Exponents Are Scary!

I will not post the complete solution, but will take you through my thinking process while solving this problem (what I say below wasn't my exact thinking process, but I have forgotten some of my original ideas and motivations, especially the motivation for the penultimate step)

I have noted something important when solving exponential diophantine equations. This important observation is as follows:

In an exponential diophantine equation, it often so happens that when the exponents are greater than some finite number k, then both sides of the equation is divisible by some k-th power

If you didn't understand what I just said, don't worry. I will walk you through my thinking process.

First thing to do as always is parity checking. If you take mod of both sides of the equation (first modulo 2, then 3), you will get some information about the parity of the exponents. But if you try to proceed using this, you might get stuck like I did(I am not saying that you won't be able to proceed, but I couldn't). So you have to start looking at the other options here.

Rearrange the equation to get $2^a+3^b=6^c-1$ . Here, the left hand side must be divisible by $5$ since the right hand side is divisible by $5$ . You will later note that this doesn't help much, though it does give you some insight as to the parity of the exponents.

Finally, being frustrated, you take $c=1,2,3,4,5$ and calculate some values of $a$ and $b$ . I did this upto $7$ ( it wasn't time-consuming). You will note that you aren't getting any solution for $c\ge3$ .

So what's so special about $c\ge 3$ . For starters, you can note that for $c\ge 3$ , the RHS is divisible by $8$ (and also by $27$ , but that is a much bigger number). Now note that if the exponent of $2$ i.e. $a$ is also greater than or equal to $3$ , then we must have $8|3^b+1$ . But the powers of $3$ mod $8$ are $3,1,3,1,3,1\cdots$ . Hence $3^b+1\equiv 4,2\pmod 8$ . So $8$ cannot divide $3^b+1$ . Hence at least one of $a$ and $c$ must be less than $3$ . So we have the following two cases:

1) $a<3$

2) $c< 3$

The rest of the solution is left as an exercise to the reader.

FINAL THOUGHTS:

The idea about $c\ge 3$ played the crucial role in this problem. I am not sure what my exact motivation was for this idea, but I know that experimentation made me think that something was up. There was definitely some other idea involved, but I am unable to recall that at the moment.

Try to use the idea that when some exponent is greater than some fixed value $k$ (in our case, $k=3$ ),then both sides of the equation must be divisible by the k-th power of an integer(in our case, that was $2^3$ ). This idea may come useful later on in more advanced problems.

Other ideas that I had that I couldn't use include LTE,further parity checking etc..

$We ~have~6^1=6.........6^2=36.........6^3=216.........6^4=1296. \\3^1=3.....3^2=9...... 3^3=27...........3^4=81.........3^5=243.........3^6=739. \\ 2^1=2...2^2=4...2^3=8...2^4=16..2^5=32..2^6=64..2^7=128...2^8=256.\\\color{#D61F06}{(1,1,1) }\text{ is a clear solution. No other value can give 6.} \\c=2,~~6^2=36.~~36-1=35.~~~35-3=32 \implies \color{#3D99F6}{ (5,1,2)},\\b=2, 2^{\huge?} \nmid 26.~~~~~~b=3, 2^3+3^3+1 =36...\color{#EC7300}{(3,3,2)}~no ~~other~~ possibility.\\c=3, 6^3-1=215,....215<2^8.....215-2^7= 87....3^{\huge ?}\nmid 87\\215-2^6= 151....3\nmid 151. ~~~\text{No combination satisfy the given}\\ equation ~any~ further.~~~~~~~~~~~~~~~~~~~Only~three~triples.$

LHS of above equation should be divisible by 3. So 'a' should be odd. Further 6^c always ends with digit 6, so on observing possible unit digit of sum in LHS We get another condition which on combining with above one we get a=b(mod4) and both a,b are odd»» Case 1: a=4a'+1,b=4b'+1. Equation becomes 2.16^a +3.81^b +1 = 6^c. Now observe that LHS can never be divisible by 16(81=1 mod 16,LHS = 16z + 4) so c<4. We get very same conclusion from case 2 a=4.a'+3,b=4.b'+3. Now we can easily see solutions for c= 1,2,3 for (a,b,c) as (1,1,1);(5,1,2);(3,3,2)