Does anyone have a solution? Thanks!

An outline without the details: First you realise that if you take $(mod a_n)$, $a_{n+1}\equiv c(mod a_n)$, which motivates one to show the following: $a_{n+d}\equiv a_d(mod a_n)$. However this is not really useful itself, so perhaps something stronger is needed, that limits the prime power of factors in $a_i$ dividing $a_n$, from which we discover: $a_{n+d}\equiv a_d(mod a_n^2)$, finally we just need $a_n$ to be large enough (more than product of previous terms), to obtain that it has a primitive prime factor

First, note that $a_{n+1} =a_n(a_n-2c)^2+c^2a_n+c\geqslant a_n^2+c>a_n^2$ for all $n$. It's not hard to prove that: for every prime number $p$ that $p\mid a_n$ for some $n$, there exists index $i_0$ that $p\mid a_{i_0}$ and $p\nmid a_j$ for all $1\leqslant j<i_0$, and For all $n$ that $p\mid a_n$, we have $i_0\mid n$ and $\nu_p (a_n)=\nu_p (a_{i_0})$ (checking modulo $p^{2\nu_p (a_{i_0})}$.) Note that this also implies $a_m\mid a_n$ for all $m\mid n$. Now, back to the problem, suppose for every prime $p\mid a_n$, there exists $1\leqslant i<n$ that $p\mid a_i$. If there exists prime number $r$ that $\nu_r (n)\geq 2$, we get that $d\mid \frac{n}{r}$ for all proper divisor $d$ of $n$. This gives $\nu_p (a_{\frac{n}{r}}) =\nu_p (a_n)$ for all $p\mid a_n$. So, $a_{\frac{n}{r}} \geqslant a_n$. But it's clear that the sequence is strictly increasing, hence contradiction. If there's no such prime divisor $r$, i.e. $n$ can be written as product of distinct prime numbers. If $n$ is prime, we get that $\nu_p (a_1)=\nu_p (a_n)$ for all $p\mid a_n$. So, $a_1\geqslant a_n$, contradiction. Now, we're left with the case that $n$ has two distinct prime factors, say they're $u$ and $v$. We get that $d\mid \frac{n}{u}$ or $d\mid \frac{n}{v}$ for all proper divisor $d$ of $n$. Hence, for all $p\mid a_n$, either $\nu_p (a_{\frac{n}{u}}) =\nu_p (a_n)$ or $\nu_p (a_{\frac{n}{v}}) =\nu_p (a_n)$. This gives $$a_n\leqslant \mathrm{lcm} (a_{\frac{n}{u}} ,a_{\frac{n}{v}} )=\frac{ a_{\frac{n}{u}}a_{\frac{n}{v}}}{\gcd (a_{\frac{n}{u}},a_{\frac{n}{v}})}\leqslant a_{\frac{n}{u}}a_{\frac{n}{v}}.$$But we've $a_{i+1}> a_i^2$ for all $i$, so $a_n>a_{\frac{n}{u}}a_{\frac{n}{v}}$, this contradict the above inequality, done.

First, note that $c$ always divides $a_i$, so define the sequence $b_i=a_i/c$, so the condition becomes $b_{n+1}=c^2b_n(b_n^2-4b_n+5)+1$. We wish to show that $b_i$ is divisible by a new prime. For a prime $p$ which divides the sequence, consider $k$, the minimal index such that $p|b_k$. Then, $b_{k+1}\equiv c^2\cdot 0+1\equiv 1\equiv b_1\pmod p$. As $\{b_i\}$ are defined based on only the previous term, we have that the residues mod $p$ are cyclic with minimum period $k$, and $p|b_k\implies p|b_{ik}$. Actually, if $p|b_k$, consider the sequence $\pmod{p^{\nu_p(b_k)+1}}$. We have that $$b_{k+1}\equiv c^2b_n^3-4c^2b_n^2+5c^2b_n+1\equiv 5c^2b_n+1\pmod{p^{\nu_p(b_k)+1}}$$$$b_{k+2}\equiv c^2(5c^2b_n+1)(-10c_n^2b_n+2)+1\equiv 2c^2+1\equiv b_2\pmod{p^{\nu_p(b_k)+1}}$$So, $b_{ik}\equiv b_k\pmod{p^{\nu_p(b_k)+1}}$, so we actually have that $\nu_p(b_k)=\nu_p(b_{k+1})$. Hence, if we have $b_n$ has no new prime factors, we must have that $$b_n=\mathrm{lcm}_{d|n,d<n}(b_d)\le \prod_{d|n,d<n} b_d\le (b_{n/2})^{n/2}$$However, note that the recursion is cubic. Hence, $b_i=O(b_{i-1}^3)$, so $b_n$ is of the order $(b_{n/2})^{3^{n/2}}\gg (b_{n/2})^{n/2}$. So, we have a contradiction and $b_i$ must have a new prime.

