A viewer suggested dream.

[@leickrobinson5186]

An easier way to solve: dividing m+n=mn by either m or n, we see m|n and n|m, which means that m=n and so 2m=m^2 and thus m=n=2.

[@NoisqueVoaProduction]

Why can you divide by m or n and still be sure everything will be a natural?

[@leickrobinson5186]

@@NoisqueVoaProduction We can rearrange the equation to get n=mn-m. If we divide by m, we see n/m = n-1. Since the rhs of the equation is an integer, the lhs must be as well, and so m|n (m divides n). Likewise, by symmetry, we get n|m as well.

[@NoisqueVoaProduction]

@@leickrobinson5186 oh, yeah. That's actually really simple. 1\m is a sum of integers, therefore it is a integer. Cool

[@secretperson8681]

Thats so cool bud. thanks

[@nickyhekster2974]

Exactly what I thought!

[@bprpfast]

If we allow complex numbers, then we can have (-1)*i=i^(-1) 😆

[@MrSlimeOfSlime]

wait, isnt that a black pen?

[@terrybickerton1455]

Can someone explain this to me? I swear (-1)^i= e^(-pi) and i^(-1)= -i e^(-pi) doesn't equal -i

[@dqrk0]

@@terrybickerton1455 its -1 TIMES i

[@terrybickerton1455]

@@dqrk0 ohh lol thanks. im clearly blind

[@3snoW_]

6:20 - "if m = 1 or 2" OR "n = 1", that's what you should have said. You need at least one of the 2 conditions to be true, not both at the same time. You pointed out that when m > 3 (first condition is false) and n = 1 (second condition is true) then m*1 = m^1 = m, so (m,1) is a solution. There is also the case where m=2 (first condition is true) and n=2 (second condition is false) and 2*2 = 2^2 = 4, so (2,2) is also a solution.

[@md2perpe]

Another way of solving m+n = mn is to rewrite it as (m-1)(n-1) = 1 which has integer solutions m-1 = n-1 = 1, i.e. m = n = 2.

[@udic01]

Me too...

[@fatemehparvizi6981]

If we put m and n zero which is an integer it works. It means that it is true if we put x zero and a and b any integers except zero. And the other solution is a=b=1 , x=2 For any integers.

[@md2perpe]

@@jkid1134 m-1 = n-1 = 0 doesn't imply m=n=0. It's however true that m=n=0 is a solution to m+n = mn. That corresponds to x=0 and a,b free (excluding a=b=0, since 0^0 isn't welldefined).

[@jkid1134]

@@md2perpe fixed my error, thanks. Too early to be doing this, haha

[@vinc17fr]

Yes, completing the product is what Michael often does, even when not necessary. It is surprising that he didn't do that here.

[@knvcsg1839]

That intuition of trying for m>=3 is really something that we need to get here. The remaining is so easy.

[@OOobstkuchenOO]

I think the easiest way to solve this is to notice that the problem can be rewritten as (x^a - 1)(x^b - 1) = 1 and noting that, everything being integers, this immediately implies either x^a=x^b=2 or x^a=x^b=0 since the only factorizations of 1 are 1*1 and -1*-1.

[@birbwithscarf]

How can x^a+x^b be rewritten as (x^a-1)(x^b-1)=1?

[@khoiduongminh5111]

@@birbwithscarf x ^ a * x ^ b - x ^ a - x ^ b + 1 = 1

[@birbwithscarf]

@@khoiduongminh5111 sorry, i understand your equation but how do you rewrite it as (x^a - 1)(x^b - 1) = 1?

[@Danicker]

@@birbwithscarf It's not that x^a+x^b can be rewritten, it's the entire equation x^a+x^b=x^(a+b) that can be written. In case that still doesn't make sense here is a step-by-step explanation: x^a + x^b = x^(a+b) x^a + x^b = x^a * x^b 0 = x^a*x^b - x^a - x^b 1 = x^a*x^b - x^a - x^b + 1 1 = (x^a - 1)(x^b - 1) by factorisation Then, as OP pointed out, since x, a and b are natural numbers, x^a - 1 and x^b - 1 must both be integers, which leads to 1, 1 and -1, -1 as the only solutions.

