international math olympiad | a^b-b^a=17 | exponential tricks | math olympiad preparation

[@calamitates77]

Isn't a = 18, b = 1 a (very obvious) solution too?

[@haydencook9298]

there are infinitely many solutions, given that a,b are real numbers

[@benjaminvatovez8823]

@@haydencook9298 So why has this been solved like they were integers?

[@haydencook9298]

@@benjaminvatovez8823 They just wanted a specific solution, not general solution

[@benyseus6325]

@@haydencook9298that doesn’t make sense, they solved by inspection and missed 18 and 1 as also valid solutions, if a and b are positive integers. Rigorous solution is required to solve for all values

[@haydencook9298]

@@benyseus6325 note that i said there were infinitely many REAL solutions.

[@benyseus6325]

Solution by inspection is boring and cheating and you end up missing solutions (for example 18 and 1 are also possible solutions if a and b are positive integers). You need to show your work and do a rigorous solution by solving for both a and b.

[@siamandmolani]

Great❤

[@Math_with_aram]

Thanks

[@benjaminvatovez8823]

Thank you for the video. I don't understand why you supposed that a,b are both even when they are clearly not as (a^b-b^a) is odd.

[@mab9316]

Beautiful

[@Math_with_aram]

Thank you!

[@ambassadorkees]

I don't see it proven that this is the only solution.
After all, even with this answer, there's a 3rd and 4th power in the problem.
x-y>0 doesn't require both being positive and x>y, it can also be both negative and x

[@BR-lx7py]

@2:00 why are x and y necessarily integers?

[@Math_with_aram]

This is just a placeholder for less writing and its not necessary

[@Math_with_aram]

The paper becomes less crowded