can you solve this exponential equation?

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[jimskea]

Why does m/n have to belong to the set {2, 3, 4}? Isn't m/n a (positive) rational > 1, not a natural number?

[rainerzufall]

8:50 Not a word about why x can't be a non-integer between 2 and 4 ?

[SuezireKaka]

By the fundamental theorem of arithmetic, we can assume that m=n^k. So we can rewrite the problem n^(kn)=n^(n^k-n). Taking the logarithm of both sides, we get n=1 or kn = n^k-n. From the second equation, we get k+1 = n^(k-1). If n >= 4, then k+1 >= 4^(k-1) but it's impossible when k > 1, and when k = 1, 2 ≠ 1, which means n <=3. Now we can only get n=2, k=3 and n=3, k=2 (notice that we can use the fundamental theorem of arithmetic again). Therefore, all possible pairs of (m, n) are (1, 1), (8, 2), (9, 3).

[salerio]

@9:50 you exclude 0 as being a member of the set natural numbers. Peano's first axiom states that 0 is a natural number

[TheBrutalDoomer]

7:00 well, we cannot use induction for a non-natural variable, though the point is pretty clear. Comparing derivatives could be a simple approach.
m/n having to be integer is also obvious, yet hasn't been stated beforehand.

[Charliethephysicist]

From the fundamental theorem of arithmetic, m^n=r^s implies m=a^{s'} and r=a^{n'} where s'=s/gcd(n, s), n'=n/gcd(n, s) and some natural number a. Substituting these for m and n then equating the exponents, we can solve the problem.

[miraj]

My solution involved more cases:
Case 1: Assume m = n. Then m^n = n^0 = 1 ==> m = n = 1. Furthermore, only assuming m = 1 implies n = 1 and only assuming n = 1 implies m = 1. So for all future cases, we can assume both m and n are at least 2.
Case 2: Assume n > m. Then the RHS is a natural number bigger than 1 raised to a negative exponent i.e. not a natural number. But the LHS is a natural number so no sol.
Case 3: Assume n < m. In other words, we can set m = n + k for some natural number k. So our equation now becomes (n+k)^n = n^k. This further breaks into 2 cases:
Case3a: Assume k <= n. (n+k)^n >= n^n + k^n >= n^k + k^n > n^k so no sol.
Case 3b: Assume k > n. In other words, we can set k = n + j for some natural number j. So our equation now becomes (2n+j)^n = n^(n+j). Dividing both sides by n^n gives us (2 + (j/n))^n = n^j. The LHS can only be a natural number if j is a multiple of n. From there, it is helpful to simply start enumerating multiples of n:
Let j = n ==> 3^n = n^n. Take both sides to the 1/n power ==> n = 3. From there, the original equation becomes 27m^3 = 3^m. It's not too hard to find that m = 9 is a solution. You can use calculus or maybe induction to show that is the only solution.
Let j = 2n ==> 4^n = n^(2n). Take both sides to the 1/n power ==> 4 = n^2 ==> n = 2. From there, the original equation becomes 4m^2 = 2^m. It's not too hard to find that m = 8. Similar to above, there is only 1 solution.
Let j = 3n ==> 5^n = n^(3n). Take both sides to the 1/n power ==> 5 = n^3 ==> no solution. From here, you run into the same problem as j gets ever bigger. Namely that the LHS grows far too slowly to keep up with the RHS and would require n to take on some value between 1 and 2, which isn't a natural number.

[veselindimov]

Here's a very simple explanation of why X=m/n can't be a non-integer.

Consider the equation X=n^(x-2), equation (*). Suppose X is not an integer and suppose X=p/q, where p and q are relatively prime (it could be that (m, n)=(p, q), but not necessarily). There are 2 cases:
1) the q-th root of n is an integer (so n^X is an integer). Then the LHS of equation (*) is non-integral, while the RHS is integral.
2) the q-th root of n is not an integer. It can't be a rational number, because the base n is an integer, therefore it has to be irrational. So then the RHS of (*) is irrational and the LHS is rational.

Contradiction in both cases. Hope this helps :)

[journeymantraveller]

Can't x be 1 < x < 5? Need to show x is an integer.

[byronwatkins]

x is rational; not necessarily natural. It seems to me that all rational numbers between 2 and 5 are possible...

[simonhenry]

At 4.10 the inequality you are writting could be false if m < 2n (the exponent would be negative). You are also assuming x is an integer. this is clear if m>= 2n as x = n^{m/n - 2}, nut not if m <2n.

[applmak]

Why must $n|m$? Would it be possible for $\frac{m}{n} \in \mathbb{Q}$?

[jursamaj]

0:30 The problem being diophantine implies that n^(m-n) is a natural number. m and m^n being natural don't imply that.

[samueldevulder]

8:30 (approx, ) Hold on... Why would m/n (aka x) be an integer ?

[martincohen]

Can't use induction since x is not an integer. Have to use analysis with f(x)=2^x-4x.

[Alan-zftt]

Stunning math work! Only comment I feel urged to make is: love what you are doing here and wonder is (n, m) = (1, 1) also a worthy paradigm shift?

[ZekeRaiden]

Isn't m=n=0 also a valid solution? 0^0 = 0^(0-0) = 1.

Edit: For anyone replying to this with "0^0 is undefined/indefinite/etc.": It is only indefinite/undefined/whatever if you're talking about the limit of two functions, f(x)^g(x), where the limit as x->c is zero for both. Evaluated as a _number, _ 0^0 = 1. It has to. Otherwise, e^x is not equal to its Taylor series expansion at 0.

[zacharysmith]

Wow I'm rusty, how are we getting the second implication where the difference of m and n is at least 0?

[ReCaptchaHeinz]

Hey! Some years ago I stated a somewhat similar problem:

Find all triples m, n, k st mⁿ = (m+k)^(n-k), k≠0

The simplest is 2, 4, 2 but i recall finding an infinite set of solutions.

Is was 2018 and I offered a melon as a prize 😂

[arekkrolak]

I see in your country 0 is not natural ;)