[@birbwithscarf]

@@Danicker Ohh ok, thank you so much!

[@goodplacetostop2973]

4:03 Michael’s homework 6:52 Good Place To Stop

[@qzwxecrvtbynumi256]

3 weeks ago?

[@pppp-ij8tk]

@@qzwxecrvtbynumi256 😳

[@goodplacetostop2973]

@@qzwxecrvtbynumi256 THE TIME WIZARD

[@Oleg-mv6cx]

I hate you

[@timetraveller2818]

@@qzwxecrvtbynumi256 bruh he is just in the 4 th dimension it is a common integral in relativity and uncertainty principle

[@Koisheep]

2:38 Didn't expect to be reminded of Hölder's inequality but here we are

[@h4z4rd28]

XD

[@sanderscheel]

I think I got recomended this video because I searched for Hölder's inequality. '-8-'

[@useruser-ht8tw]

Koishi!

[@user-engahmed]

I hope to put a translation into Arabic, please, please🙏🥺

[@TrinoElrich]

Ah, yes, good ol' conjugate exponents lol

[@StarnGz]

0:07 I come in every video and expect the sentence: 'and by natural number, I mean positive integers' lol

[@garrettthompson3286]

gotta assert that 0 isn't natural every video for dominance

[@egillandersson1780]

It is a very usefull clarification, because if 0 belongs to N or not depends on the authors. None is wrong, because it's just a convention.

[@benjaminvatovez8823]

@@egillandersson1780 Is it the norm in English? In French from Belgium, N always includes 0 as positive integers do (we don't say non-negative integers in French. If we want to exclude 0, we say "strictly positive".)

[@egillandersson1780]

@@benjaminvatovez8823 I don't know : I'm a french-speaking Belgian. I learned the same as you.

[@notar2123]

@@benjaminvatovez8823 In Bosnia (and I can vouch it's the same in all of former Yugoslavia) I've been taught that N does not include 0.

[@nipunnohria186]

Another way for second: Case1- n=1 m=anything Case 2 m,n !=2 Suppose p is a prime divisor of n and its power is k Power of p in m^n-1=k So k=>n-1 Since p^k divides n P^n-1=

[@goodplacetostop2973]

HOMEWORK : Find the set of all possible values that can be attained by the expression (ab + b²)/(a² + b²), where a and b are positive real numbers. Express your answer in interval notation. SOURCE : 2012 HMMT - Guts Round

[@debayuchakraborti1963]

that looks like that imo p6

[@anthonypham7563]

Let a = b*x, where x is a positive real number. Then the expression becomes E = (b^2*x + b^2)/(b^2*x^2 + b^2) = (x + 1)/(x^2 + 1). We notice that E is always positive, and E becomes arbitrarily close to 0 for large x, so 0 is the lower open bound. The function is analytical over the positive numbers, so we can simply take the derivative of E wrt x and set it to 0. Thus, E' = 0 = (x^2 + 1 - 2x^2 - 2x)/(x^2 + 1)^2 => x^2 + 2x - 1 = 0 => x = -1 +/- sqrt(2). Note that we are not concerned with x = -1 - sqrt(2) since it is a negative solution. Finally, we evaluate E at the endpoints and the extreme points: E(0) = 1, E(-1 + sqrt(2)) = sqrt(2)/(4 - 2sqrt(2)) = (sqrt(8) + 2)/4 = (sqrt(2) + 1)/2 > 1. E is continuous for x > 0, so the set of all possible values of E is (0, (sqrt(2) + 1)/2].

