If a and b are both positive integers, then 2^a and 2^b will be positive even integers, so 2^a + 2^b will be a positive even integer.
However, 5^c will never be a positive even integer for any valid value of c.
Hence, a and b cannot both be positive.
If a < -1 or b < -1 , then 2^a + 2^b will have the value of a proper fraction whose denominator is a power of 2, and 5^c will never be such a value for valid values of c.
Hence, a and b must both be greater than or equal to -1.
If a is positive and b is negative (or vice-versa), then 2^a + 2^b will have the value of a proper fraction whose denominator is a power of 2, and 5^c will
never be such a value for valid values of c.
Hence, a and b cannot have opposing signs. Therefore, if a or b is positive then the other must be 0.
Without loss of generality, let's assume a <= b. Hence, the only valid values for a are -1 and 0 (if a > 0 then both a and b would be positive).
Case 1: a = -1
If a is -1, b cannot be positive (since a and b cannot have opposite signs), nor can b be zero (since 2^a + 2^b would equal 3/2, a value 5^c can never be). Hence, if a=-1, then b can only be -1. As it turns outs 2^(-1) + 2^(-1) = 1 = 5^0, so one solution is (-1,-1,0).
Case 2: a = 0
If a is 0, then our equation becomes 2^0 + 2^b = 5^c, or 1 + 2^b = 5^c, or 5^c - 1 = 2^b. Effectively, we need to find all values of c such that 5^c -1 is a power of two.
Note that you can factor the ( 5 - 1) from the polynomial 5^c - 1. In other words, we can write 5^c - 1 = ( 5 - 1) * Z, where Z = 5^(c-1) + 5^(c-2) + ... + 5^1 + 1. Hence, 5^c - 1 = 4Z. If c is odd, then Z is odd (since it is the sum of c odd integers). The only odd value of Z for which 4Z is a power of two is Z = 1, which only occurs when c=1. When a=0 and c=1, b must equal 2, so one more solution is (0,2,1).
If c is even (say c = 2d for some d), then we have 5^(2d) - 1 = 2^b, or (5^d + 1) ( 5^d - 1) = 2^b. Since 2^b is a power of two, both (5^d + 1) and (5^d - 1) must also be powers of two. The only powers of two that differ by 2 are 2 and 4. Hence 5^d + 1 = 4 and 5^d - 1 = 2, but there is no integer d that satisfies those equations. Thefore c cannot be even, and we are done with solutions.
Conclusion
The only 3 integral solutions are (-1,-1,0), (0,2,1), and (2,0,1).