Question

3. Prove that, for any n + 1 integers, x0, x1, x2, . . . , xn,

there exist two integers xi and xj with i 6= j such that xi − xj is
divisible by n.

Answer 1

The proof utilizes the pigeonhole principle to demonstrate that among n+1 integers, there will be at least two with the same remainder when divided by n, making their difference divisible by n.

Step-by-step explanation:

The question is asking to prove the pigeonhole principle in the context of number theory, particularly that in any set of n+1 integers, there will always be two distinct integers, xi and xj, such that xi − xj is divisible by n. This can be proven using a direct argument based on the principle of pigeonholes. If you take any n+1 integers, there are exactly n possible remainders when you divide them by n, namely, 0, 1, 2, ..., n-1. Since you have n+1 numbers, by the pigeonhole principle, at least two of them must have the same remainder when divided by n. If you subtract these two numbers, their difference will be a multiple of n, thus proving the statement.