Let $({{x}_{n}})$ be an integer sequence such that $0\le {{x}_{0}}<{{x}_{1}}\le 100$ and
$${{x}_{n+2}}=7{{x}_{n+1}}-{{x}_{n}}+280,\text{  }\forall n\ge 0.$$
a) Prove that if ${{x}_{0}}=2,{{x}_{1}}=3$ then for each positive integer $n,$ the sum of divisors of the following number is divisible by $24$   $${{x}_{n}}{{x}_{n+1}}+{{x}_{n+1}}{{x}_{n+2}}+{{x}_{n+2}}{{x}_{n+3}}+2018.$$
b) Find all pairs of numbers $({{x}_{0}},{{x}_{1}})$ such that ${{x}_{n}}{{x}_{n+1}}+2019$ is a perfect square for infinitely many nonnegative integer numbers $n.$ 

### Part (a)

1. **Lemma:** If \( n \in \mathbb{N} \) such that \( 24 \mid n + 1 \), then the sum of its positive divisors \( \sigma(n) \) is divisible by \( 24 \).

   **Proof:** Note that if \( 24 \mid n + 1 \), then \( n \equiv 2 \pmod{3} \) and \( n \equiv 7 \pmod{8} \). Thus, \( n \) is not a perfect square. Then 
   \[
   \sigma(n) = \sum_{d \mid n, d^2 < n} \left(d + \frac{n}{d}\right) = \sum_{d \mid n, d^2 < n} \left(\frac{d^2 + n}{d}\right).
   \]
   Since \( d^2 \equiv 1 \pmod{3} \) and \( d^2 \equiv 1 \pmod{8} \) for all positive divisors \( d \) of \( n \), \( d^2 + n \equiv 0 \pmod{24} \). Thus, \( \sigma(n) \equiv 0 \pmod{24} \).

2. Now we need to show that for all positive integers \( n \),
   \[
   (x_{n}x_{n+1} + x_{n+1}x_{n+2} + x_{n+2}x_{n+3} + 2018) + 1 \equiv 0 \pmod{24} \iff x_{n}x_{n+1} + x_{n+1}x_{n+2} + x_{n+2}x_{n+3} \equiv -3 \pmod{24}.
   \]

3. If we take \( (x_n) \pmod{8} \), then we get \( 2, 3, 3, 2, 3, 3, \ldots \), and for all \( n \) we have
   \[
   x_{n}x_{n+1} + x_{n+1}x_{n+2} + x_{n+2}x_{n+3} \equiv 2 \cdot 3 + 3 \cdot 3 + 3 \cdot 2 \equiv -3 \pmod{8}.
   \]

4. If we take \( (x_n) \pmod{3} \), then we get \( 2, 0, 2, 0, 2, 0, \ldots \), and for all \( n \) we have
   \[
   x_{n}x_{n+1} + x_{n+1}x_{n+2} + x_{n+2}x_{n+3} \equiv 2 \cdot 0 + 0 \cdot 2 + 2 \cdot 0 \equiv 0 \pmod{3}.
   \]

5. Thus, combining the results, we get
   \[
   x_{n}x_{n+1} + x_{n+1}x_{n+2} + x_{n+2}x_{n+3} \equiv -3 \pmod{24}
   \]
   for all \( n \).

### Part (b)

1. **Lemma:** \( x_{n+1}^2 - x_{n}x_{n+2} - 280x_{n+1} = C \) for some integer constant \( C \).

   **Proof:** 
   \[
   x_{n+1}^2 - x_{n}x_{n+2} - 280x_{n+1} = x_{n+1}(x_{n+1} - 280) - x_{n}(7x_{n+1} - x_n + 280)
   \]
   \[
   = x_{n+1}(7x_{n} - x_{n-1}) - x_{n}(7x_{n+1} - x_{n} + 280) = x_{n}^2 - x_{n-1}x_{n+1} - 280x_{n}
   \]
   \[
   = \ldots = x_1^2 - x_0x_2 - 280x_1 = C.
   \]

2. Now,
   \[
   x_{n+1}^2 - x_{n}(7x_{n+1} - x_n + 280) - 280x_{n+1} = C
   \]
   \[
   \iff (x_{n+1} + x_{n} - 140)^2 = 9(x_{n+1}x_{n} + 2019) + C + 1429.
   \]

3. If \( x_{n+1}x_{n} + 2019 \) is a perfect square, then
   \[
   x_{n+1} + x_{n} - 140 + 3\sqrt{x_{n+1}x_{n} + 2019} \mid C + 1429.
   \]
   Notice that \( (x_n) \) is an increasing unbounded sequence, so there exists an integer \( m \) such that
   \[
   x_{n+1} + x_{n} - 140 + 3\sqrt{x_{n+1}x_{n} + 2019} > C + 1429
   \]
   for all \( n \geq m \). So if \( x_{n+1}x_{n} + 2019 \) is a perfect square for infinitely many \( n \),
   \[
   C + 1429 = 0 \iff (x_{n+1} + x_{n} - 140)^2 = 9(x_{n+1}x_{n} + 2019)
   \]
   for all \( n \).

4. Then,
   \[
   (x_1 + x_0 - 140)^2 \geq 2019 \cdot 9 > 134^2,
   \]
   thus,
   \[
   |x_0 + x_1 - 140| \geq 135.
   \]
   Since \( x_0 + x_1 - 140 < 59 < 135 \), \( x_0 + x_1 - 140 < 0 \) and
   \[
   140 - x_0 - x_1 \geq 135 \iff x_0 + x_1 \leq 5.
   \]
   From \( 2x_0 < x_0 + x_1 \leq 5 \), we get \( x_0 = 0, 1, 2 \).

5. Remember that
   \[
   C = x_1^2 - x_0x_2 - 280x_1 = x_1^2 - x_0(7x_1 - x_0 + 280) - 280x_1 = x_0^2 + x_1^2 - 7x_0x_1 - 280(x_0 + x_1) = -1429.
   \]
   This implies
   \[
   x_0^2 + x_1^2 \equiv 6 \pmod{7}.
   \]
   If \( x_0 = 0 \), \( x_1^2 \equiv 6 \pmod{7} \), which is impossible.
   If \( x_0 = 1 \), \( x_1^2 \equiv 5 \pmod{7} \), which is impossible.
   If \( x_0 = 2 \), since \( x_0 + x_1 \leq 5 \) and \( x_0 < x_1 \), \( x_1 = 3 \). In this case,
   \[
   C = x_0^2 + x_1^2 - 7x_0x_1 - 280(x_0 + x_1) = 4 + 9 - 42 - 280(5) = -1429.
   \]

Therefore, \( (x_0, x_1) = \boxed{(2, 3)} \) is the only pair such that \( x_{n}x_{n+1} + 2019 \) is a perfect square for infinitely many \( n \).