Factorize 8051 in your head.

## - Answer -
If we find $P \in \mathbb{Z}[x]$ such that $8051=P(a)-P(b)$, we would have the factor $a-b$. For $P=x^{2}$, we have $8051=90^{2}-7^{2}$. Therefore, $8051=97 \cdot 83$, both of which are prime.

## 3 Sequences

A sequence of elements from a set $E$ is a function $u: \mathbb{N} \rightarrow E$. We generally write $u_{n}$ instead of $u(n)$. The sequence is denoted $\left(u_{n}\right)_{n \in \mathbb{N}}$ or simply $\left(u_{n}\right)$. In the following, unless otherwise stated, we take $E=\mathbb{C}$.

## - 1. Classical Sequences. -

An arithmetic sequence is a sequence that satisfies a recurrence relation of the form $u_{n+1} = u_{n} + r$. The number $r$ is called the common difference of the sequence. We have $u_{n}=u_{0}+n r$, and

$$
\sum_{k=0}^{n} u_{k}=\frac{\left(u_{0}+u_{n}\right)(n+1)}{2}
$$

A geometric sequence is a sequence that satisfies a recurrence relation of the form $u_{n+1} = q \cdot u_{n}$. The number $q$ is called the common ratio of the sequence. We have $u_{n}=q^{n} \cdot u_{0}$, and if $q \neq 1, \sum_{k=0}^{n} u_{k} = u_{0} \cdot \frac{q^{n+1}-1}{q-1}$.

Given a sequence $(u_{n})$ defined by a recurrence relation, we can sometimes find an explicit expression (i.e., an expression of $u_{n}$ in terms of $n$) by reducing it to a sequence of the form $v_{n}=f(u_{n})$ (where $f$ is a bijection of the set $E$, for example, an affine transformation) for which we know an explicit formula.