1. Originally Posted by snowtea

The basic idea is you are using proof by contradiction.
You know for a fact that
$|S(a_0)| \leq |S(a_1)| \leq |S(a_2)|\leq\dots\leq |S(a_{2^n+1})|\leq2^n$

Now, assume that for no $m$ does $a_m\equiv a_{m+1}$.
This means the inequalities must be strict:
$|S(a_0)| < |S(a_1)| < |S(a_2)| < \dots < |S(a_{2^n+1})| \leq 2^n$

But this lead to a contradiction because this forces $|S(a_0)| < 0$.

This means the initial assumption was incorrect, and there must exist an $m$ s.t. $a_m\equiv a_{m+1}$.
Could you tell me how can i say that..i dont remember very well the sequences.
Again thank you both..a loooooooooooot

2. $|S(a_0)| < |S(a_1)| < |S(a_2)| < \dots < |S(a_{2^n+1})| \leq 2^n$
Since $|S(a_i)|$ are integers, we must have $|S(a_i)| + 1 \leq |S(a_{i+1})|$.
From this, you can use induction to prove:
$|S(a_0)| + 2^n + 1 \leq |S(a_{2^n+1})| \leq 2^n$
so $|S(a_0)| \leq -1$
size of a set cannot be negative (contradiction).

3. Basically, you start with a nonnegative number (the number of elements in $S(a_0)$). Then you make $2^n + 1$ steps. In each step, you increase the number. (All numbers are integers.) If after $2^n+1$ steps you have a number that is at most $2^n$, then you can win the Fields medal.

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