One of the cases has to be true, not all 3. (as you show, they're mutually exclusive for positive integers) i.e. "either" is important in the parent comment.
Which is why I indicated that it would be a misreading of the problem.
The original problem is a little ambiguously worded. You could say "one of their numbers is the sum of the other two" and it would be a little clearer.
> The original problem is a little ambiguously worded.
No it isn't. If it said "the sum of any two of the numbers is equal to the third", that would be a contradiction. What it says is "the sum of two of the numbers is equal to the third".
There's a certain mind that either doesn't realize they're sidestepping the problem and turning it into a editing review, or realizes it, and doesn't understand why it seems off-topic/trivial to others.
What's especially strange here is, they repeatedly demonstrate if you interpret it that way, the problem is obviously, trivially, unsolvable, in a way that a beginner in algebra could intuit. (roughly 12 years old, at least, we started touching algebra in 7th grade)
I really don't get it.
When I've seen this sort of thing play out this way, the talking-down is usually for the benefit of demonstrating something to an observer (i.e. I am smart look at this thing I figured out; I can hold my own when the haters chirp; look they say $INTERLOCUTOR is a thinker but they can't even understand me!), but ~0 of that would apply here, at least traditionally.
So A + B = C and A + C = B. But we know that A + B = C, so we can replace C with (A + B). So we know that A + A + B = B.
So 2A + B = B. Or 2A = 0.
And this holds any way you slice it.
Even if you were to try and brute force it.
A = 1
B = 2
Then C = 3. But A + C has to equal B. That's 1 + 3 = 2? That's not true.
I don't see a case where you can add to the sum of two numbers one of the numbers and get the other number.
I'm guessing that's a misreading of the problem. Because it looks like the third number is the sum of the first two.