I can't believe that this is N7!

CLAIM 1. $a_{(i,j)}=(a_i,a_j)$ for all positive integers $i,j$. Proof. Suppose $i>j$.Let $i=aj+b$ where $0\leq b\leq j-1$ then if $n$ is a fixed integer, $$a_{n+1}=a_n^3-4c\cdot a_n^2+5c^2\cdot a_n+c\equiv c=a_1\pmod{a_n}$$Hence for all integers $k$ we have $a_{n+k}\equiv a_k\pmod{a_n}$. Hence $a_i\equiv a_b\pmod{a_j}$. By Euclidean algorithm we have $$a_{(i,j)}=(a_i,a_j)$$as desired. $\blacksquare$ CLAIM 2. If $n|i$ then $a_i\equiv a_n\pmod{a_i^2}$ Proof. Notice that $$a_{n+1}=5c^2\cdot a_n+c\pmod{a_n^2}$$Hence \begin{align*} 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\\ &\equiv (15c^4-40c^4+25c^4)a_n+2c^2+c\\ &\equiv 2c^2+c\\ &\equiv a_2\pmod {a_i^2} \end{align*}Hence for all integers $k\geq 2$ we have $a_{n+k}\equiv a_k\pmod{a_n^2}$ Hence $a_{i}\equiv a_n\pmod{a_n^2}$ as desired. $\blacksquare$ Now suppose there exists an integer $m$ such that for all $p|a_m$, there exists $1\leq i\leq m-1$ such that $p|a_i$. Let $p_1,p_2,...,p_k$ be the prime factors of $m$. And let $q_i=\frac{m}{p_i}$ for all $1\leq i\leq k$。 CLAIM 3. Let $p$ be a prime factor of $a_m$, then $p|a_{{q_i}}$ for some $1\leq i\leq m-1$. Moreover $v_p(a_m)=v_p(a_{q_i})$ Proof. Suppose not, then there exists $1\leq i\leq n$ such that $p|a_n$. From CLAIM 1, we have $p|a_{(m,n)}$. Notice that $(m,n)\neq m$, hence there exists $1\leq i\leq k-1$ such that $(m,n)|a_{q_i}$, which implies $p|a_{q_i}$. Contradiction. Now from claim 2 we have $$a_m\equiv a_{q_i}\pmod{a_{q_i}^2}$$which clearly implies $v_p(a_m)=v_p(a_{q_i})$$\blacksquare$ CLAIM 4. $a_m\geq (a_{q_i})^{2^{p_i}}$ Proof. Notice that for $(c,n)\neq(1,2)$ we can easily show using some algebra that $$a_{n+1}=a_n^3-4c\cdot a_n^2+5c^2\cdot a_n+c\geq a_n^2$$Therefore if $(c,q_i)\neq (1,2)$ then the inequality easily follows. If $c=1$ and $q_i=2$, then $m\geq 4$. Since $a_4=183\geq 3^4=a_2^4$, we are also done. $\blacksquare$ Now from CLAIM 3 we have $$a_m<\prod_{i=1}^ka_{q_i}$$From claim 4 we have $$1<\sum_{i=1}^k\frac{1}{2^{p_i}}<\sum_{i=1}^{\infty}\frac{1}{2^i}=1$$contradiction.

