rafaello wrote: Determine all infinite sequences $(a_1,a_2,\dots)$ of positive integers satisfying \[a_{n+1}^2=1+(n+2021)a_n\]for all $n \ge 1$. $a_n\ge 1$ $\implies$ $a_{n+1}^2>n$ and so sequence is not upperbounded. So $\exists N_0$ such that $a_{N_0+1}>a_{N_0}$. $a_{n+2}^2-a_{n+1}^2=(1+(n+2022)a_{n+1})-(1+(n+2021)a_n)$ $=(n+2021)(a_{n+1}-a_n)+a_{n+1}$ So $a_{n+1}>a_n$ implies $a_{n+2}>a_{n+1}$ So $a_n$ is strictly increasing for $n>N_0$. $a_{n+1}\ge a_n+1$ $\implies$ $a_{n+1}-(n+1)\ge a_n-n$ and so $a_n-n$ is non decreasing for $n>N_0$ $a_{n+1}\ge a_n+1$ $\implies$ $a_{n+1}^2\ge a_n^2+2a_n+1$ $\implies$ $1+(n+2021)a_n\ge a_n^2+2a_n+1$ $\implies$ $a_n-n\le 2019$ So $a_n-n$ is non decreasing and upperbounded and so constant from a given point. So $a_n=n+c$ $\forall n>N_1$ for some $N_1$ Plugging this in $a_{n+1}^2=1+(n+2021)a_n$, we get $c=2019$ And so $a_n=n+2019$ $\forall n>N_1$. But easy downstream induction using $a_n=\frac{a_{n+1}^2-1}{n+2021}$ shows then : $\boxed{a_n=n+2019\quad\forall n\in\mathbb Z_{>0}}$, which indeed fits.

This problem was proposed by me.

@Tintarn very nice problem! Let $2021=k+2$. We will prove that the only solution is $a_n=n+k$. Take $n$ such that $n+k+2$ is a prime $p$. Then, $p \mid (a_{n+1})^2-1$, hence $a_{n+1} \equiv \pm 1 \pmod p$, which implies that $a_{n+1} \geq p-1$ (it's evident that $a_{n+1} \neq 1$ since this would imply that $a_n=0$). Therefore, $a_{n+1} \geq n+k+1$ for at least one $n$. It's evident that this implies that $a_n \geq n+k$ for all $n$, since by the given relation $a_n \geq n+k \Rightarrow a_{n+1} \geq n+k+1$ and $a_{n-1} \geq n+k-1$. Let $b_n=a_n-n$. Then, $b_n \geq k$ for all $n$. In addition, $$(a_n+1)^2-a_{n+1}^2=(a_n+1)^2-(1+(n+k+2)a_n)=a_n^2-(n+k)a_n=a_n(a_n-n-k) \geq 0,$$therefore $a_{n+1} \leq a_n+1$, implying that $b_{n+1} \leq b_n$ for all $n$. Hence the sequence $b_n$ is decreasing and satisfies $b_n \geq k$ for all $n$, therefore it is eventually constant. Thus, there exists a $t$ such that $b_{n}=b_{n+1}$ for all $n \geq t$, which implies that $a_{n+1}=a_n+1$ for all $n \geq t$. Substituting into the original relation we get that $a_n=n+k$ for all $n \geq t$, and now an easy two-sided induction finishes the problem. In conclusion, it's easy to check that $a_n=n+k$ satisfies the given equation. For our problem, $k=2019$ so $a_n=n+2019$.

n= p-2021 ,p>2021 prime number. ap-2020>p

Assume $x$ such that $a_{x+1} > a_x$. $a_{n+2}^2-a_{n+1}^2 = (n+2021)(a_{n+1}-a_n)+a_{n+1}$ so $a_{n+1} > a_n$ implies $a_{n+2} > a_{n+1}$ so for every $n > x$ we have $a_n$ is increasing sequence. $1+(n+2021)a_n\ge a_n^2+2a_n+1 \implies a_n-n\le 2019$ also we have $a_{n+1}\ge a_n+1 \implies a_{n+1}-(n+1)\ge a_n-n$ so $a_n - n$ must be constant from a point on means $a_n = n + k$. Plugging this with main equation we have $k = 2019$ so from a point on $a_n = n+2019$. assume that point and use induction backward we have $a_n = n + 2019$ for all natural $n$.