@user21820 not sure how to start, any hint? @user21820

it's just that this seemed like a trivial counterexample to a claim in a book I was reading, and it was. Of course, it is obvious enough. The imprecision could be fixed with what came after, fortunately

@BigSocks Ahh... so it was in relation to something else that made you think you're going nuts. No wonder. =)

@RaphX Well you can look at the binary representation of n. First prove that adding a power of 2 would increase the number of ones in the binary representation by at most 1. So if n in binary has more than 4 ones then there is no solution. If it has 4 ones, clearly only 1 solution. If 3 ones, 1 of the ones must be from 2 powers of two. If 2 ones, ...

There is nothing much to this; it's just that 10 was quite small. You can see that with such analysis you can easily deal with larger n like 1040.