[@goodplacetostop2973]

SOLUTION *(0, (1+√2)/2]* Suppose that k = (ab + b²)/(a² + b²) for some positive real a, b. We claim thatklies in (0, (1+√2)/2]. Let x = a/b. We have that (ab + b²)/(a² + b²) = (a/b + 1)/((a/b)² + 1) = (x + 1)/(x² + 1). Thus, x + 1 = k(x²+ 1), so the quadratic kt² − t + k − 1 = 0 has a positive real root. Thus, its discriminant must be nonnegative, so 1² ≥ 4(k−1)(k) ⇒ (2k−1)² ≤ 2, which implies (1−√2)/2 ≤ k ≤ (1+√2)/2. Since x > 0, we also have k > 0, so we know that k must lie in (0, (1+√2)/2]. Now, take any k in the interval (0, (1+√2)/2]. We thus know that 1² ≥ 4k(k−1), so the quadratic kt² − t + k − 1 = 0 has a positive solution, (1+√(1−4k(k−1)))/2k. Call this solution x. Then k(x²+ 1) = x + 1, so (x + 1)/(x² + 1) = k. If we set a = x and b = 1, we get that (ab + b²)/(a² + b²) = k. Thus, the set of all attainable values of (ab + b²)/(a² + b²) is the interval (0, (1+√2)/2].

[@goodplacetostop2973]

@@aashsyed1277 Yeah

[@pullamallama2669]

Wait doesn’t using 2x2=2^2 work as well? That is, with mn=m^n

[@karankamat2087]

the problem with the second problem is we've proved we can't have m>=3 AND n>=2 I.e we can't have both together but that does not mean m

[@danielbranscombe6662]

Yes, I believe he forgot that the restriction of n>=2 only applies when m>=3, so for m=1 and m=2 he has to consider any possible n to be complete. So to complete it. m=1 gives 1*n=1^n, n=1 which is already part of the solution given m=2 gives 2n=2^n n=2 giving the extra solution of m=n=2

[@DawidEstishort]

@@karankamat2087 Yeah, it's a classic mistake of assuming \(A+B)\A+\B

[@srijanbhowmick9570]

@@DawidEstishort Can you explain that notation ?

[@DawidEstishort]

​@@srijanbhowmick9570 \ "negation" + "and" (usually it is written as ^ but I forgot I could write that) "if and only if" "It is not true that both A and B are true, if and only if A is not true and B is not true." That is false because it should be "It is not true that both A and B are true, if and only if A is not true or B is not true."

[@SV-yo6nq]

this should be called 'the middle schooler's dream'

[@scribblespoon1919]

fever dream

[@iabervon]

If a, b, c, and n are positive integers, n^a + n^b can't equal n^c if n > 2. If (wlog) a≥b, n^a

[@davidseed2939]

error at 6:26 if m>=3 and n>=2 has been eliminated. then the remaining possible solutions are m=1,2 OR n=1 in fact by checking we find the solutions are n=1 for any m or m=2= n^(1/(n-1)) and the only integer value of n for that expression is 2

[@ichigonixsun]

1:31 "So since they're natural numbers, none of them are zero." TRIGGERED

[@icfj77]

For the equation m*n=n+m, if we consider gauss integers, i.e. , complex numbers like z=a+bi, with a and b integer, we have not only that m; n=1; 1 and m; n=2; 2, but also that m=1+i and n=1-i (or m=1-i and n=1+i as well).

[@jerrysstories711]

The title made me thing someone dreamed about this problem. Math dreams always feel so weird. I dreamed once that pi was suddenly equal to 4,. Everything was all wonky, and I was having a hard time getting to work.

[@Umbra451]

when pi=4, circles become squares

[@jerrysstories711]

@@Umbra451 Nah, they were still circles, the underlying geometric of the universe was just difference suddenly, such that the diameter and circumference had a different ratio. It was really weird.

[@frodoswift70]

I dreamed a derivation of the Pythagorean theorem, which I had never previously derived

[@jbtechcon7434]

@@frodoswift70 when you woke up, did it turn out to be a valid derivation?

[@EebstertheGreat]

@@Umbra451 Actually, if the value of pi is not 3.14159..., rectangles can't exist at all.

[@udic01]

The method of (m-1)(n-1)=1 also gives answer to the homework question (m,n in Z). Because either m-1=n-1=1(which we already seen the solution in the video) or m-1=n-1=-1 and the second option gives us x=0 and a and b are free.

[@demenion3521]

exactly my thoughts!

