This is a problem from the JMBO 1998 competition. It was surprisingly difficult even though I had a n idea of what to do. The main strategy was to create as many constraints to narrow the options.
I love your ending message: "those who stop learning stop living". I was always into math through High School, but in college I went into another field but wound up dropping out, and I wound up being depressed with where I ended up in life, but I always came back to math by RU-vid, such as your channel and others', and I really feel like keeping learning has improved my mental wellbeing.
i don't rly understand why (x-y)/y shouldn't be a fraction, for example you can raise 27 to the power 4/3 and still get an integer (81 in this case) edit: okay i've made the fix 1) you can show that x and y can be expressed as m^p and m^q, where m, p and q are integers and p > q 2) plug it into the eq and get m^(py) = m^(q(x-y) (p+q)*y = q*x (p+q) * m^q = q * m^p m^(p-q) = (p+q) / q = 1 + p/q 3) from here we get that 1 + p/q is an integer(cuz m^(p-q) is) and therefore p is a multiple of q, and that means x is a power of y
I had the same idea with 16 = 8 ^(4/3) All we can say is that k > 1 the middle section is correct and doesn't depend on k being an integer, so the conclusion n
BEFORE WATCHING: Because x and y are positive integers, x^y is a positive integer. y^(x-y) must also be a positive integer, and there are two cases to consider here: Case 1: y = 1. If y = 1, then y^(x - y) = 1. x^1 = 1, so x = 1. SOLUTION FOUND: (1, 1) Case 2: x >= y. This is where it gets complicated. If x = y then y^(x - y) = 1 and by extension x^y = 1, and so actually x = 1 or y = 0. y is positive so it can't be 0, so the only other case: x = 1 then means 1 = y^(1 - y), and y = 1. already found that solution, moving on. all of the above means x > y > 1. consider the form of our equation: x^n = y^m. because all of the terms are integers, then log_y(x^n) = m, and thus log_y(x) = m/n which is rational. let b / a = m / n, where b / a is in simplest form. we can rewrite the above as log_k(x) / log_k(y) = b / a and set k^b = x and k^a = y for some positive k > 1. If b/a is in simplest form then k will turn out to be an integer. PF: given that b/a is in simplest form, suppose k isn't an integer but k^b = x. Then k is some radical of x. Let c be the smallest integer where k^c is also an integer. Then c divides b. c cannot be 1 because k isn't an integer. the same logic applies with k^a = y. c then also divides y, but then b/a wouldn't be in simplest form because they share a common factor. so by contradiction, if b/a is in simplest form then k must be an integer. back to the main quest: k^b = x, k^a = y because x > y, b > a. because y > 1, a > 0. Now the equation is (k^b)^(k^a) = (k^a)^(k^b - k^a) k^(bk^a) = k^(ak^b - ak^a) bk^a = ak^b - ak^a (b+a)k^a = ak^b 1 + b/a = k^(b - a) b > a, and k is an integer, so k^(b - a) is an integer. 1 + b/a must also be an integer, and therefore b/a must be an integer. call that integer n. b > a means n > 1. then log_y(x) = b/a => x = y^(b/a) => x = y^n Now consider the main equation yet again: (y^n)^y = y^(y^n - y) y^yn = y^(y^n - y) yn = y^n - y (safe because y > 1) n = y^(n - 1) - 1 (safe because y > 0) (n + 1) = y ^ (n - 1) (n + 1)^(1/(n - 1)) = y (safe because n > 1. Now we can start plugging in values of n. n = 2: y = 3, x = 9 9^3 = 3^(9 - 3) = 729 n = 3: y = 2, x = 8 8^2 = 2^(8 - 2) = 64 for n = 4 and above... watch this. for n = 4, n + 1 < 2^(n - 1), 5 < 8. for all n >= 4, if 2^(n - 1) > n + 1, then... 2^n = 2 * 2^(n - 1) > 2 * (n + 1) = 2n + 2 > n + 2 (n + 1) < 2^(n - 1) for all n >= 4 by induction. But also n + 1 > 1. So, 1 < n + 1 < 2^(n - 1) 1 < (n + 1)^(1/(n - 1)) < 2 There are no integers between 1 and 2, so if n >= 4 then there are no solutions. The only solutions (x, y) are (1, 1), (8, 2), and (9, 3)
