Solving an Exponential Diophantine Equation

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[walterufsc]

Another excellent video. At the beginning of the problem, I did it like this (which is equivalent to your method, of course):
2^3a + 2^2b = 2^5c
(2^3a)/(2^5c) + (2^2b)/(2^5c) = 1
2^(3a - 5c) + 2^(2b - 5c) = 1

For equality, the only possibility (for integer exponents) is 1/2 + 1/2 = 1 ==> 2^(-1) + 2^(-1) = 1
Hence: 3a - 5c = 2b - 5c = -1
Therefore 3a = 2b = 5c - 1.

[youuber]

Beautiful. Sir, you are a prodigy. Not in math (well maybe, those exercises are too simple to tell), but in teaching math. You make it both very easy to understand and really fun to do. Thank you so much. People like you really help mankind to move forward. I wish you all the best.

[scottleung]

I didn't solve this the first time around, but I have a few observations. First of all, since the left side of the equation deals with multiples of 2 and 3 in the powers, it's obvious we should focus on multiples of 6. Second, based on Syber's solution, it seems that only multiples of 6 ending in a 4 whose tens digit is also even (24, 84, 144, etc) will do the trick. Thus, there are infinitely many integer solutions that satisfy this equation. Finally, I see a pattern in the ordered triples. If 24 is the power of interest, the ordered triple would be (a, b, c)=(8, 12, 5). Similarly for 84 it would be (28, 42, 17) and for 144, (48, 72, 29). Clearly a increases by 20, b increases by 30, and c increases by 12. I'm not sure how to make any algebraic sense behind these remarks, but if there's a way let me know!

[rakenzarnsworld]

a = 8, b = 12, c = 5
a = 18, b = 27, c = 11
a = 28, b = 42, c = 17
a = -2+10k, b = -3+15k, c = -1+6k (k is an integer)

[almazchati]

When 3a = 2b, the left hand side become 2^(3a+1), so, 3a+1 = 5c, a=6n/3, b=6n/2, c = (6n+1)/5. Integer values

[olivierklepper]

Super, pure entertainment! Thank you 💝

[Matematika_more_math]

Awesome thanks that was a great explanation

[mcwulf]

The real problem is finding the parametric solution for a, b, c. The first bit is simple because the only way of adding powers of 2 together to get a power of 2 is if they are the same. Easy to prove.

[rajeshbuya]

Kudos too to the power of Integer! Or must I simply say - Kudos 2 ^ Z!!!

[carlyet]

this took me two goes 3a + 1 = 5c & 2b + 1 = 5c ... my first thoughts was 3a == 2b so(error) 6a = 5c etc Nice problem

[wonghonkongjames]

Sir I think there's something wrong with this problem ： if we choose 2 to the power of 3a to be the smaller one - we can choose the other one too just an assumption - and then factorize it and then divide on both sides It happens to me that one side is even and the other side is odd！ Which is a contadiction
That's because from here it's clear that 3a must be equal to 2b clearly they arent

[mohitg]

I used the AM GM and HM inequality and got -2>= 3a+2b-10c further I wasn't able to solve.

[wonghonkongjames]

Sir sorry to my wrong solution After detail consideration I had found that they could be equal when 3a=24 2b=24 and 5c=25 that is a b and c are 8 12 and 5 respectively Such mistake occurred in overseeing the larger integers possibility a common error

[natanytzhaki]

You gotta say the joke next time, that's why I love your videos because you explain it with fun 😊

[rajeshvaghela]

the fact i learnt that sum of power 2 is not power of 2.

[isaacdagwom]

This question is out of my league. They don't teach this kind of questions or maths in Nigeria. I'm so ashamed

[alphaomega]

Dazzling. But you managed to say "2B" many more times than usual with the excuse of basing the problem on powers of 2. Could you try another base number next time? Please? Or banish the variable "B" going forward? Why expose yourself to temptations that you obviously cannot resist?