[@joaquingutierrez3072]

1:24. My approach: m + n = mn m = mn - n m = (m - 1)n m/(m - 1) = n So (m - 1) divides both m and m - 1 So (m - 1) divides m - (m - 1) = 1 So, as (m - 1) non-negative we must have m - 1 = 1 m = 2 Substituing we get 2 + n = 2n 2 = n n = 2 This approach works when n and m are integers too. But in that case we need to consider the case m - 1 = -1 NOTE: Sorry, I did not consider m - 1 = 0. In that case n + 1 = n 1 = 0, a contradiction

[@chungyanwong7293]

I have another similar dream too but I don't know how to solve this: When does (x+y)^n = x^n +y^n for all x,y,n∈ℝ. I know equation satisfies when either -{n=0 | y= 2-x} or- { n=1 | x, y∈ℝ} or {x=0 | y, n∈ℝ} or {y=0 | x, n∈ℝ} but how to write a proof for n>1? P.S. I also know that for n

[@romilgoel4191]

I am not sure but.. expanding via binomial theorem gonna help ?

[@chungyanwong7293]

@@romilgoel4191 Thanks for a reply, that's where I was stuck. After Bionomial Theorem and testing different n values, I got into polynomials of all degrees beyond the fifth one and I don't know if I should use Galois Theory or something for solving polynomials beyond fifth degree

[@Haalita21]

I’m not sure if your first solution is valid as LHS = 1 but RHS = 2. Otherwise, other than the trivial solution of n=1, you have an infinite solutions of n>1 when xy=0, i.e. either or both of x and y = 0 and that isn’t too difficult to see from the binomial expansion of (x+y)^n.

[@chungyanwong7293]

@@Haalita21 Oh wait, you're right about my first solution!! Because (x +y)^0 = 1 and x^0 +y^0 = 2. My Math error. Thank you for pointing out :)

[@michaelcampbell6922]

{n is odd | x=-y, y∈ℝ} satisfies the equation

[@cicik57]

Using diophant equasion factorisation trick: x^a + x^b - x^a x^b -1 = -1 x^a(1-x^b) - 1 ( 1 - x^b ) = - 1 (x^a - 1)(x^b - 1 ) = 1 in natural numbers: x^a =2 and x^b = 2 or x^a =0 and x^b = 0 (what is impossible) so x = 2, a = 1, b = 1 is only solution ??

[@江勖豪]

For the second question, I have some additional thoughts. For m>=3 and n>=2, m^(n-1)>= 3^(n-1) > n. So, while m=1 ,m=2 or n=1, m*n=m^n is solved. While n=1, m is free. While m=1, n=1. While m=2, n=1 or 2.

[@michaelaristidou2605]

0 is a natural number, so simply adjust the question to be true for all positive naturals. Also m=2, n=2, is also a solution for the 2nd version of the problem.

[@phiefer3]

0 is not a natural number actually, the first set of numbers that 0 belongs to is whole numbers.

[@mryip06]

natural numbers may include 0 depending places.

[@trishanmondal7813]

I suggest let there exists a prime p>2 that devide x The largest exponent of P thta should devide LHS is a and it should be (a+b) for RHS . It is possible iff b=0 but that doesn't lead to the set of natural number . . Only case : only p =2 devide x X≈2^r From here rest is easy

[@titassamanta6885]

m|(m+n)=> m|n (m and n are not 0) Smilarly n|m. Hence m=n. This gives 2m = m^2 => m is 2 or 0. Though we assumed that m and n are not zero yet m=n=0 is a solution (if problem were to be solved for all integer values of m and n)

[@tomkerruish2982]

An alternate route (equivalent, but not using fractions) is noting that m + n = mn implies (1 - m)(1 - n) = 1; consequently, (1 - m) and (1 - n) are either both +1 or -1. The second choice gives us m = n = 2; the first choice yields m = n = 0, which is valid for x = 0 and a, b any positive integers, which are precisely the additional solutions resulting from using Z rather than N.

