### 2^{x} + 2^{y} = k

2^{x} + 2^{y} = k where x, y and k are positive integers
and k is a constant. The question's that for different values of k how many
solution does our equation have? For any odd k there is no solution and
for positive powers of 2 there is one. We want a general solution
for this equation.

#### Solution by Susam Pal

We will take this fact for granted: Every positive integer can be uniquely expressed as the sum of distinct powers of 2. In other words, every positive integer has a unique binary representation.

Now consider the case where x = y. Let m = x = y. Then k = 2^{m} +
2^{m} = 2^{(m+1)}. Further, if k = 2^{(m+1)} where m > 1, x =
y = m is a solution. There cannot be a solution in which x != y because that would
result in another binary representation of k apart from the existing representation of
2^{(m+1)} which would contradict the uniqueness of the binary representation of k. Therefore there must be exactly 1 solution when k is a positive power of 2.

Now consider the case where x != y. Let x = m and y = n. Then k = 2^{m} +
2^{n} = 2^{n} + 2^{m}. Thus both (x = m, y = n) and (x = n, y
= m) are solutions. There cannot be a solution in which x = y because that would lead
to k = 2^{(x+1)} which violates the uniqueness of the binary representation of k. Also, there cannot be another solution in which either x or y does not belong to {m, n} because that would too violate the uniqueness of the binary representation of k. We conclude that there must be exactly 2 solutions when k is a sum of two distinct positive powers of 2.

For any other positive integer k, no solutions are possible. We can show this by
contradiction. Let k be such that it is neither a positive power of 2 nor a sum of two
distinct positive powers of 2. If we assume that a solution exists for such k, then we
have two positive integers x and y such that 2^{x} + 2^{y} = k. If x = y, it contradicts the fact that k is not a positive power of 2. If x != y, it contradicts the fact that k is a sum of two distinct positive powers of 2.

Thus, we can conclude that the equation has exactly 1 solution when k is a positive power of 2, 2 solutions when k is a sum of two distinct positive powers of 2, no solutions otherwise.

Thanks,
Susam