Thank you for your wonderful and blessed videos, which I enjoy very much. For this problem, the reason why k = x/y must be an integer, which is important to the solution, was not obvious to me. But I think this is so because if not, k = n/m for some positive integers n and m with hcf(n, m)=1. Then the equation k = y^(k-2) becomes n/m = y^(n/m-2) hence (n/m)^m = y^(n-2m) where n-2m > 0 as k>2 RHS is an integer raised to a positive integer power and hence an integer. LHS is a non-reducing fraction raised to a positive integer power and hence still a non-reducing fraction and so not an integer. This is a contradiction. Hence k must be an integer. Again, thank you for your lovely videos and i look forward to seeing the next one. God bless you ❤
Thanks for the video! I think the fastest way to arrive at the solution is recognizing that since the RHS is a power of 'y' (and must be an integer since the LHS clearly is), also the LHS must be a power of 'y'. This means that 'x' itself must be a power of 'y'. If you substutite 'x=y^n' into the original equation it simplifies to 'y^(n-1)=n+1'. This limits the possibilities a lot since the left-hand side grows generally much faster than the right-hand side. You can easily find upper bounds for 'y' and 'n' after which there cannot be any solutions to this equation. What remains is manual checking of the remaining few options for 'y' and 'n'.
@user-pl7tr9dv6l : In a case like x^2 = 9^1, the RHS is really a power of 9. This DOES NOT imply that 'x' itself must be a power of 'y'. Your assertion is true ONLY when 'y' is a prime number (or a product of prime numbers each raised to the power of 1).
Challenging problem with a nice solution but I wonder if (x-y)/y could be a fraction since an integer raised to a rational power >1 would increase the number meaning that the inequality x>y would hold true
Before watching video x and y both are +ve integer so x^y is integer, which implies x>=Y, Else right side is non integer. 1) x = y then right side is 1, that makes only one case for left side (1^1) X AND Y both are 1 (x, y) = (1,1) 2) x > y Because both x and y is integer, and left side is x^y then right side should also be x^y but it is y^(x-y) That means y^k = x We can rewrite eqn as y^(ky) = y^(x-y) ky = x-y (k+1)y = x = y^k k = 1 - - > 2y = y (not possible) k = 2 - - > 3y = y^2 - - > y=3 & x = 9 (straightforward) k = 3 - - > 4y = y^3 - - > y = 2, x = 8 (solvable / straightforward) k = 4 If we take y = 2 then also 5y is less than 2^4 = 16, for k = 5 also 6y = 12 is less than 2^5 = 32 For all y > 2 and k >= 4 also same inequality that RHS (y^k) is greater than LHS ((k+1)y) holds and difference between RHS and LHS Will be bigger and bigger as value of k and y increases. When we take y = 1 then LHS is always higher than RHS for any value lf k So we can conclude that for k >= 4 all solution of (k+1)y = x = y^k is between 1 and 2 which are fractions or irrational numbers hence no more integer solutions. Therefore 3 possible solutions are (x, y) = (1,1), (8,2) and (9,3)
Loved all the constraints !!!! And love you !!! Being on a rough patch seeing your videos soothes the mind remind of my classes and the university I was free so I like to think of that time !!!!!
there is an error at 5:50. x being bigger than y is no explanation for k being integer, 8 = 4^1.5 is a counterexample. So it needs a bit more at this part.