[@Sora1the]

For the second question, wouldn't m=2 and n= 2 works as well? 2*2 = 2^2 = 4

[@goduck-x6u]

1:34 Easier way: m + n = mn ==> (m-1)(n-1) = 1, so m = n = 2

[@leonardovalente9772]

i am seriously upset he didn't talk about (2,2) as a solution to the second problem

[@alfredtrietsch215]

Yes, at 5:52 narrowing down to m>=3 is not motivated

[@Fine_Mouche]

only (2;2) in N, but for m in N and m in R where are infinites countable other solutions for m = 1/3 we have n = 1/(3/2) , for m = 1/4 with have n = 1/(4/3), for m = 5, we have n = 1/(4/5) and so on. And for both n and m in R, there are infinite uncountable solutions which satisfy 1/m + 1/n = 1, it's just n = 1 / 1 - 1/m

[@lucashoffses9019]

6:30 shouldn’t the condition be m=1,2 *or* n =1, and not and? m=2 and n=2 works as a solution.

[@mrphlip]

My working on this one was similar, but went like: m + n = mn n = m(n-1) m = n/(n-1) m = 1 + 1/(n-1) m-1 = 1/(n-1) LHS is an integer, so RHS is an integer, so n = 0 or 2 (I always forget that this channel typically defines ℕ as starting at 1... but then this also serves as the extension to ℤ) This gives our solutions as m=n=0 and m=n=2. so our solutions are: x^a = x^b = 2 ⇒ x = 2; a = b = 1 x^a = x^b = 0 ⇒ x = 0; a, b ≠ 0

[@benedictrodil4931]

Also, a ≠ -b

[@mrphlip]

@@benedictrodil4931 Good point, I missed that. Actually, thinking about it I think the condition is a, b > 0

[@benedictrodil4931]

@@mrphlip Ah yes they are, I also missed that too

[@balajik8473]

m = n/n-1, n and n-1 are always co primes, so that's the simplest form of rational p/q, for it to be integer, denominator should be 1 -> n-1=1

[@davidgustavsson4000]

mn = m^n log(mn) = n log(m) log(n) = (n-1) log(m) Trivially true for n = 1 If m = n ≠ 1, n = 1+1 = 2 If n>1 (n-1)>log(n), and if m>2 log(m)>1 so lhs

[@mathOgenius]

And I thought I was dreaming

[@anupagarwal4843]

In the follow up question, m and n can both be equal to 2. this would be another solution. I think you have missed this one.

[@liorean]

I've had math textbooks in Swedish and English, by authors from Sweden, UK, US and Australia, and formulae and tables books from three different publishers or universities, and every single one of them distinguishes positive integers, ℤ+, from natural numbers, ℕ, in that the latter includes zero. I just learned that not everybody use these definitions. Why the difference?

[@peterdecupis8296]

in modern Set theory N is the minimal inductive set {0,1,2...}, i.e. the minimal infinite ordinal; thus its starting element is the null set, i.e. zero! This is consistent with the classic Peano's Axiomatic Foundation of N, where the presence of zero is explicitly stated by the first axiom. Z is commonly defined as the Set of translations in N (i.e. is a quotient set of N×N with respect to a peculiar "shift-equivalence" relation); therefore, rigorously speaking, Z+ is not the same set as N, whereas there is an isomorphic bijection between them; in the case of point, the isomorphism maintain ordering and fundamental operations (sum, multiplication, and exponentiation). Unfortunately there is no universality in some conventional symbols: e.g. for some authors Z+ does contain the zero (i.e. the null-translation in N), for others it doesn't

[@andrewparker8636]

Note to self - RTFQ . I thought we were solving for a,b in N and x in R. Which is interesting. And you can classify solutions, but can't compute them closed form.

[@rbnn]

For the first problem, assume wlog b>=a and divide by x^a to get 1+x^(b-a)=x^b. x>1 by inspection, and if b>a then taking both sides mod x yields a contradiction. Hence a=b, and 1+1=x^b, so x^b=2, a=b=1, x=2.

[@calyodelphi124]

mn=m^n is also true if m=n=2 but that is the only solution that exists for the case n>1. For all other solutions n must = 1, and then m can be any natural number.