Solved with Pujnk Define $b_n = \frac{a_n}{c}$ so that $b_1 = 1$ and we have $b_{n+1} = c^2b_n(b_n^2 - 4b_n + 5) + 1$ Claim 1: $b_{m+n} \equiv b_m \pmod{b_n}$ Proof: We will induct on $m$. The base case is easy since $b_{m+1} = c^2b_m(b_m^2 - 4b_m + 5) + 1 \equiv 1 = b_1 \pmod b_m$. Now, for the inductive step, suppose its true until $k-1$, then we have $b_{m+k} = c^2b_{m+k-1}(b_{m+k}^2 - 4b_{m+k}+5)+1 \equiv c^2b_{k-1}(b_{k-1}^2 - 4b_{k-1}+5)+1 = b_k \pmod {b_m}$ and so the claim holds. $\square$ Claim 2: $b_{m+n} \equiv b_m \pmod{b_n^2}$ for $m \ge 2$ Proof: Literally same as the previous claim except the calculations are very annoying. See that $b_{m+1} = c^2b_m(b_m^2 - 4b_m + 5)+1 \equiv 5c^2b_m+1 \pmod {b_m^2}$ So, $b_{m+2} = c^2b_{m+1}(b_{m+1}^2 - 4b_{m+1}+5)+1 \equiv c^2(5c^2b_m+1)((5c^2b_m+1)^2 - 4(5c^2b_m+1) + 5)+1$ $= c^2(5c^2b_m+1)(25c^4b_m^2 + 1 + 10c^2b_m - 20c^2b_m + 1)+1$ $\equiv 2c^2(5c^2b_m+1)(1-5c^2b_m)+1$ $= 2c^2(1-25c^4b_m^2)+1$ $\equiv 2c^2 + 1 \pmod {b_m}$ The induction step is the same, suppose its true till $k-1$, then we have $b_{m+k} = c^2b_{m+k-1}(b_{m+k}^2 - 4b_{m+k}+5)+1 \equiv c^2b_{k-1}(b_{k-1}^2 - 4b_{k-1}+5)+1 = b_k \pmod {b_m^2}$, which proves the claim. $\square$ Now, we are almost done. Note that if for a prime $p$, $b_i$ is the smallest index such that $p | b_i$, then we can have $p | b_j$ if and only if $p | j-i$ by Claim 1 and further the $v_p$ cannot increase by Claim 2. So, if we show that $b_n$ is larger than the product of $b_i$ with which is has a common factor, then we are done since then it must have a new prime in it. But by induction, we can just show that $a_n > a_{n-1}a_{n-2}...a_2a_1$, and so $b_n$ must indeed have a new prime, meaning $a_n$ must have a new prime, as desired. $\blacksquare$

ThE-dArK-lOrD wrote: It's not hard to prove that: for every prime number $p$ that $p\mid a_n$ for some $n$, there exists index $i_0$ that $p\mid a_{i_0}$ and $p\nmid a_j$ for all $1\leqslant j<i_0$, and For all $n$ that $p\mid a_n$, we have $i_0\mid n$ and $\nu_p (a_n)=\nu_p (a_{i_0})$ (checking modulo $p^{2\nu_p (a_{i_0})}$.) How to prove this ?

MatBoy-123 wrote: How to prove this ? Read any of the other solutions?, if your question is why must we have $p | a_n \implies i_0 | a_n$, then suppose that $n \equiv r \pmod {i_0}$, $r \neq 0$ then since we have $a_{m+n} \equiv a_m \pmod {a_n}$, we have $p | a_r$ with $r < i_0$, which is impossible by definition.

hajimbrak wrote: 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 $n\ge 1$. Prove that for each integer $n \ge 2$ there exists a prime number $p$ dividing $a_n$ but none of the numbers $a_1 , \ldots , a_{n -1}$ . Proposed by Austria Positing for storage:

Let $b_n=\frac{a_n}{c}$ for positive integers $n$, and let $b_0=0$. Then, we get the recurrence \[b_{n+1}=c^2\left(b_n^3-4b_n^2+5b_n\right)+1.\]Notice that $b_n$ is a positive integer for all positive integers $n$. The key claim is as follows: Claim: For any prime $p$, if there exists a minimal positive integer $k$ such that $p \mid b_k$, then $\nu_p(b_k) \geq \nu_p(b_n)$ for any positive integer $n$. Proof: Let $r=\nu_p(b_k)$, and let $b_k=qp^r$. We calculate $b_{k+1} \equiv 5c^2qp^r+1 \pmod{p^{r+1}}$ and $b_{k+2} \equiv 2c^2+1=b_2 \pmod{p^{r+1}}$. Thus, $b_k$ is periodic $\!\!\!\mod{p^{r+1}}$ when $k \geq 2$, so $p^{r+1} \nmid b_n$ for all positive integers $n$, proving the claim. If there exists a positive integer $n \geq 2$ contradicting the problem statement, then we must have $\nu_p(b_n) \leq \max(\nu_p(b_1),\nu_p(b_2),\ldots,\nu_p(b_{n-1}))$ for all primes $p$ by the claim, which means \[b_n \leq b_1b_2 \cdots b_{n-1}.\]However, this can be verified to be impossible (the LHS is $\Theta(b_n)$ while the RHS is $\Theta\left(\sqrt{b_n}\right)$), so we are done.

