## Il Buono, il Brutto, il Cattivo

### Il Buono (nice and easy)

Prove that in a subset of size \(n+1\) of the set \(\{1, 2, ..., 2n\}\) there are two numbers such that one divides the other.

### Il Cattivo (beautiful and hard)

Prove that in a sequence of \(r\times s+1\) distinct numbers, there is either a monotonically increasing sub-sequence of length \(r+1\) or a monotonically decreasing sub-sequence of length \(s+1\).

### Il Brutto (just easy)

Prove that in a subset of size n+1 of the set \(\{1, 2, ..., 2n\}\) there are two relatively prime numbers (i.e. numbers whose gcd is 1).

### Il Non Collegato (another easy one)

Prove that a degree 7 polynomial with integer coefficients, that receives the values +1 or -1 on 7 integer points is irreducible over the integers.

Here is some inspirational music to help you solve these riddles:

## Spoiler Alert - Solution Ahead!

### Il Buono Solution

For any \(n \in \mathbb{N}\) write \(n = 2^{s_n}b_n\) such that b is odd. Define \(f:\mathbb{N}\rightarrow\mathbb{N}\) such that \(f(n) = b_n\). Now, \(f(\{1, 2, ..., 2n\}) \subset \{1, 3, ..., 2n - 1\}\), so \(|f(\{1, 2, ..., 2n\})| \le n\). So f on any subset of size n+1 of the set \(\{1, 2, ..., 2n\}\), will contain a collision, i.e. x and y s.t. \(f(x) = f(y)\). This implies that \(b_x = b_y = b\). Which implies that \(x = 2^{k}b, y = 2^{j}b\), which finally implies that either x divides y or y divides x. \(\blacksquare\)

### Il Brutto Solution

In a subset of size n+1 of the set \(\{1, 2, ..., 2n\}\) there must be two consecutive numbers (i.e. numbers whose difference is 1), and those must be relatively prime. \(\blacksquare\)

### Il Non Collegato Solution

Let P be a polynomial as in the question. Assume by contradiction that P factors into \(Q_1\) and \(Q_2\) non trivially. That means that \(min(deg(Q_1), deg(Q_2)) \le 3\). WLOG assume \(deg(Q_1) \le 3\). So \(Q_1\) receives the values +1 or -1 on 7 integer points. That means that \(Q_1\) receives +1 on at least 4 points, or it receives -1 on at least 4 points. WLOG assume it receives +1 on at least 4 points. Than the polynomial \(Q'_1 = Q_1 - 1\) is equal to 0 on 4 distinct points, in contradiction to \(deg(Q'_1) = deg(Q_1) \le 3\). \(\blacksquare\)