[@jamirimaj6880]

3:22 is that really a symbol for contradiction?

[@happygimp0]

1:33 What, why? Why shouldn't they be 0? It would work with 0 and 0 is a natural number.

[@zeropotential6830]

Hi dear Micheal Penn I made a problem using my childhood experience when i was palying with my elder brother's calculator. Problem:a^n= x^99 Where a and n are integer. Find the value of a.For value a n should be maximum.

[@mrhatman675]

I did it like this for the second problem m^n-1=n suppose n≠1 this means m^n-1 can be written in the form m×m×m... n-1 times=n which means n|m and so we can write m=n×k which means k×m×m...=1 but since k and m are integers and 1 is a pime number and in this case the only number that devides it is 1 and can be written only in the form of 1×1×1... so it means k=1 and m=1 thus n=1 which can t be true so our assamtion is wrong so n has to be 1

[@srprantor]

Cynthia loves Pokemon and she wants to catch them all. In Victory Road, there are a total of 80 Pokemon. Cynthia wants to catch as many of them as possible. However, she cannot catch any two Pokemon that are enemies with each other. After exploring around for a while, she makes the following two observations: 1.Every Pokemon in Victory Road is enemies with exactly two other Pokemon. 2.Due to her inability to catch Pokemon that are enemies with one another, the maximum number of the Pokemon she can catch is equal to n. What is the sum of all possible values of n?

[@skywing786]

If a pokemon , say mimikyu , is enemies with another pokemon , say pikachu, does that imply that pikachu is enemies with mimikyu too ?

[@Abcd-hs4cv]

If you divide both sides to x^a+b you get 1/x^a + 1/x^b=1 and only 1/2 + 1/2 equals to 1 so x=2 a, b =1 in natural numbers. And in all real numbers solution is a=b=logx 2.

[@Guessyyyy6270]

For decimal numbers etc. We have X = sqrt(2) and a & b = 2

[@l.3ok]

For n = 2 we have that n = m^{n-1} implies that 2 = m and we also have a solution.

[@divisix024]

I would solve m+n=mn in N by factoring it as (m-1)(n-1)=1, so we must have m=n=1. The matter is similar if we’re in Z, just that we may have m=n=-1 as well

[@shiveshpratapsingh3501]

For integer solutions (m,n) we have (0,0) and (2,2) as the only solutions

[@damianbla4469]

For the second problem (m*n = m^n) we have also infinitely many solutions of the form n=1, m=arbitrary constant.

[@irrelevant_noob]

... but that's what was already described in the video, so what do you mean "ALSO"? o.O

[@mrvnoble]

Trivially, in the integers, x=0 gives solutions for all a,b. I haven't done any work beyond that yet.

[@242math]

watching and learning bro, you are a master at this

[@irrelevant_noob]

... a master who can't negate a conjunction? xD

[@tcoren1]

If x os at least two, and wlog a>b (I'll treat a=b and x=1 later) then: x^(a+b)=x^a*x^b>x^a*2^1=x^a+x^a>x^a+x^b Contradiction x=1: 1=1+1 contradiction a=b: (x^a)^2=2(x^a) x^a=2 x=2 a=b=1

[@ishansingh2391]

Try this Find all pairs (a,b) of real numbers such that whenever p is a root of x^2 + ax+b. =0 then p^2 - 2 is also a root of the equation. Source regional mathematics olympiad : India

[@mansoorali7951]

1) Since m+n is a multiple of m (i.e, mn), which is only possible if n is a multiple of m. But m+n is also a multiple of n (i.e, nm), which means that m is a multiple of n. Since m is a multiple of n and n is a multiple of m => n=m. Thus m+n =2m=m^2 => m=n=2 2) mn=m^n => m^(n-1)=n For n=1; LHS=RHS = 1 for every value of m. For n=2; m=2 => the result holds for m=n=2 also For n>3; m=ⁿ⁻¹√n, which is an irrational number for n>3 contradicting the fact that m is a natural number. Hence the relation doesn't hold for n>3. Thus the relation holds; If n=1 (for any value of m) OR If m=n=2.