My solution CLAIM 1: $gcd(a_n,a_m) \mid a_{gcd(n,m)}$ for all positive intergers $n,m$ Proof Suppose that $n>m$, then for all $t \mid gcd(a_n,a_m)$ we have \[a_{n}=a_{n-1}^3-4c\cdot a_{n-1}^2+5c^2\cdot a_{n-1}+c=f(a_{n-1})=f(a_{n-2}^3-4c\cdot a_{n-2}^2+5c^2\cdot a_{n-2}+c)=f(f(a_{n-2}))= \ldots = f^{n-m}(a_m) \equiv f^{n-m}(0) \equiv f^{n-m-1}(c) \equiv a_{n-m} \pmod t \]Then $t \mid a_{n-m}$, and from Euclidean algorithm, we have $t \mid a_{gcd(n,m)} \implies gcd(a_n,a_m) \mid a_{gcd(n,m)}$ CLAIM 2: $a_n \equiv a_k \pmod {{a_k}^2}$ for all $k \mid n$ and $k \ge 2$, therefore $v_p(a_n) = v_p(a_k)$ for all $k \mid n$ and $k \ge 2$ Proof We see this from induction. First we have $$ a_{k+1} \equiv 5c^2a_k+c \pmod {{a_k}^2} $$Then we implies that \begin{align*} a_{k+2}&\equiv(5c^2\cdot a_k+c)^3-4c\cdot(5c^2\cdot a_k+c)^2+5c^2\cdot(5c^2\cdot a_k+c)+c\\ &\equiv (15c^4-40c^4+25c^4)a_k+2c^2+c\\ &\equiv 2c^2+c\\ &\equiv a_2\pmod {a_k^2} \end{align*}Following from induction, we have $a_{k+i} \equiv a_i \pmod {a_k^2}$ Thus $a_k \equiv a_{k+k} \equiv a_{k+k+k} \equiv \ldots \equiv a_{ \frac{n}{k} \cdot k} \equiv a_n \pmod {a_k^2}$ From CLAIM 1 and CLAIM 2, we have $gcd(a_n,a_m) = a_{gcd(n,m)}$ Now come back to the main problem, suppose the contrary that there doesn't exist a prime number $p$ dividing $a_n$ but none of the numbers $a_1 , \ldots , a_{n -1}$; which means for every prime $p \mid a_n$, there exists $ 1 \le i \le n-1$ such that $p \mid a_i$ Hence $p^{v_p(a_n)} \mid a_{gcd(n,i)}$, then we implie that $a_n \le a_{n-1}a_{n-2}a_{n-3} \ldots a_2 a_1$ (*) On the other hand, let's consider $g(a_n)=f(a_n)-a_n^2= a_n^3-(4c+1)a_n^2+5c^2a_n+c$, we have $$g'(x)=3a_n^2-2 \cdot (4c+1)a_n +5c^2 = 0 \Leftrightarrow a_n= \frac{1}{3}\left( {4 \pm \frac{{(c + 4)}}{{\sqrt {{c^2} + 8c + 1} }}} \right) $$Also notice that \[{\left( {\frac{{c + 4}}{{\sqrt {{c^2} + 8c + 1} }}} \right)^2} = \frac{{{c^2} + 8c + 16}}{{{c^2} + 8c + 1}} = 1 + \frac{{15}}{{{c^2} + 8c + 1}} \in \left( {1,\frac{5}{2}} \right] \to \frac{1}{3}\left( {4 \pm \frac{{(c + 4)}}{{\sqrt {{c^2} + 8c + 1} }}} \right) \in \left[ {\left\lceil {\frac{1}{3}(4 - \sqrt {\frac{5}{2}} )} \right\rceil ,\left\lfloor {\frac{1}{3}(4 + \sqrt {\frac{5}{2}} )} \right\rfloor } \right] = \left[ {0,1} \right]\]Hence we have $g'(a_n)>0 \implies g(a_n) \ge g(0) =c >0 \implies a_{n+1} > a_n^2$ for all $n$ Thus $a_{n+1} >a_n.a_n >a_n.a_{n-1}^2 >a_n.a_{n-1}.a_{n-1}> \ldots > a_na_{n-1}a_{n-2} \ldots a_1$, which gives a contradiction to (*)