We have X and Y are positive integers, then x^y=y^(x-y) means x=ky where k is a positive integer. (ky)^y=y^(ky-y) (ky)^y=y^(y(k-1)) ky=y^(k-1) ky^(1-(k-1))=1 k=1 & y=1 (bc k=1 not 2) or y^(2-k)=1/k Then x=y=1✓ or y=k & 2-k=-1 k=3 and y=k=3 (x=ky) so x=9 & y=3✓ S={(1,1);(9,3)}
6:00 - this is incorrect as stated. An integer y to a fraction power can be greater than y. For example, 4^(3/2)=8 > 4. You need to use some additional/another argument here.
Wolfram Alpha gives: {{x == -8, y == -2}, {x == 1, y == -1}, {x == 1, y == 1}, {x == 8, y == 2}, {x == 9, y == 3}} which reduces to {{x == 1, y == 1}, {x == 8, y == 2}, {x == 9, y == 3}} when x and y need to be positive
1:20 that is not true. You can get a negative exponent on the RHS to get a fraction and also get a fraction on the LHS if y is negative. An integer raised to an integer can definitely give you a fraction if the exponent is negative.
question: at 3:13, shouldn't x=y=0 also be a solution as then by inserting values we would get 0^0 = 0^0-0 => 0^0 = 0^0 and since LHS = RHS we can say x=y=0 is also a solution i am not good at math at all and u are the teacher so i could be wrong, also haven't watched the whole video yet. edit: ok just realized the question asks for positive values for x and y, but if it didn't, would x=y=0 be a solution?
The power 0^0 at working only on the real line, as here, is not defined. Other thing is the definition of 0^0 = 1 in, e.g., binomial theorem. But let's suppose for a moment (even though it's wrong) it isn't a problem that 0^0 = 1. Then it also can't be a solution since x, y are positive integers, as stated in the video. 0 is not positive but neutral.
This is a fun video, but the timing of that horn at 5:42 as you moved across the board was phenomenal. (Editing in - I may have realized that you possibly added that in yourself but it made me laugh so I'm leaving this comment lol)
Not an easy problem for a 15 year old - I'll try to solve it like my 15 year old self would: You can say that it is homogeneous (dimensionally consistent) because dimensional analysis won't lead to a contradiction - e.g. assume metres for x and y. So x = ky and, because x and y are +ve integers, k must be a positive integer. (ky)^y = y^((k-1)y) = (y^(k-1))^y => ky = y^(k-1) if k = 1, x = y = 1 if k = 2, 2y = y, no solution if k = 3, 3y = y^2 => y = 3, x = 9 if k = 4, 4y = y^3 => y = 2, x = 8 if k = 5, 5y = y^4, no solution ... if k = 8, 8y = y^7, no solution and no solution for any k > 4
@@PrimeNewtons I think the solution is basically correct and I enjoyed watching, but there is something missing with the logic as @Viki13 said in the comments " but I wonder if (x-y)/y could be a fraction since an integer raised to a rational power >1 would increase the number meaning that the inequality x>y would hold true" and someone gave an example where the exponent is not an integer: 8 = 4^(3/2).
There’s a quick and easy proof: Take mod x of both sides: 0 = y^(x - y) mod x => y = 0 mod x Take mod y of both sides: x^y = 0 mod y => x = 0 mod y Therefore, x = y. Plugging that back in: x^x = x^(x-x) => x^x = x^0 = 1 The only positive integer solution for x^x = 1 is x = 1. Therefore (x=1, y=1) is the only solution.
If y=1, then y raised to any power is 1. Thus x^y must equal 1, and either x=1 or y=0. y is not equal to zero, so x must be 1 and therefore not greater than y.
@@nidoking042 (1,1) was the first solution. He wrote "if X >y then y >=2" when he was considering the case X>y(since X cannot be less than y). But If X>y and X cannot be 1, this means that X is >=2
@@LovePullups If you think you're disagreeing with me, you're not. You're literally saying exactly the same thing. x > y implies that both numbers are greater than one, because if either number is one then both are one. I agree that the statement should have been made in the video, but it's true nevertheless.
Maths helps you to reason, use all options to get to the solution. Exactly what we do in life, you brainstorm before you undertake any project. This is where maths comes in.
