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.
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.
@@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
@@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.)
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 😂
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.
@@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.
@@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.
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.
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).
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.)
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.
@@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.
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.
@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
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.❤️❤️❤️❤️❤️❤️❤️
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
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
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 :)
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 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.
@@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
@@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.
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.
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?
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.
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.
@@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.
@@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.
@@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
@@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
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
I have always felt more confident in proofs finding a horrendous irracional number as a solution than finding "no solution".
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.
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.
true. Thx for the one-liner
Only if you explicitly exclude 0 and 1
@@RexxSchneider and 0 and -1, and -8 and -9, if you are OK with complex numbers
Using the Catalan's conjecture for this problem is like using a nuclear weapon to kill a mosquito. :)
@@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
8:48 Anyone else watching the Olympics opening ceremony?
no...😅😊😊
Me, it’s very exciting, looking at ppl waving flags for an hour
oh, really?...i just got home from work...i'll watch the replayed film of it, tomorrow🥰🥰🥰thank you💜💜💜💜💜
Its nice to see that these actually are solvable...
If we let ourselves go over the integers we get -1^-1 +1 = 0^2.
Natural numbers* bc integers include -1 and 0
@@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.
And there can’t be any others because if m < -1 then m^m is not an integer.
Not integers only Natural numbers ie positive integers.
P
@@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.)
Based on the title and the thumbnail, I guessed there’s no solutions!
hey, I did not expect you to be here for some reason :)
@@abdullahyousef3596 hey, Abdullah, didn't expect to see you here! Hope you are good!
@@d-rex7043 I am good alhamdolilah, thanks for asking.
it’s just as important to prove things are impossible as it is to prove that they are
Hi,
For fun:
0:15 : "or so on and so forth",
4:29 : "ok, let's get rid of this and".
You say that you're not doing modular arithmatic, but isn't even/odd just modulo 2? :P
Nice observation. Here are your 3 points.
Yep I guess so…..
Yep...
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 😂
It'd probably be great for YT comment "engagement" numbers, though. 🙃
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.
@@OscarCunningham 0^0=1. I never understood why until watching a video recently, I think by redpenblackpen.
@@nottaexpert2690 As a limit, it is true. We must be careful to state that x tends towards 0+.
@@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.
The ending of this video was two-tally awesome! Thanks so much for making all of these great videos and sharing.
It is. I dare say, it is even the most satisfying to prove, there are no possible solutions :-)
I think there is one if you define 0^0 as 0 and take n=1
@@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.
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.
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).
Satisfying? Yes! Because that is all that is needed.
This is also just a special case of Mihăilescu's theorem: the only two consecutive perfect powers are 8 and 9 :D
A non-number theory number problem!
Thank you, professor!
6:15 I thought Michael would fall on stage.
The even case was fun but the odd case was hard work...
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.)
Mass to the mass power plus one equals moles squared
A trivial solution is
m=-1 & n=0
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.
@@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.
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.
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
@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
@@jamescollis7650 yep
Hi Dr. Penn!
Let's start a war: m^m=0, n=1
No more Vertex Algebras?
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!
the ending made me cry 😭
it certainly is
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.❤️❤️❤️❤️❤️❤️❤️
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
I think its true
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
Man, you really wrote or that on a computer keyboard, I salute you.
@@abdullahyousef3596 Ah as a software engineer I'm used to it... :D
I know it was impossible because of the Catalan conjecture (which was proven by the way).
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?
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 :)
Sir ! What's the 20th term in the expansion of the multinomial (2x--3y+4z--5z) ^20 in the dictionary order ?
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.
m=0 gives you 0^0 on the LHS which is undefined, so m=0 is not a solution
@@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.
@@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
@@wyboo2019 Michael has consistently used the non-zero definition of natural number. Though he is not consistent in always stating this...
@@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.
Can someone explain the n-1=a^m step because gcd=1. Thanks
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.
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?
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.
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.
@@BroseMusic Precision in thinking needs to be your starting point.
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.
@@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.
why would you destroy our hopes...
there is a stronger theorem. for any n≥3,x²-yⁿ=1 has no positive integer solution.
This have a one unic solution
That's not a stronger theorem. It does not involve a natural number raised to the power of itself.
@@omp199 what?
@@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.
@@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
Controversial answer: m=0, n=1
I am the guy who does odd case first
Hello mike. Can you please make video editorials on IMO problems?
Solutions are only satisfying if you could answer it without help
9 minutes to show "no solution" 😫
9 minutes to *_prove_* no solution! 😜
@@trelligan42 just kidding, not so serious 😂
2^2 +1 = sqrt of 5^2
you mean 2^2+1 = 5?
@@ieaturanium574 yes that’s what I said
(Sqrt of 5)^2 = 5
but it could also be:
2^2+1 = f⁻¹(f(5))
where f(x) is any invertible function
@@ieaturanium574 what’s that
@@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
so much problem solving.. but he cant solve his balding lol jokes but nice videos
First ²
Second ² 😛
@@cadaver123 ln (first ) 😼
@@charly6052 Second! (factorial😁)
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
It's satisfying cuz its a clever solution using pretty simple ideas
@@mudkip_btw may be ideas are not so hard, but its hard to understand how this objects, equation works inside after this methods
Lmao he referred to ur comment after he changed the title lol