m^k is greater than 2
    i.e m^(m-k)-m^k >2. So, no such natural numbers m and n exist.

  • @BroseMusic
    @BroseMusic 3 роки тому +2

    If you subtract one from both sides and factor n^2 - 1, you get m^m = (n+1)(n-1). Could we then make the argument that because m^m only has factors of m, (n+1) and (n-1) must both simultaneously be powers of m which is impossible therefore there are no solutions?

    • @RexxSchneider
      @RexxSchneider 3 роки тому +5

      I don't think so. If m is prime, your logic would hold, but 6^6 has a lot of factors that aren't 6. It's also not impossible for (n+1) and (n-1) to be simultaneously powers of m. Take n=2 and m=3 and you'll see that both 3 and 1 are powers of 3. Of course it doesn't satisfy the m^m requirement, but you need to be precise when making your arguments.

    • @BroseMusic
      @BroseMusic 3 роки тому

      Ah, thanks for catching the error in my logic! I'm about to begin as a math major and I'm trying my best to build some semblance of intuition.

    • @omp199
      @omp199 3 роки тому +1

      @@BroseMusic Precision in thinking needs to be your starting point.

    • @BroseMusic
      @BroseMusic 3 роки тому +1

      I'd like to think that's what I'm trying to develop. I'm working out the kinks in my logic by trying to work through these problems and the feedback I get from others steers me in the right direction.

    • @omp199
      @omp199 3 роки тому +1

      @@BroseMusic That's great. :)
      One thing you can try is that when you think you have spotted a pattern or a rule - such as "all factors of m^m are powers of m" - you can get out a piece of paper - perhaps the back of an envelope, or something - and make a chart.
      In this case, your first column could contain m, the second m^m, and the third the list of factors of m^m.
      So your chart would end up looking something like this:
      m|m^m|factors
      1|1|1
      2|4|1, 2, 4
      3|27|1, 3, 9, 27
      4|256|1, 2, 4, 8, 16, 32, 64, 128, 256
      5|3125|1, 5, 25, 125, 625, 3125
      6|46656|1, 2, 3, 4, ...
      ... and then you wouldn't have to continue any further, because you'd have spotted that 6^6 has factors that aren't 6 or powers thereof.
      Actually, you could have stopped as soon as you saw that 2 was a factor of 4^4.
      Constantly looking for patterns and then testing to see if those patterns continue to hold is one way to build your mathematical intuition. You'll gradually get better and better at being able to intuit what sorts of patterns look like promising lines of enquiry and what sorts don't.

  • @johnxina650
    @johnxina650 3 роки тому

    why would you destroy our hopes...

  • @fansuli427
    @fansuli427 3 роки тому

    there is a stronger theorem. for any n≥3,x²-yⁿ=1 has no positive integer solution.

    • @eduardomalacarne9024
      @eduardomalacarne9024 3 роки тому

      This have a one unic solution

    • @omp199
      @omp199 3 роки тому

      That's not a stronger theorem. It does not involve a natural number raised to the power of itself.

    • @fansuli427
      @fansuli427 3 роки тому

      @@omp199 what?

    • @omp199
      @omp199 3 роки тому

      @@fansuli427 The equation in the video involves a natural number (denoted m) raised to the power of itself.
      Your equation doesn't.
      Therefore, your statement about your equation tells us nothing about the equation in the video.

    • @yurkoflisk
      @yurkoflisk 3 роки тому

      @@omp199 No, it immediately tells there are no solutions for m^m + 1 = x^2 with m >= 3, because if there was, y = n = m >= 3 with appropriate x would clearly satisfy the equation x^2 - y^n = 1

  • @franzliszt4545
    @franzliszt4545 3 роки тому

    Controversial answer: m=0, n=1

  • @adityaekbote8498
    @adityaekbote8498 3 роки тому +1

    I am the guy who does odd case first

  • @manialikhani4758
    @manialikhani4758 3 роки тому

    Hello mike. Can you please make video editorials on IMO problems?

  • @myguy1374
    @myguy1374 3 роки тому +3

    Solutions are only satisfying if you could answer it without help

  • @fatihsinanesen
    @fatihsinanesen 3 роки тому

    9 minutes to show "no solution" 😫

    • @trelligan42
      @trelligan42 3 роки тому +3

      9 minutes to *_prove_* no solution! 😜

    • @fatihsinanesen
      @fatihsinanesen 3 роки тому

      @@trelligan42 just kidding, not so serious 😂

  • @uggupuggu
    @uggupuggu 3 роки тому

    2^2 +1 = sqrt of 5^2

    • @ieaturanium574
      @ieaturanium574 3 роки тому

      you mean 2^2+1 = 5?

    • @uggupuggu
      @uggupuggu 3 роки тому

      @@ieaturanium574 yes that’s what I said
      (Sqrt of 5)^2 = 5

    • @ieaturanium574
      @ieaturanium574 3 роки тому

      but it could also be:
      2^2+1 = f⁻¹(f(5))
      where f(x) is any invertible function

    • @uggupuggu
      @uggupuggu 3 роки тому

      @@ieaturanium574 what’s that

    • @ieaturanium574
      @ieaturanium574 3 роки тому

      @@uggupuggu thats a more general description of what you wrote. basically you did an operation (square root) and then the inverse operation to that (power of 2) i just noted you might as well do any operation and undo it. the number is still 5 and 5 is still not a perfect square

  • @chrissy93x
    @chrissy93x 3 роки тому

    so much problem solving.. but he cant solve his balding lol jokes but nice videos

  • @charly6052
    @charly6052 3 роки тому +3

    First ²

    • @cadaver123
      @cadaver123 3 роки тому +1

      Second ² 😛

    • @charly6052
      @charly6052 3 роки тому +2

      @@cadaver123 ln (first ) 😼

    • @cadaver123
      @cadaver123 3 роки тому

      @@charly6052 Second! (factorial😁)

  • @Monolith-yb6yl
    @Monolith-yb6yl 3 роки тому

    Its not satisfynibg becase i don't understand how to apply this type of solution to other problems and i cannot learn the ideas why they applied here. Looks like an accident

    • @mudkip_btw
      @mudkip_btw 3 роки тому

      It's satisfying cuz its a clever solution using pretty simple ideas

    • @Monolith-yb6yl
      @Monolith-yb6yl 3 роки тому

      @@mudkip_btw may be ideas are not so hard, but its hard to understand how this objects, equation works inside after this methods

    • @srijanbhowmick9570
      @srijanbhowmick9570 3 роки тому

      Lmao he referred to ur comment after he changed the title lol