Hensel lifting but you realize the vp stack you're lifting is a literal feather and the question died before you could have any fun. Notice by function properties, the condition $p \mid a_n$ is periodic modulo some "order" of $p$. Furthermore, notice if we compose the polynomial when using $b_n = \frac{a_n}{c}$ and $P(x) = c^2(x^3-4x^2+5x)+1$ looking solely at linear and constant coefficients, we get that the next linear term is a multiple of the previous linear term. So when plugging in $a = 1, b = 0$ for $ax + b$, notice that by the second iteration, $a = 0$ so the linear term is forever 0. Even when we are looking at the first iteration, $P(0) = c$ and the polynomial's linear term, $5c^2$, vanishes mod $p$. Notice that $v_p$ cannot increase when iterating a polynomial on zero without a linear term and with a constant term that p divides, where that polynomial is $P(x)$ composed until it has a constant term that vanishes mod $p$. Hence the sequence cannot "lift". So assuming contradiction, you must have $a_n \mid \text{lcm}(a_1, a_2, \ldots ,a_{n-1})$. Trivial bounding finishes.

Let $b_n = \frac{a_n}{c}$. We receive $b_1 = 1, b_{n+1} = c^2b_n(b_n^2-4b_n+5)+1$. Let $b_k$ be the first term in $b_n$ such that $p|b_k$. Trivially, $b_{n+k} = b_{n} \pmod p$. Therefore the only terms that are divisible by $p$, are the terms $b_m$ where $k|m$. We will now show that $v_p(b_m)$ is constant as $m$ varies over all multiples of $k$. Taking, the sequence $\pmod{p^{r+1}}$ where $r = v_p(b_m)$, we receive that $b_{k+2} = b_{2} \pmod{p^{r+1}}$. Therefore the sequence repeats, implying that $v_p(b_m)$ is constant. We will now continue FTSOC. Assume there exists a $b_i$ that does not introduce a new prime. Because there doesn't exist a prime $q$ such that $v_q(b_i)$ that is strictly greater than $v_q(b_k)$ for all $k$, except $i$, $b_i$ must divide $\prod_{n=1}^{i-1} b_n$. This implies $b_i \le \prod_{n=1}^{i-1} b_n$. However the LHS is a polynomial in terms of $c$ with a larger degree than the RHS, so this is absurd. $\blacksquare$

Similar to other solutions (disappointing N7). Let $b_n=\frac{a_n}c$ (this is due through easy induction we get $c \mid a_n$). And so we have \[b_{n+1}=c^2b_n^3-4c^2b_n^2+5c^2b_n+1\]Now let $\text{ord}(p)$ be the minimal $i>0$ such that $p \mid b_{\text{ord}(p)}$ (if it exists) then see that \[p \mid b_j \iff \text{ord}(p) \mid j\]Let $e=\nu_p(b_{\text{ord}(p)})$. Now see that $b_{\text{ord}(p)+1} \equiv 5c^2b_{\text{ord}(p)}+1 \pmod {p^{e+1}}$ and so \begin{align*} b_{\text{ord}(p)+2} & \equiv c^2(15c^2b_{\text{ord}(p)}+1)-4c^2(10c^2b_{\text{ord}(p)}+1)+5c^2(5c^2b_{\text{ord}(p)}+1)+1 \\ & =2c^2+1=b_2 \pmod {p^{e+1}} \end{align*}Now since $p \nmid b_1$ we get that \[b_{m \cdot \text{ord}(p)} \equiv b_{\text{ord}(p)} \pmod {p^{e+1}} \implies \nu_p(b_{m\cdot\text{ord}(p)})=e \text{ }(\clubsuit)\]And now say for some $k$ we have \[\text{rad}(b_k) \mid b_1\dots b_{k-1}\]And now because of $\clubsuit$ and the fact that $k \geq 2$, we get that \[b_k \mid b_1\dots b_{k-1} \implies b_1\dots b_{k-1} \geq b_k>b_{k-1}^2>b_{k-1}b_{k-2}^2 > \dots \dots \geq b_{k-1}\dots b_1\]And this is because $b_n>b_{n-1}^2$ for all $n \geq 2$, so we get a contradiction.