Problem 1688

1. Define the sequence and initial conditions:
Let $c \ge 1$ be an integer. Define a sequence of positive integers by $a_1 = c$ and
$$a_{n+1} = a_n^3 - 4c \cdot a_n^2 + 5c^2 \cdot a_n + c$$
for all $n \ge 1$.

2. **Establish the periodicity modulo $a_n$:**
We claim that for all integers $k$, $a_{n+k} \equiv a_k \pmod{a_n}$. This can be shown by induction:
- Base case: For $k = 1$, we have $a_{n+1} \equiv c \pmod{a_n}$.
- Inductive step: Assume $a_{n+k} \equiv a_k \pmod{a_n}$. Then,
$$a_{n+k+1} = a_{n+k}^3 - 4c \cdot a_{n+k}^2 + 5c^2 \cdot a_{n+k} + c \equiv a_k^3 - 4c \cdot a_k^2 + 5c^2 \cdot a_k + c = a_{k+1} \pmod{a_n}.$$
Thus, $a_{n+k+1} \equiv a_{k+1} \pmod{a_n}$.

3. **Show that $a_i \equiv a_j \pmod{a_{(i,j)}}$:**
By the Euclidean algorithm, for any positive integers $i$ and $j$, we have $a_i \equiv a_b \pmod{a_j}$ where $i = aj + b$ and $0 \leq b < j$. Hence, $a_{(i,j)} = (a_i, a_j)$.

4. **Prove that if $n \mid i$, then $a_i \equiv a_n \pmod{a_i^2}$:**
Notice that
$$a_{n+1} = 5c^2 \cdot a_n + c \pmod{a_n^2}.$$
Therefore,
$$a_{n+2} \equiv (5c^2 \cdot a_n + c)^3 - 4c \cdot (5c^2 \cdot a_n + c)^2 + 5c^2 \cdot (5c^2 \cdot a_n + c) + c \pmod{a_n^2}.$$
Simplifying, we get:
$$a_{n+2} \equiv 2c^2 + c \equiv a_2 \pmod{a_n^2}.$$
Hence, for all integers $k \ge 2$, $a_{n+k} \equiv a_k \pmod{a_n^2}$. Thus, $a_i \equiv a_n \pmod{a_n^2}$.

5. **Assume there exists an integer $m$ such that for all $p \mid a_m$, there exists $1 \leq i \leq m-1$ such that $p \mid a_i$:**
Let $p_1, p_2, \ldots, p_k$ be the prime factors of $a_m$. Let $q_i = \frac{m}{p_i}$ for all $1 \leq i \leq k$.

6. **Claim that if $p$ is a prime factor of $a_m$, then $p \mid a_{q_i}$ for some $1 \leq i \leq m-1$:**
Suppose not, then there exists $1 \leq i \leq n$ such that $p \mid a_n$. From the periodicity, we have $p \mid a_{(m,n)}$. Notice that $(m,n) \neq m$, hence there exists $1 \leq i \leq k-1$ such that $(m,n) \mid a_{q_i}$, which implies $p \mid a_{q_i}$. This is a contradiction.

7. **Show that $a_m \equiv a_{q_i} \pmod{a_{q_i}^2}$:**
From the periodicity, we have:
$$a_m \equiv a_{q_i} \pmod{a_{q_i}^2}.$$
This clearly implies $v_p(a_m) = v_p(a_{q_i})$.

8. **Prove that $a_m \geq (a_{q_i})^{2^{p_i}}$:**
Notice that for $(c,n) \neq (1,2)$, we can show using algebra that:
$$a_{n+1} = a_n^3 - 4c \cdot a_n^2 + 5c^2 \cdot a_n + c \geq a_n^2.$$
Therefore, if $(c, q_i) \neq (1,2)$, the inequality follows. If $c = 1$ and $q_i = 2$, then $m \geq 4$. Since $a_4 = 183 \geq 3^4 = a_2^4$, we are also done.

9. Derive the final contradiction:
From the previous claims, we have:
$$a_m < \prod_{i=1}^k a_{q_i}.$$
From the inequality, we have:
$$1 < \sum_{i=1}^k \frac{1}{2^{p_i}} < \sum_{i=1}^{\infty} \frac{1}{2^i} = 1,$$
which is a contradiction.