Let us proceed by contradiction and assume that there exists a prime number $p$ that divides all integers $u_{n}$.
First, since $\varphi(1)=\varphi(2)=1$ and $\varphi(\mathfrak{n})$ is even for all $\mathfrak{n} \geqslant 2$, we note that $u_{1} \equiv a_{0}+a_{1}(\bmod 2)$ and that $u_{2} \equiv a_{0}(\bmod 2)$. Since $a_{0}$ and $a_{1}$ are coprime, the integers $\mathfrak{u}_{1}$ and $\mathfrak{u}_{2}$ cannot both be even, so $p \geqslant 3$.
Let $\delta$ then be the largest integer such that $p$ does not divide $a_{\delta}$. For any integer $n \geqslant 1$, we denote by $T_{n}$ the $\delta$-tuple $(\varphi(n), \varphi(n+1), \ldots, \varphi(n+\delta-1))$, considered modulo $p$. Since $u_{n} \equiv 0(\bmod p)$, it follows that

$$\varphi(n+\delta) \equiv-a_{\delta}^{-1} \sum_{k=0}^{\delta-1} a_{k} \varphi(n+k) \quad(\bmod p)$$

Thus, if we denote by $\lambda:(\mathbb{Z} / \mathrm{p} \mathbb{Z})^{\delta} \mapsto(\mathbb{Z} / \mathrm{p} \mathbb{Z})^{\delta}$ the function defined by

$$\lambda:\left(x_{0}, x_{1}, \ldots, x_{\delta-1}\right) \mapsto\left(x_{1}, x_{2}, \ldots, x_{\delta-1},-a_{\delta}^{-1} \sum_{k=0}^{\delta-1} a_{k} x_{k}\right)$$

we have $\mathbf{T}_{\mathfrak{n}+1}=\lambda\left(\mathbf{T}_{\mathfrak{n}}\right)$. By the pigeonhole principle, there exist two integers $k, \ell \geqslant 1$ such that $\mathbf{T}_{\mathrm{k}}=\mathbf{T}_{\mathrm{k}+\ell}$. We then have $\mathbf{T}_{\mathbf{n}}=\mathbf{T}_{\mathrm{n}+\ell}$ and $\varphi(\mathrm{n}) \equiv \varphi(\mathrm{n}+\ell)(\bmod p)$ for all $n \geqslant k$.
We then prove the following lemma: for all integers $a \geqslant 1$ and $b \geqslant 3$, there exists a prime number $q>a$ such that $q \not \equiv 1(\bmod b)$. Indeed, the integer $m=\max \{a, b\}!-1$ is congruent to -1 $(\bmod b)$, so it must have a prime factor $q \not \equiv 1(\bmod b)$, and $q$ cannot divide $\mathrm{m}+1=\max \{\mathrm{a}, \mathrm{b}\}!$, so $\mathrm{q}>\mathrm{a}$.
By applying this lemma twice, we deduce that there exist two prime numbers $q$ and $r$ such that $q, r \not \equiv 1(\bmod p)$ and $q>r>k+\ell$. But then $q$ is coprime with $\ell$, so $q^{\varphi(\ell)} \equiv 1(\bmod \ell)$, and $\varphi\left(q^{\varphi(\ell)} n\right) \equiv \varphi(n)(\bmod p)$ for all $n \geqslant k$.
Since $q>r \geqslant k$, we deduce that

$$\begin{array}{rlr} (q-1) q^{\varphi(\ell)} & \equiv \varphi\left(q^{\varphi(\ell)+1}\right) \equiv \varphi(q) & \equiv q-1 \quad(\bmod p) \\ (r-1)(q-1) q^{\varphi(\ell)-1} & \equiv \varphi\left(r q^{\varphi(\ell)}\right) & \equiv \varphi(r) \end{array}$$

Since $q, r \not \equiv 1(\bmod p)$, it follows that $q^{\varphi(\ell)} \equiv(q-1) q^{\varphi(\ell)-1} \equiv 1(\bmod p)$, so $q^{\varphi(\ell)-1} \equiv 0$ $(\bmod p)$, which is impossible. We have thus obtained the desired contradiction, which concludes the proof.

Note: We can also appeal to Dirichlet's theorem, which states that for all integers $a$ and $b$ that are coprime, there are infinitely many prime numbers $q \equiv a(\bmod b)$. There then exist prime numbers $q_{0}>q_{1}>\ldots>q_{d}>p+d$ congruent to $1(\bmod p)$. Then, if we choose $k \in\{0,1\}$ such that $p$ does not divide $a_{k}$, there also exists a prime number $r$ such that $r \equiv 2(\bmod p)$ and $r \equiv k-i\left(\bmod q_{i}\right)$ for all $i \neq k$. Then $p$ does not divide $u_{r-k}$.

Exercise 7. Let $n \geqslant 2$ be an integer. Clara has a board of size $3 n \times 3 n$, similar to a chessboard. She has just invented a new piece, the leopard, which she can move as follows: by moving it one square up, one square to the right, or one square diagonally, down and to the left. Clara placed her leopard on one of the squares of the board, then moved it so that it never passes through the same square twice, until it returns to its starting square.
What is the maximum number of moves Clara could have made in this way?