Given is the equation $x^2+y^2-axy+2=0$ where $a$ is a positive integral parameter.

$i.$Show that,for $a\neq 4$ there exist no pairs $(x,y)$ of positive integers satisfying the equation.

$ii.$ Show that,for $a=4$ there exist infinite pairs $(x,y)$ of positive integers satisfying the equation,and determine those pairs.
### Part (i): Show that for \( a \neq 4 \), there exist no pairs \((x,y)\) of positive integers satisfying the equation.

1. Suppose there exist \( a \neq 4 \) such that there exist \((x,y) = (u_1, v_1)\) that satisfy the equation \( u_1^2 + v_1^2 - a u_1 v_1 + 2 = 0 \).

2. Assume \((u, v)\) is a pair of positive integers with the smallest value of \( u + v \) that satisfies the equation. Then \( u \) is one of the roots of the quadratic equation in \( x \):
   \[
   x^2 - (av)x + (v^2 + 2) = 0
   \]

3. By Vieta's formulas, the other root of this quadratic equation is:
   \[
   av - u = \frac{v^2 + 2}{u}
   \]
   Since \( u \) and \( v \) are positive integers, \(\frac{v^2 + 2}{u}\) must also be a positive integer.

4. Therefore, \((x, y) = \left( \frac{v^2 + 2}{u}, v \right)\) is also a solution. Since \((u, v)\) was chosen to minimize \( u + v \), we must have:
   \[
   \frac{v^2 + 2}{u} \geq u
   \]

5. If \( u \geq v + 1 \), then:
   \[
   \frac{v^2 + 2}{u} \geq v + 1
   \]
   This implies:
   \[
   v^2 + 2 \geq u(v + 1)
   \]
   Since \( u \geq v + 1 \), we have:
   \[
   v^2 + 2 \geq (v + 1)(v + 1) = v^2 + 2v + 1
   \]
   This simplifies to:
   \[
   2 \geq 2v + 1
   \]
   Which is impossible for positive integers \( v \).

6. Therefore, \( u \) must equal \( v \). Substituting \( u = v \) into the original equation:
   \[
   u^2 + u^2 - a u^2 + 2 = 0 \implies 2u^2 - a u^2 + 2 = 0 \implies (2 - a)u^2 + 2 = 0
   \]
   For this to hold, \( a \) must be 4, which contradicts the assumption \( a \neq 4 \).

Thus, for \( a \neq 4 \), there exist no pairs \((x, y)\) of positive integers satisfying the equation.

### Part (ii): Show that for \( a = 4 \), there exist infinite pairs \((x, y)\) of positive integers satisfying the equation, and determine those pairs.

1. Let \( a = 4 \). The equation becomes:
   \[
   x^2 + y^2 - 4xy + 2 = 0
   \]

2. We need to show that there are infinite pairs \((x, y)\) of positive integers satisfying this equation. Consider the sequence defined by:
   \[
   a_1 = 1, \quad a_2 = 3, \quad a_{n+2} = 4a_{n+1} - a_n
   \]

3. We will show that \((x, y) = (a_n, a_{n+1})\) satisfies the equation for all \( n \in \mathbb{Z}^+ \).

4. Base cases:
   - For \( n = 1 \), \((x, y) = (a_1, a_2) = (1, 3)\):
     \[
     1^2 + 3^2 - 4 \cdot 1 \cdot 3 + 2 = 1 + 9 - 12 + 2 = 0
     \]
   - For \( n = 2 \), \((x, y) = (a_2, a_3) = (3, 11)\):
     \[
     3^2 + 11^2 - 4 \cdot 3 \cdot 11 + 2 = 9 + 121 - 132 + 2 = 0
     \]

5. Inductive step: Assume \((a_n, a_{n+1})\) satisfies the equation. We need to show \((a_{n+1}, a_{n+2})\) also satisfies the equation:
   \[
   a_{n+2} = 4a_{n+1} - a_n
   \]
   Substituting into the equation:
   \[
   a_{n+1}^2 + a_{n+2}^2 - 4a_{n+1}a_{n+2} + 2 = 0
   \]
   Using the recurrence relation:
   \[
   a_{n+2} = 4a_{n+1} - a_n
   \]
   We have:
   \[
   a_{n+1}^2 + (4a_{n+1} - a_n)^2 - 4a_{n+1}(4a_{n+1} - a_n) + 2 = 0
   \]
   Simplifying:
   \[
   a_{n+1}^2 + 16a_{n+1}^2 - 8a_{n+1}a_n + a_n^2 - 16a_{n+1}^2 + 4a_{n+1}a_n + 2 = 0
   \]
   \[
   a_{n+1}^2 + a_n^2 - 4a_{n+1}a_n + 2 = 0
   \]
   Which is the same as the original equation, confirming the inductive step.

Thus, for \( a = 4 \), there exist infinite pairs \((x, y)\) of positive integers satisfying the equation, given by the sequence \((a_n, a_{n+1})\).

The final answer is \( \boxed{ (a_n, a_{n+1}) } \) where \( a_1 = 1, a_2 = 3, a_{n+2} = 4a_{n+1} - a_n \).