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[@Ravindermathematicstutor]

Very nice solutions

[@mks-learning-easy]

Thank you 😊

[@ironhawk929]

25 - 4 = 21 √m/2 = 2 m = 16

[@prime423]

This is easier to solve with observation,not perspiration!!5 squared minus two squared is 21.10 seconds not 10 minutes.Mathletes should use "Tricks" only when needed !!

[@taci9118]

And what is a 21 and 1

[@tychovos4085]

To be complete also develop x=21 and y=1

[@seondojin]

m is supposed to be a positive integer. Otherwise we have two answers. One is m=16 the other is m=(4ln11/ln5)^2.

[@vuyyurisatyasrinivas3849]

Good solution. Not come at a time

[@user-dq3uh6ee5w]

16.

[@angelorsfreitas]

Where is write that (x+y) and (x-y) has to be integer???

[@davidbrisbane7206]

It's not. It just happens (luckily) for this solution approach that they are integers.

[@user-mx8sj1nc6v]

I found the derivative and convinced myself that the function is increasing. [even by just looking at the function it is simple to be convinced that it is an increasing function] m=16 easy to guess. Your way is obviously more matamatical.

[@xgx899]

Actually, your solution is shorter and more rigorous. Combination of guessing and uniqueness proof is perfectly ``mathematical".

[@marcgriselhubert3915]

f: m ---> (sqrt(5))^(sqrt(m)) - (sqrt(2))^(sqrt(m)) is strictly increasing on R+, so if the equation f(m) = 21 has a solution then this solution is unique. Now as 25 - 4 = 21 it is evident that m = 16 is a solution. Finally m = 16 is solution and it is unique. That's all !!!

[@SuperAnangs]

It's only another cuadratic equation

[@davidbrisbane7206]

(√5)^(√m) - (√2)^(√m) is clearly an ever increasing function as m increases from 1 to infinity. Therefore there is at most one real value of m such that (√5)^(√m) - (√2)^(√m) = 21. By observation, m = 16 satisfies the equation and it is therefore the only value that does when m ≥ 1. If m < 0, then there are no real solutions. If m = 0, then there is no solution either. Consider 0 < m < 1. So, 1 < (√5)^(√m) < √5 and 1 < (√2)^(√m) < √2 So (√5)^(√m) - (√2)^(√m) < 21 So there are no solutions to be found here. So, m = 16 is the only real solution to the given equation.

[@Jedidias_68]

Muy complicado tu solución Es más sencillo igualar en vez de 21 a 5^2 - 2^2 y por bases iguales en ambos miembros se despeja "m" de una forma rápida

[@user-fr3xf9gj7p]

Собственно, задача простая, как большинство Диофантовых. Тут сразу видно, что m делится на 4 и должно быть полнись квадратом . Это или 16 или 36. функция слева -возрастающая и 36 - много. Всё.

[@sitarsbi]

Manakanakku m=16

[@krishnaraolingam4812]

U=2 m=16

[@MathSync]

“Fantastic explanation! I love how you break down complex concepts. I do similar math tutorials and challenges on my channel. Check it out for a fresh perspective on math problems!”

[@mks-learning-easy]

Thank you 😊. Sure I will check it out!

[@JangirBK]

It should be mentioned here that solve this for integer value of m

[@Vrazhino_Pruhin]

Это проще решит "в уме" методом банального подбора

[@gouravsoni-ug3jy]

By hit & trial we know 25-4=21 so m=16

[@user-xn6nd9hy5o]

It has an infinite number of Solutions.

[@userbermer]

Но ведь может быть такой вариант: x=11, y=10. Как решать в таком случае?

[@zxb_4d]

Вагон ошибок

[@SuperAnangs]

x=16, solved in 4 sec

[@mijazukant]

really? is that the time it took you to slide the cursor directly to the end of the video ? if not, please explain what you could have possibly thought in the short period of 4 sec, after reading that and saying "yep thats 16" 😂😂😂

[@SuperAnangs]

@@mijazukant first tried by 2 then I got the wrong answer, then 4 But I aware then that it's in root number So, I multiply 4 with itself to get the solution

[@williamspostoronnim9845]

А что если данное решение не единственное?

[@user-dq3uh6ee5w]

Решение единственное, что видно из условия.

[@maximcoroli8306]

это в каком-то смысле подбор. А если бы не целые числа были? А если бы не 5, 2, 21, а сотни или тысячи? Так вообще можно было 1 мин 38 сек записать, что 5^u - 2^u = 25 - 4 => 5^u - 2^u = 5^2 - 2^2 => u=2, и не тратить ещё 6 минут Не знаю, как сейчас, но когда я учился в школе, такое решение не зачли бы.