Subfield: Number Theory


Question: Let $c \geqslant 1$ be an integer. Define a sequence of positive integers by $a_{1}=c$ and

$$
a_{n+1}=a_{n}^{3}-4 c \cdot a_{n}^{2}+5 c^{2} \cdot a_{n}+c
$$

for all $n \geqslant 1$. Prove that for each integer $n \geqslant 2$ there exists a prime number $p$ dividing $a_{n}$ but none of the numbers $a_{1}, \ldots, a_{n-1}$.

Solution:
Let us define $x_{0}=0$ and $x_{n}=a_{n} / c$ for all integers $n \geqslant 1$. It is easy to see that the sequence $\left(x_{n}\right)$ thus obtained obeys the recursive law

$$
x_{n+1}=c^{2}\left(x_{n}^{3}-4 x_{n}^{2}+5 x_{n}\right)+1\tag{1}
$$

for all integers $n \geqslant 0$. In particular, all of its terms are positive integers; notice that $x_{1}=1$ and $x_{2}=2 c^{2}+1$. Since

$$
x_{n+1}=c^{2} x_{n}\left(x_{n}-2\right)^{2}+c^{2} x_{n}+1>x_{n}\tag{2}
$$

holds for all integers $n \geqslant 0$, it is also strictly increasing. Since $x_{n+1}$ is by (1) coprime to $c$ for any $n \geqslant 0$, it suffices to prove that for each $n \geqslant 2$ there exists a prime number $p$ dividing $x_{n}$ but none of the numbers $x_{1}, \ldots, x_{n-1}$. Let us begin by establishing three preliminary claims.

Claim 1. If $i \equiv j(\bmod m)$ holds for some integers $i, j \geqslant 0$ and $m \geqslant 1$, then $x_{i} \equiv x_{j}\left(\bmod x_{m}\right)$ holds as well.

Proof. Evidently, it suffices to show $x_{i+m} \equiv x_{i}\left(\bmod x_{m}\right)$ for all integers $i \geqslant 0$ and $m \geqslant 1$. For this purpose we may argue for fixed $m$ by induction on $i$ using $x_{0}=0$ in the base case $i=0$. Now, if we have $x_{i+m} \equiv x_{i}\left(\bmod x_{m}\right)$ for some integer $i$, then the recursive equation (1) yields

$$
x_{i+m+1} \equiv c^{2}\left(x_{i+m}^{3}-4 x_{i+m}^{2}+5 x_{i+m}\right)+1 \equiv c^{2}\left(x_{i}^{3}-4 x_{i}^{2}+5 x_{i}\right)+1 \equiv x_{i+1} \quad\left(\bmod x_{m}\right),
$$

which completes the induction.

Claim 2. If the integers $i, j \geqslant 2$ and $m \geqslant 1$ satisfy $i \equiv j(\bmod m)$, then $x_{i} \equiv x_{j}\left(\bmod x_{m}^{2}\right)$ holds as well.

Proof. Again it suffices to prove $x_{i+m} \equiv x_{i}\left(\bmod x_{m}^{2}\right)$ for all integers $i \geqslant 2$ and $m \geqslant 1$. As above, we proceed for fixed $m$ by induction on $i$. The induction step is again easy using (1), but this time the base case $i=2$ requires some calculation. Set $L=5 c^{2}$. By (1) we have $x_{m+1} \equiv L x_{m}+1\left(\bmod x_{m}^{2}\right)$, and hence

$$
\begin{aligned}
x_{m+1}^{3}-4 x_{m+1}^{2}+5 x_{m+1} & \equiv\left(L x_{m}+1\right)^{3}-4\left(L x_{m}+1\right)^{2}+5\left(L x_{m}+1\right) \\
& \equiv\left(3 L x_{m}+1\right)-4\left(2 L x_{m}+1\right)+5\left(L x_{m}+1\right) \equiv 2 \quad\left(\bmod x_{m}^{2}\right),
\end{aligned}
$$

which in turn gives indeed $x_{m+2} \equiv 2 c^{2}+1 \equiv x_{2}\left(\bmod x_{m}^{2}\right)$.

Claim 3. For each integer $n \geqslant 2$, we have $x_{n}>x_{1} \cdot x_{2} \cdots x_{n-2}$.

Proof. The cases $n=2$ and $n=3$ are clear. Arguing inductively, we assume now that the claim holds for some $n \geqslant 3$. Recall that $x_{2} \geqslant 3$, so by monotonicity and (2) we get $x_{n} \geqslant x_{3} \geqslant x_{2}\left(x_{2}-2\right)^{2}+x_{2}+1 \geqslant 7$. It follows that

$$
x_{n+1}>x_{n}^{3}-4 x_{n}^{2}+5 x_{n}>7 x_{n}^{2}-4 x_{n}^{2}>x_{n}^{2}>x_{n} x_{n-1},
$$

which by the induction hypothesis yields $x_{n+1}>x_{1} \cdot x_{2} \cdots x_{n-1}$, as desired.



Now we direct our attention to the problem itself: let any integer $n \geqslant 2$ be given. By Claim 3 there exists a prime number $p$ appearing with a higher exponent in the prime factorisation of $x_{n}$ than in the prime factorisation of $x_{1} \cdots x_{n-2}$. In particular, $p \mid x_{n}$, and it suffices to prove that $p$ divides none of $x_{1}, \ldots, x_{n-1}$.

Otherwise let $k \in\{1, \ldots, n-1\}$ be minimal such that $p$ divides $x_{k}$. Since $x_{n-1}$ and $x_{n}$ are coprime by (1) and $x_{1}=1$, we actually have $2 \leqslant k \leqslant n-2$. Write $n=q k+r$ with some integers $q \geqslant 0$ and $0 \leqslant r<k$. By Claim 1 we have $x_{n} \equiv x_{r}\left(\bmod x_{k}\right)$, whence $p \mid x_{r}$. Due to the minimality of $k$ this entails $r=0$, i.e. $k \mid n$.

Thus from Claim 2 we infer

$$
x_{n} \equiv x_{k} \quad\left(\bmod x_{k}^{2}\right)
$$

Now let $\alpha \geqslant 1$ be maximal with the property $p^{\alpha} \mid x_{k}$. Then $x_{k}^{2}$ is divisible by $p^{\alpha+1}$ and by our choice of $p$ so is $x_{n}$. So by the previous congruence $x_{k}$ is a multiple of $p^{\alpha+1}$ as well, contrary to our choice of $\alpha$. This is the final contradiction concluding the solution.