This solution is based on all of (a,b,c) being powers of 2. The strategy is to assign (a,b) so that each term of LHS is equal and then adjust the exponents so they are divisible by the exponent on RHS.
Set a=2^(2n) and b=2^(2n-1). Then LHS reduces to 2^((2n)(2n-1) + 1). This isn't divisible by (2n+1) but a little manipulation with the congruence k*(2n)(2n-1) + 1 = 0 mod(2n+1) yields the solution k = n.
So a=2^(n*2n), b=2^(n*(2n-1)), c=2^(2n^2 - 2n + 1) is a solution for all n. Increasing k by a multiple of (2n+1) gives another solution. Since the multiples are infinite, so are the solutions.
For example, for a^5 + b^6 = c^7, set a=2^(3*6) and b=2^(3*5) to get 2^91 = (2^13)^7.
The next solution is a=2^(10*6), b=2^(10*5) which gives 2^301 = (2^43)^7.
Posted by xdog
on 2015-09-27 17:23:30