To prove that (3a+4b, 4a+4b) = 1, we need to show that the greatest common divisor of 3a+4b and 4a+4b is 1.

Let's assume that there exists a common divisor d > 1 of 3a+4b and 4a+4b. This means that d divides both 3a+4b and 4a+4b.

We can write 3a+4b as (3a+4b) = (3a+4b) - (4a+4b) = -a.

Similarly, we can write 4a+4b as (4a+4b) = (4a+4b) - (3a+4b) = a.

Therefore, d divides both -a and a. This implies that d divides their sum, which is 0.

Since d divides 0, it must also divide any multiple of 0. In particular, d divides 0*a = 0.

However, d is assumed to be greater than 1, so it cannot divide 0.

This contradiction shows that our assumption was incorrect. Therefore, there is no common divisor greater than 1 for 3a+4b and 4a+4b.

Hence, (3a+4b, 4a+4b) = 1.