If we had that m/n isn't integer, we would have that m is less than 2n. This means that the exponent of n is less than the exponent of m, which is a contradiction since m>n.
(m/n)=n^[(m/n)-2] 1. If (m/n)2, then the right hand side is a positive rational power of an integer. Positive rational powers of integers are either irrational or integers, so the right hand side is either an integer or irrational. However, the left hand side is rational. The only way they can be equal is if they are equal to an integer.
(m/n)=n^[(m/n)-2] 1. If (m/n)2, then the right hand side is a positive rational power of an integer. Positive rational powers of integers are either irrational or integers, so the right hand side is either an integer or irrational. However, the left hand side is rational. The only way they can be equal is if they are equal to an integer.
@@Keithfert490 You are right with "either irrational or integer", but that's a whole theorem. At least, he should say, it's "trivial". But he didn't mention this fact at all!
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
I came here to say this. I did something slightly different, I solved for n = (k+1)^(1/(k-1)) and observed that the function on the RHS is decreasing so there are only a small number of values of k to check. I then checked values of k rather than n, but the result was the same.
I disagree with your assumption of use of the theorem. You are quickly implying that m/n is an integer without plea or sake of an easy proof. But still on still your solution is great!
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.
@@forcelifeforce That is the only value that it can consistently have, and in set theory and integer exponentiation--which is what matters in this context--it _is_ universally agreed to be 1. No other value matters.
@@TomFarrell-p9z I always forget that different groups define it differently. The way I was taught, 0 is a natural number. It seems Professor Penn falls in the camp that excludes 0.
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 :)
This is brilliant ❤️ You explained this like a computer scientist. The prove by contradiction had an underlying assumption to prove against and had a pair of sets of complementary arguments that where contradictory. So the assumption had to be overturned
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.
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^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.
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.
(m/n)=n^[(m/n)-2] 1. If (m/n)2, then the right hand side is a positive rational power of an integer. Positive rational powers of integers are either irrational or integers, so the right hand side is either an integer or irrational. However, the left hand side is rational. The only way they can be equal is if they are equal to an integer.
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
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 😂
Lowercase m,n is an unfortune choice of letters, especially with your handwriting, Michael. The sound is also similar. Perhaps it is just me, English is not my native language.
It’s just the standard convention where when you are working with Natural numbers your arbitrary variables are typically n and m. (And then if you are summing or multiplying over a range of Naturals then the variable used in the sum/product is k or j ranging from, say, 1 to n.). So while m and n aren’t necessarily actually the best choice of letters in terms of clarity depending on handwriting and sounding similar when spoken, it’s just kind of tradition that people seem to use them. 🤷♂️
If you're a "0^0=1" chad it works just fine apart from the arbitrary "solutions must be natural" restriction. If your a "0^0 is undefined" virgin then you can't have that solution.
Wdym part from? 1^2 = 1 Also not exactly a coincidence. We’re looking at an equation where natural number variables satisfy an exponential equation. While it can’t be guaranteed by this reasoning, it shouldn’t be surprising to find that one solution is a natural exponent of the other. Good on you for asking questions, though! That’s always important in mathematics.
5:46 why does x have to be integer?
I also didn't get it. Why not check 37/13?
If we had that m/n isn't integer, we would have that m is less than 2n. This means that the exponent of n is less than the exponent of m, which is a contradiction since m>n.
(m/n)=n^[(m/n)-2]
1. If (m/n)2, then the right hand side is a positive rational power of an integer. Positive rational powers of integers are either irrational or integers, so the right hand side is either an integer or irrational. However, the left hand side is rational. The only way they can be equal is if they are equal to an integer.
@@alexeykamenev1348 How can any power of 37 and 13 be equal?
@nicolacircella5259 If m/n=37/13 then m>2n so I don't know what your point is supposed to be
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?
(m/n)=n^[(m/n)-2]
1. If (m/n)2, then the right hand side is a positive rational power of an integer. Positive rational powers of integers are either irrational or integers, so the right hand side is either an integer or irrational. However, the left hand side is rational. The only way they can be equal is if they are equal to an integer.
@@Keithfert490 Thank you for the careful explanation.
@@Keithfert490 You are right with "either irrational or integer", but that's a whole theorem. At least, he should say, it's "trivial". But he didn't mention this fact at all!
@rainerzufall42 agreed that he should've addressed it for sure!
My question exactly
8:50 Not a word about why x can't be a non-integer between 2 and 4 ?
