China | A Nice Algebra Problem | Math Olympiad | Radical Problem 

good video good solution

It would have been helpful to mention that the square roots should be positive integers. For example, we can get solutions involving logarithms.

The key in solving problem of this type is to manipulate the left hand side as the difference of 2 squares. (k^u)-(k^v)=n where k and n are constant while u and v are unknowns.

2⁶ - 2² = 60 √2¹² - √2⁴ = 60 √x = 12 => x = 144 √y = 4 => y = 16

@ 2:40 / 11:08 There's no need to introduce an additional variable as k 2^(m) - 2^(n) = 60 2^(m + n - n) - 2^(n) = 60 2^(n + m - n) - 2^(n) = 60 [2^(n) * 2^(m - n)] - 2^(n) = 60 2^(n) * [2^(m - n) - 1] = 60 2^(n) * [2^(m - n) - 1] = 4 * 15 2^(n) * [2^(m - n) - 1] = 2^(2) * 15 → by identification n = 2 2^(m - n) - 1 = 15 2^(m - n) = 16 2^(m - n) = 2^(4) m - n = 4 m = 4 + n → recall: n = 2 m = 6

Before watching: Alright, so the first thing I did was find what powers of 2 (2^a and 2^b) can, when one is subtracted from the other, give you 60. The answer? 2^6 (64) - 2^2 (4) = 0. So, our first term must equal 64, and our second must equal 4. √2 = 2^(1/2), and (a^m)^n = a^(mn) So our initial terms can be rewritten as 2^((√x)/2) - 2^((√y)/2) = 60. From what I wrote above, this means that (√x)/2 = 6 -> √x = 12 -> x = 144 And (√y)/2 = 2 -> √y = 4 -> y=16 The solution is X=144, Y = 16.

64-4=60 √2^(√x)=64=√2^(6×2)=√2^12 √2^(√y)=4=2^2=√2^4 √x=12 則x=144 √y=4 則y=16

3a-2a=a => a=60 3a=3*60=180 => V2^Vx=180 => Vx = ln(180)/ ln(V2) => Vx= 5,1929.../ 0,34657....= 14,9837... x= Vx^2 = 224,51.... 2a=2*60=120 => V2^Vy=120 => Vy = ln(120)/ ln(V2) => Vy= 4,78749.../ 0,34657...= 13,81.... y= Vy^2 = 190,82....

I feel like the solution to this problem you can always find it through trial and error.

Ответ - 3.Обожаю африканских беженцев.

Look at the original equation!!Thats all folks.

X= 144, Y= 16.

Do you need to require the condition that the exponents are natural numbers?

Ici on trouve une solution mais quand on considère 2^n(2^k-1)=60 on considère que n et k sont des entiers mais on y apporte pas la preuve. Il faudrait prouver qu'en dehors des entiers on ne puisse pas avoir l'égalité 2^n(2^k-1)=60 ou alors prouver que la solution trouvée est unique soit un seul couple possible pour(x,y) ce qui n'est pas le cas. Pour l'instant je me casse les dents sur l'unicité et je ne suis pas davantage arrivé à prouver que n et k (utilisés dans la vidéo) ne peuvent être qu'entiers même si j'en ai la conviction profonde.