1. The problem is to find the number of unordered pairs $(x,y)$ where $x$ and $y$ are integers satisfying the equation $$x^2 - y^2 = 49.$$ 2. Rewrite the equation using the difference of squares factorization: $$x^2 - y^2 = (x-y)(x+y) = 49.$$ 3. Since $x$ and $y$ are integers, both $(x-y)$ and $(x+y)$ must be integers. Let $a = x-y$ and $b = x+y$, so $ab = 49$. 4. The integer factor pairs of 49 are: $(1,49), (7,7), (49,1), (-1,-49), (-7,-7), (-49,-1)$. 5. For each pair $(a,b)$, solve for $x$ and $y$: $$x = \frac{a+b}{2}, \quad y = \frac{b - a}{2}.$$ 6. Check which pairs $(a,b)$ yield integer $x$ and $y$: - $(1,49)$: $x=\frac{50}{2}=25$, $y=\frac{48}{2}=24$ (valid) - $(7,7)$: $x=\frac{14}{2}=7$, $y=\frac{0}{2}=0$ (valid) - $(49,1)$: $x=\frac{50}{2}=25$, $y=\frac{-48}{2}=-24$ (valid) - $(-1,-49)$: $x=\frac{-50}{2}=-25$, $y=\frac{-48}{2}=-24$ (valid) - $(-7,-7)$: $x=\frac{-14}{2}=-7$, $y=\frac{0}{2}=0$ (valid) - $(-49,-1)$: $x=\frac{-50}{2}=-25$, $y=\frac{48}{2}=24$ (valid) 7. The valid integer solutions are $(25,24), (7,0), (25,-24), (-25,-24), (-7,0), (-25,24)$. 8. Since the problem asks for unordered pairs, $(x,y)$ and $(y,x)$ are considered the same. Check if any pairs are duplicates under unordered condition: - $(25,24)$ and $(24,25)$ but $(24,25)$ is not in the solution set. - $(7,0)$ and $(0,7)$ but $(0,7)$ is not in the solution set. 9. So the unordered pairs are: $ \{(25,24), (7,0), (25,-24), (-25,-24), (-7,0), (-25,24)\}$. 10. Note that $(25,-24)$ and $(-25,24)$ are distinct unordered pairs because the absolute values differ in sign positions. 11. Therefore, the total number of unordered integer pairs $(x,y)$ satisfying the equation is $$\boxed{6}.$$