VMO

Lemma 1. If a positive integer $n$ satisfies $24|n+1$ then the sum of its positive divisors $\sigma (n)$ is divisible by 24.

Proof. Indeed, if $d$ is a divisor of $n$ then $\frac{n}{d}$ is also a divisor of $n$. Because $n\equiv 2(\mathrm{mod}3)$ so it cannot be a perfect square, which means the sum of its divisors can be divided into pairs of the form $$d+\frac{n}{d}=\frac{d^2+n}{d}.$$ Note that $n\equiv 2(\mathrm{mod}3)$ and $d^2\equiv 1(\mathrm{mod}3)$ so the sum above is divisible by 3. On the other hand, $n\equiv 7(\mathrm{mod}3)$ and $d\equiv 1,3,5,7(\mathrm{mod}8)$, which implies $d^2\equiv 1(\mathrm{mod}8)$ so the above sum is also divisible by 8. Since $(3,8)=1$ then the above sum is divisible by 24 and the lemma is proved. $\square$

a. Back to our problem, denote $y_n=x_n{x}_{n+1}+x_{n+1}{x}_{n+2}+x_{n+2}{x}_{n+3}$, we need to prove that $\sigma (y_n+2018)$ is divisible by 24. We also have $2018\equiv 2(\mathrm{mod}24)$ so according to the lemma, we need to show $y_n\equiv -3(\mathrm{mod}24)$.

Consider the period of the remainder when being divided by 3 of the sequence $(x_n)$, note that $x_{n+2}\equiv x_{n+1}-x_n+1(\mathrm{mod}3)$ we have 2, 0, 2, 0, 2, 0, ... this sequence is periodic with period 2 and $$y_n\equiv 0\cdot 2+2\cdot 0+0\cdot 2=0(\mathrm{mod}3).$$

Similarly, consider the remainder of $(x_n)$ when being divided by 8, note that $x_{n+2}\equiv -x_{n+1}-x_n(\mathrm{mod}8)$ we have $$2,3,3,2,3,3,2,3,3,\dots$$ which means this sequence is periodic with period 3 and $$y_n\equiv 2\cdot 3+3\cdot 3+3\cdot 2=5(\mathrm{mod}8).$$ It follows that $(y_n+3)$ is both divisible by 3, and 8, so $y_n\equiv -3(\mathrm{mod}24)$ and a) is proved.

b. Now, we prove the following lemma.

Lemma 2. Consider the integer sequence $(z_n)$ satisfying $z_{n+2}=a{z}_{n+1}-z_n+b$ then the following quantity is constant $$z_{n+1}^2-z_n{z}_{n+2}-b{z}_{n+1}\text{ for all }n\ge 0.$$ Proof. Indeed, we have the following transformation $$\begin{array}{rl}{z}_{n+1}^2-z_n{z}_{n+2}-b{z}_{n+1}& =z_{n+1}(z_{n+1}-b)-z_n(a{z}_{n+1}-z_n+b)\\ & =z_{n+1}(a{z}_n-z_{n-1})-z_n(a{z}_{n+1}-z_n+b)\\ & =z_n^2-z_{n-1}{z}_{n+1}-b{z}_n.\end{array}$$ The above equality holds for all $n\ge 0$ so $z_{n+1}^2-z_n{z}_{n+2}-b{z}_{n+1}=z_1^2-z_0{z}_2-b{z}_1=c$ where $c$ is a constant. $\square$

Thus, there exists $C\in \mathbb{Z}$ such that $x_{n+1}^2-x_n{x}_{n+2}-280x_{n+1}=C$. We have $$\begin{array}{rl}{x}_{n+1}^2-x_n(7x_{n+1}-x_n+280)-280x_{n+1}& =C\\ x_{n+1}^2+x_n^2-7x_{n+1}{x}_n-280(x_{n+1}+x_n)& =C\\ (x_{n+1}+x_n-140{)}^2& =9(x_{n+1}{x}_n+2019)-9\cdot 2019+C+140^2\\ u_n^2& =v_n^2+C+1429,\end{array}$$ where $u_n=x_{n+1}+x_n-140$, $v_n=3\sqrt{x_{n+1}{x}_n+2019}$ for all $n\ge 0$.

Since $(x_n)$ is an increasing integer sequence so it is unbounded, thus $(u_n)$ is increasing and unbounded. It is also clear that if $x_n{x}_{n+1}+2019$ is a perfect square then $v_n\in {\mathbb{Z}}^{+}$.

Hence, $u_n+v_n|C+1429$ for infinite values of $n$. Clearly, this case only happens when $C+1429=0$ so $$(x_{n+1}+x_n-140{)}^2=9(x_{n+1}{x}_n+2019),\forall n\in {\mathbb{Z}}^{+}.$$ We have $(x_0+x_1-140{)}^2\ge 2019\cdot 9>44^2\cdot 3^2=132^2$ so $|140-x_0-x_1|\ge 133$, but $0\le x_0<x_1<101$ then $140-(x_0+x_1)\ge 133$, i.e. $x_0+x_1\le 7$. We also have $$C=x_1^2+x_0^2-7x_1{x}_0-280(x_1+x_0)=-1429.$$ Notice that $x_1^2+x_0^2\le 49$ so $-1429=C<49-280(x_1+x_0)$, which implies $x_0+x_1\ge 5$. By direct checking, the case $x_1+x_0=7$ and $x_1+x_0=6$ has no solution. So $x_0+x_1=5$, which implies $x_0{x}_1=6$ so it's easy to see that $x_0=2,x_1=3$.

Therefore, $(x_0,x_1)=(2,3)$ is the only satisfying pair. $\square$