Find all Integer Solutions

We find all integer solutions a, b, and c, to an equation involving complex numbers:

(1 + (2 + sqrt{3})i)^a = b(1 + (2 - sqrt{3})i)^c

00:00 Modulus and denesting
02:35 Arguments
04:47 Comparing arguments
07:53 Simplifying the modulus
09:17 Comparing the modulus
12:13 Rational vs irrational
14:42 Case (i) conclusion
16:51 Case (ii) conclusion

[worldnotworld]

"The first thing we're going to do is make things easier for ourselves". :-)

[scudlee]

I feel like you could probably consolidate the two cases somewhat, since for b>0 c=5a+12(2k) and b<0 c=5a+12(2k+1), so in essence either way c=5a+12j, where when j is even it's the positive case, and when j is odd it's the negative case.
If you pull out a factor of 2 (and an appropriate power of (-1) for the b term) from the two separate solutions, you can probably write them to just be one solution:
(a, b, c)=(2n, (-1)^n*4^(2n), -2n)

[kasiphia]

Very interesting problem and solution. May I ask where you found the problem or how you came up with it?

[cparks]

7:55 At this point, you could remove the cases by stating that c=5a+12k for some integer k. This throws some information away, but it's easily correctable by verifying answers later. You then replace b with |b|.

Overall, this was a really cool problem.

[riadsouissi]

It would have been faster to start by equating the modulus (or the square of modulus) which immediately leads to c=-a then the rest is quick. No need to use the argument.