Let $(a,b)=(a_n,a_{n+1}),\forall n\in\mathbb{N}$ all be positive interger solutions that satisfies
$$1\leq a\leq b$$ and $$\dfrac{a^2+b^2+a+b+1}{ab}\in\mathbb{N}$$
And the value of $a_n$ is [b]only[/b] determined  by the following recurrence relation:$ a_{n+2} = pa_{n+1} + qa_n + r$ 
Find $(p,q,r)$.
1. **Given Conditions and Initial Setup:**
   We are given that $(a, b) = (a_n, a_{n+1})$ for all $n \in \mathbb{N}$ are positive integer solutions satisfying:
   \[
   1 \leq a \leq b
   \]
   and
   \[
   \frac{a^2 + b^2 + a + b + 1}{ab} \in \mathbb{N}
   \]
   Let:
   \[
   k = \frac{a^2 + b^2 + a + b + 1}{ab}
   \]
   Then:
   \[
   a^2 + b^2 + a + b + 1 = kab
   \]

2. **Forming the Quadratic Equation:**
   Rearrange the equation:
   \[
   b^2 + (1 - ka)b + a^2 + a + 1 = 0
   \]
   This is a quadratic equation in \(b\). Let \(b'\) be the other root of this quadratic equation. By Vieta's formulas, we have:
   \[
   b + b' = ka - 1
   \]
   and
   \[
   bb' = a^2 + a + 1
   \]

3. **Properties of \(b'\):**
   Since \(b'\) is an integer, we need \(b'\) to be a positive integer and \(b' \leq a\). This allows us to continue the recursion until \(a = b = 1\).

4. **Recurrence Relation:**
   From the equation \(b + b' = ka - 1\), we can express \(b\) in terms of \(b'\):
   \[
   b = ka - b' - 1
   \]
   Given the recurrence relation \(a_{n+2} = pa_{n+1} + qa_n + r\), we need to determine \(p\), \(q\), and \(r\).

5. **Initial Values and Determining Coefficients:**
   Consider the initial values \(a_1 = 1\), \(a_2 = 1\), and \(a_3 = 3\). We can use these to find \(p\), \(q\), and \(r\):
   \[
   a_3 = pa_2 + qa_1 + r
   \]
   Substituting the values:
   \[
   3 = p \cdot 1 + q \cdot 1 + r
   \]
   Simplifying, we get:
   \[
   p + q + r = 3
   \]

6. **Further Analysis:**
   From the properties of the quadratic equation and the recurrence relation, we can deduce:
   \[
   q = -1 \quad \text{and} \quad r = -1
   \]
   Substituting these into the equation:
   \[
   p - 1 - 1 = 3 \implies p = 5
   \]

Thus, the values of \(p\), \(q\), and \(r\) are:
\[
(p, q, r) = (5, -1, -1)
\]

The final answer is \( \boxed{ (p, q, r) = (5, -1, -1) } \)