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
I came here to say this. I did something slightly different, I solved for n = (k+1)^(1/(k-1)) and observed that the function on the RHS is decreasing so there are only a small number of values of k to check. I then checked values of k rather than n, but the result was the same.
Doesn't your approach assume n is prime? How can that be preemptively true?
I disagree with your assumption of use of the theorem. You are quickly implying that m/n is an integer without plea or sake of an easy proof.
But still on still your solution is great!
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.
Seems natural to me, though it appears many definitions appear not to equate natural numbers with whole numbers.
0^0 is not agreed to equaling 1.
@@forcelifeforce That is the only value that it can consistently have, and in set theory and integer exponentiation--which is what matters in this context--it _is_ universally agreed to be 1. No other value matters.
@@TomFarrell-p9z I always forget that different groups define it differently. The way I was taught, 0 is a natural number. It seems Professor Penn falls in the camp that excludes 0.
@@ZekeRaidenyeah it’s often an American vs European split.
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 :)
This is brilliant ❤️
You explained this like a computer scientist.
The prove by contradiction had an underlying assumption to prove against and had a pair of sets of complementary arguments that where contradictory.
So the assumption had to be overturned
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.
m/n being integer has to be proven, or at least being talked about...
@rainerzufall42 No, but it is pretty trivial. Threw me for a sec though too when he said "let's use induction"
@ He can do this kind of induction, if and when he proves, that m/n is integer (natural)! Not earlier!
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^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.
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.
12:43
Michael Penn but every time he says "M" or "N" he gets faster...
Pretty clean
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.
m=1, n=2 results in n^(m-n)=1/2 not being natural. It is not true to say n, m are integers implies n^(m-n) is an integer.
There was already an initial condition that m, n are elements of natural number, so we do not have to complicate things.
Why must $n|m$? Would it be possible for $\frac{m}{n} \in \mathbb{Q}$?
(m/n)=n^[(m/n)-2]
1. If (m/n)2, then the right hand side is a positive rational power of an integer. Positive rational powers of integers are either irrational or integers, so the right hand side is either an integer or irrational. However, the left hand side is rational. The only way they can be equal is if they are equal to an integer.
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
Can't use induction since x is not an integer. Have to use analysis with f(x)=2^x-4x.
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?
Wow I'm rusty, how are we getting the second implication where the difference of m and n is at least 0?
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 😂
Edit: m,n,k are positive integers
x is rational; not necessarily natural. It seems to me that all rational numbers between 2 and 5 are possible...
Lowercase m,n is an unfortune choice of letters, especially with your handwriting, Michael. The sound is also similar. Perhaps it is just me, English is not my native language.
It’s just the standard convention where when you are working with Natural numbers your arbitrary variables are typically n and m. (And then if you are summing or multiplying over a range of Naturals then the variable used in the sum/product is k or j ranging from, say, 1 to n.).
So while m and n aren’t necessarily actually the best choice of letters in terms of clarity depending on handwriting and sounding similar when spoken, it’s just kind of tradition that people seem to use them. 🤷♂️
Assuming 0 being excluded from N does (0,0) work if it had being allowed?
0^0=0^(0-0) iirc
If you're a "0^0=1" chad it works just fine apart from the arbitrary "solutions must be natural" restriction. If your a "0^0 is undefined" virgin then you can't have that solution.
0^0 is undefined
Outstanding sir! Very thanks for presentation of this nice maths
Is it a coincidence that - part from (1,1) - m is always n^2?
Part from (1,1) and (8,2), yes
Wdym part from? 1^2 = 1
Also not exactly a coincidence.
We’re looking at an equation where natural number variables satisfy an exponential equation. While it can’t be guaranteed by this reasoning, it shouldn’t be surprising to find that one solution is a natural exponent of the other.
Good on you for asking questions, though! That’s always important in mathematics.
You miss a case for m=n. m=n=0 is also a solution if you consider that 0^0 = 1
0 is not considered a natural number
M and N have to be greater than 0. It’s a condition of the problem
For me, N = [0;inf[. N*=]0;inf[
Except that 0^0 is undefined.
@@MrWarlls"Natural numbers" generally means using the peano axioms, which clearly state that 1 is the smallest natural number.
Take m=n=1
1¹ = 1⁰ = 1 😂
How x can be a fraction, its definetely an integer. If its not m and n's powers can't be equal
Bro reply pls😢
1st comment 😊