[@Nikioko]

How do you define natural numbers? Positive integers or non-negative integers?

[@LukePalmer]

2:23 "but that's impossible" is hilarious to me for some reason

[@uttankjha8888]

The dejected/disappointed look is everything

[@gordistador]

It doesnt have to be x=2. It just has to be x^a = 2. X could be any number larger than 1, as long as the exponent is the right value to turn X into 2.

[@JBRewind]

For the mn = m^n, calling it now before I finish watching: m and n must both equal 2 OR m can be anything and n is 1.

[@JBRewind]

Oh, so I found a solution he didn't. How?

[@surem8319]

Alternatively one could rewrite the expression to: m = n/(n-1) and utilise the fact that the only way n-1 can divide n is if it is equal to 1 (since n is different from 0) which gives us n = 2 immediately.

[@tzovgo]

DEFINITION: Let a [n] b = a [n-1] a [n-1] a [n-1] a ... a [n-1] a (with b a's). a [0] b = a+1. (Note that this is only defined for natural numbers.) This makes addition a [1] b, multiplication a [2] b, and exponentiation a [3] b. STATEMENT: For all k>0, i>0, n=2, m=2 solves n [k] m = n [i] m. PROOF: 2 [k] 2 = 2 [k-1] 2 = 2 [k-2] 2. In general, 2 [k] 2 = 2 [k-j] 2 as long as k > j. (At k = j, we have 2 [0] 2 = 3 =/= 2 [1] 2 = 4) Choose j = k - 1. 2 [k] 2 = 2 [1] 2 = 4. This reasoning applies exactly to 2 [i] 2 as well. Thus, 2 [k] 2 = 2 [i] 2 for all k>0, i>0.

[@Utesfan100]

n+m=mn implies (m-1)(n-1)=1. Since both are integers, m-1=n-1=+/-1. Thus m=n=2, or m=n=0, but the second is not in N.

[@federicorossi8587]

Can you please teach how to play message in a bottle?

[@AlphaPizzadog]

X = 2 A = 1 B = 1 Or if a and b don't need to be integers, an example: X = 4, A = 0.5, B = 0.5

[@kaizoisevil]

For the mn=m+n, I did simon’s favorite factoring trick. (m-1)(n-1)=1

[@AnonimityAssured]

I think we can all delight in having spotted the glaring omission of m = n = 2.

[@Wurfenkopf]

This is a strange choice of the dominion, I would be curious to find the solutions outside of the natural numbers. At least, setting a, b as natural but x real

[@mike1024.]

The opposite of m>=3 AND n>=2 is m

[@demenion3521]

the problem gets more interesting if you allow x, a, b to be negative. if you only consider m+n=m*n, then only m=n=0 is a new solution over the integers, but if a or b are negative, that makes x^a or x^b rational and not integers.

[@MrBrain4]

Here's how I solved: Write x^a and x^b in base x. Adding these when a!=b results in a sum in base x such as 10010, which is not a power of x. So for any possible solution, a=b. For x=1, we get 1+1=2, which is not correct. For x>2, we result in a sum in base x such as 2000, which is not a power of x. So for any possible solution, x=2. The only a and b that work are a=b=1, since we get in base 2, 10+10=100. For any higher a and b, x^a + x^b = x^(a+1), which is not equal to x^(a+b).

[@patrickhodson8715]

m = n = 2 is also a solution to the second question

[@theash4361]

An alternative method could be taking a

[@thelivetoad]

What about rational solutions? Diophantos' method enumerates all solutions to m + n = mn in Q.^2

[@kadoozy3752]

Am I the only one who heard a moan at 1:09 ; 3:38? :))) I think math isn't the only thing is going on there :))))

[@daltonrainer6734]

Same bro lmao

[@seroujghazarian6343]

mn=m^n also has (m,n)=(2,2) as a solution

[@titan1235813]

On the second problem, m = n = 2 is also a valid solution.

