Problem 937

[heteroing]

I found the wording "...number of distinct pairs x, y..." very confusing for a while, as I assumed it wanted me to count perhaps the number of distinct ordered pairs (which is what it says). That was until I saw it counts 1 x 4 only once, okay, sure, so the number of distinct unordered pairs (x, y). Nope, not that either, since for other reasons this would not allow for the problem's conditions to hold. I think the wording "...number of unordered pairs of distinct x, y..." or something would make a lot more sense.

[Swistakk]

+1, from the example we can conclude that there has to be a condition that $x \neq y$ (otherwise $x=y=2$ should count), which is not conveyed with the current usage of the word "distinct". I'd also suggest "unordered pairs of distinct x,y"

[nledoux]

I don't understand how p(A,z) can be equal to p(B,z) for z=1. The only pair x,y such that xy = +-1 is (1,1) because -1 is not in T due to the constraint a>0, or a=0 and b>0. A∪B = T, so -1 is not in A and not in B.
So, for me, p(A,1) = 1 and p(B,1) = 0. What am I missing?

[RobertStanforth]

I found the wording "...number of distinct pairs x, y..." very confusing for a while, as I assumed it wanted me to count perhaps the number of distinct ordered pairs (which is what it says).

Thank you for flagging this. We have discussed it internally, and agree that it is ambiguous. The wording has now been updated to:

define $p(S,z)$ to be the number of ways of choosing two distinct elements from $S$ with product either $z$ or $-z$

[RobertStanforth]

I don't understand how p(A,z) can be equal to p(B,z) for z=1. The only pair x,y such that xy = +-1 is (1,1) because -1 is not in T due to the constraint a>0, or a=0 and b>0. A∪B = T, so -1 is not in A and not in B.
So, for me, p(A,1) = 1 and p(B,1) = 0. What am I missing?

$p(A,1)$ and $p(B,1)$ are both zero, because they count (unordered) pairs of distinct numbers. The problem statement has been updated to clarify this.