Does there exist an irreducible two variable polynomial $f(x,y)\in \mathbb{Q}[x,y]$ such that it has only four roots $(0,1),(1,0),(0,-1),(-1,0)$ on the unit circle.
To determine whether there exists an irreducible two-variable polynomial \( f(x,y) \in \mathbb{Q}[x,y] \) that has only four roots \((0,1), (1,0), (0,-1), (-1,0)\) on the unit circle, we need to analyze the given polynomials and their properties.

1. **Consider the polynomial \( f(x,y) = x^2 + y^2 - 1 + xy \):**
   - This polynomial can be rewritten as:
     \[
     f(x,y) = x^2 + y^2 + xy - 1
     \]
   - To check if \((0,1), (1,0), (0,-1), (-1,0)\) are roots:
     \[
     f(0,1) = 0^2 + 1^2 + 0 \cdot 1 - 1 = 0
     \]
     \[
     f(1,0) = 1^2 + 0^2 + 1 \cdot 0 - 1 = 0
     \]
     \[
     f(0,-1) = 0^2 + (-1)^2 + 0 \cdot (-1) - 1 = 0
     \]
     \[
     f(-1,0) = (-1)^2 + 0^2 + (-1) \cdot 0 - 1 = 0
     \]
   - These points are indeed roots of \( f(x,y) \).
   - However, we need to check if there are any other roots on the unit circle \( x^2 + y^2 = 1 \). For this, we substitute \( y = \pm \sqrt{1 - x^2} \) into \( f(x,y) \):
     \[
     f(x, \sqrt{1 - x^2}) = x^2 + (1 - x^2) + x \sqrt{1 - x^2} - 1 = x \sqrt{1 - x^2}
     \]
     \[
     f(x, -\sqrt{1 - x^2}) = x^2 + (1 - x^2) - x \sqrt{1 - x^2} - 1 = -x \sqrt{1 - x^2}
     \]
   - Both expressions are zero if and only if \( x = 0 \) or \( x = \pm 1 \), which correspond to the given roots. Thus, there are no other roots on the unit circle.

2. **Consider the polynomial \( f(x,y) = x^2 + y^2 - 1 - 2x^2y^2 \):**
   - This polynomial can be rewritten as:
     \[
     f(x,y) = x^2 + y^2 - 1 - 2x^2y^2
     \]
   - To check if \((0,1), (1,0), (0,-1), (-1,0)\) are roots:
     \[
     f(0,1) = 0^2 + 1^2 - 1 - 2 \cdot 0^2 \cdot 1^2 = 0
     \]
     \[
     f(1,0) = 1^2 + 0^2 - 1 - 2 \cdot 1^2 \cdot 0^2 = 0
     \]
     \[
     f(0,-1) = 0^2 + (-1)^2 - 1 - 2 \cdot 0^2 \cdot (-1)^2 = 0
     \]
     \[
     f(-1,0) = (-1)^2 + 0^2 - 1 - 2 \cdot (-1)^2 \cdot 0^2 = 0
     \]
   - These points are indeed roots of \( f(x,y) \).
   - To check for other roots on the unit circle, substitute \( y = \pm \sqrt{1 - x^2} \) into \( f(x,y) \):
     \[
     f(x, \sqrt{1 - x^2}) = x^2 + (1 - x^2) - 1 - 2x^2(1 - x^2) = -x^2(1 - 2x^2)
     \]
     \[
     f(x, -\sqrt{1 - x^2}) = x^2 + (1 - x^2) - 1 - 2x^2(1 - x^2) = -x^2(1 - 2x^2)
     \]
   - Both expressions are zero if and only if \( x = 0 \) or \( x = \pm 1 \), which correspond to the given roots. Thus, there are no other roots on the unit circle.

Since both polynomials \( x^2 + y^2 - 1 + xy \) and \( x^2 + y^2 - 1 - 2x^2y^2 \) have only the four specified roots on the unit circle and no others, we conclude that such irreducible polynomials do exist.

The final answer is \( \boxed{ x^2 + y^2 - 1 + xy } \) or \( x^2 + y^2 - 1 - 2x^2y^2 \)