[@winterdark8657]

I have a question :"if f'(a)=f'(b)f(a)=f(b) with a,b is realnumber" is is right??? Please

[@NoahUbf]

"not because we need to, but because we can" :)

[@NikolayVityazev]

m=2, n=2 also works

[@log2306]

That's literally the only solution

[@teeweezeven]

@@log2306 I think he means for the question mn=m^n

[@NikolayVityazev]

@Teeweezeven, @Neo right, for mn=m^n

[@themanwhocarriesthesun780]

Here's how I solved it. Assume a>b (1) and divide both side by x^a we have 1 + x^(b-a) = x^b (2). Of course, x^(b-a) is a non-integer because of (1). Thus, it does not satisfy (2). This proves (1) false. Similarly, the proof for a=3, x^a >= 3^a > 2 which does not satisfy (3). So, we must have x=2 (4). From (3) and (4), we solve for a=b=1. So, the only solution is x=2, a=b=1.

[@heeraksharma1224]

Can this identity be used: (Secx)^2 + (cosecx)^2 = [(secx)^2]•[(cosecx)^2]

[@michawalus842]

I found a quicker solution: So we know that x^a + x^b = x^a * x^b We move everything to right side and add 1 to both sites 1 = x^a*x^b - x^a - x^b + 1 Notice that this is (x^a - 1)(x^b - 1) = 1 We know that both x^a - 1 and x^b - 1 are non negative integers so x^a - 1 = x^b - 1 = 1 x^a = x^b = 2 Only numbers that satisfy this equation are x = 2, a = 1, b=1.

[@michawalus842]

In this method it's easy to generalize it to all integers, only difference is that x^a - 1 and x^b - 1 can be equal to the - 1 so x^a = x^b = 0. Therefore we get more solutions in form: (x, a, b) = (0, m, n) where m and n are non zero integers.

[@Creativemathlearning]

Is there another solution?

[@juliusgutjahr1406]

6:27 but if you set m=n=2 the equation would still be true

[@cyan1294]

i think for the "homework" we would first start with m

[@Adomas_B]

Top 10 facts teachers don't want you to know about

[@teslachess7319]

Is it possible to make e with complex numbers without using e?

[@蛰扉]

m+n=mn, we can solve this by making this mn-m-n+1= 1, then (m-1)(n-1)=1. 1 can only be factored as 1*1, hence m=n=2

[@rileywaugh1022]

If x>2 can one work mod (x-1) in order to arrive at a contradiction?

[@mrhatman675]

I did it like that m+n=m×n which means m=-n÷(1-n) whih means m=-(n÷(n-1)) suppose n is 2k+1 this means m=-((2k+1)÷2k) but this can t be a an integer and we want m an integer so n=2k which means m=-(2k÷(2k-1)) and we know the only uneven number that devides all numbers and in this case our even number n is 1 so 2k-1=1 which means n=2 and m=2 so x=2 a=b=1

[@dominiquebercot9539]

J’ai essay é autrement Si a =b, on trouve x=2 Si a

[@KarlDeux]

I am afraid that 0 is a natural number. You just have to consider the case so you can skip it afterwards.

[@enigma-zb7km]

0 is a natural number, and that's an other solution if x=0, and a,b different from 0

[@walidch9162]

mn=m^n (2;2) is a solition Because the condition is n≥2 if m≥3 Then if m≤2 that doesnt mean n≤1

[@giuseppelanna]

I can't stop thinking it looks like the last Fermat's therorem

[@ZeroSleap]

Isnt m=2 and n=2 a solution to m*n=m^n 2*2=2^2

[@tonyennis1787]

At about 3:20, why did you decide 1/n >= 2/3 and not 1/n

[@schweinmachtbree1013]

if 1/3 + 1/n >= 1 then 1/n >= 2/3 by subtracting 1/3 from both sides of the inequality :)

[@tonyennis1787]

@@schweinmachtbree1013 Thanks, that part was just a little too fast for me

[@MrRyanroberson1]

you missed a solution at the end: m=n=2 again. 2+2=2*2=2^2=... so mn=m^n

[@Lucifri]

It would work in Z with x=1 and a or b =0 while the other is 1, right?