OK, all this math talk has me remembering something:
[In any set of “N” numbers (being integers), the difference between at least two of the numbers in the set can be evenly divided by “N-1”.]
For example, let’s say my set contains {1,3,4). N=3 (the number of members in my set). N-1 = 2. 3-1 (both numbers in my set) equals 2, which can be divided evenly by 2.
I’m not arguing that there’s a counter-example. I believe it. But how do you prove it?
I don’t know how to prove it, but I know how to disprove it
[1,2,3,4,5,6,7,8,9,0] N=10
7-0 = 7 N-1 = 9 7/9 doesn’t divide out
sorry
Your set is 0, 1, 2, …, 9. Therefore N=10. How about using the difference between 9 and 0 which is 9 of course. That is divisible by 10 - 1.
RESOL:I don’t think you’ve disproved it
In order for the difference between two of the numbers to be divisible by n-1, they must have different remainders when divided by n-1. If the set is all integers, then the number of potential remainders is n-1 (0, 1, … n-2). Since there are n numbers in the set, two of them must have the same remainder.
oops upon re-reading the post I noticed it didn’t say all the compinations had to work, like I originally thought.
never mind I can’t find one now…
An old favorite of mine… I used thisoneto tripupbad math teachers…
Given: X=Y
Prove: 1=2
X = Y
XX = XY (multiply both sides by X)
XX-YY = XY-YY (subtract YY from both sides)
(X+Y)(X-Y) = Y(X-Y) (factor)
(divide both sides by (X-Y))
X+Y = Y
(since we were given X = Y at the start of the problem:)
Y+Y = Y
2Y = Y
(divide by Y)
2 = 1
There is, of course, a simple, basic flaw with this. What is it?
That SHOULD have read:
‘I used this one to trip up bad math teachers’
damn sticky keyboard!!
lawoot: if X = Y, then dividing by (X - Y) is dividing by 0, which given indeterminite results.
ENugent got the answer to the OP exactly right.
I think I’ve got the core of it.
Let the set S’ consist of all of the numbers in S, mod (N-1). There are then a maximum of (N-1) distinct numbers in S’, and N total elements, so at least two elements in S’ must be identical, with a value we will call M. Now, the corresponding numbers in S must be of the form P*(N-1)+M and Q*(N+1)+M, so their difference must be (P-Q)(N+1), which is divisible by (N+1).
And I just realized that this is equivalent to what ENugent wrote already. Oops.
Bingo! I had TWO math teachers who couldn’t figure that one out… one at the COLLEGE level.
At the COLLEGE level??? I knew this one when I was 12!
(turns away, muttering to himself)
Bill
CRAP CRAP CRAP!! Finally one I can prove, and I didn’t see the thread soon enough.
Damn you, ENugent and Chronos! Damn you all to hell!!
Only kidding. 
lawoot: I’m amazed that 2 college math teachers missed a division by zero problem. WTF? But I also have to say, that’s a really nice piece of flawed reasoning. I’d never seen that before.
They musta been pulling your leg. All math teacher colleges have a required course in Encouraging Young Minds that basically consists of deception.
Y’know, that’s even worse than a college DiffEQ teacher I had who had never heard of the Golden Ratio…