1. **Problem statement:** We need to find the number of ordered positive integer pairs $(a,b)$ such that $$a^2 - b^2 = 2026.$$ 2. **Formula and approach:** Recall the difference of squares factorization: $$a^2 - b^2 = (a-b)(a+b).$$ 3. **Rewrite the equation:** Set $$x = a-b$$ and $$y = a+b.$$ Then the equation becomes $$xy = 2026,$$ where $x$ and $y$ are positive integers and $y > x$ because $a+b > a-b$ for positive $a,b$. 4. **Express $a$ and $b$ in terms of $x$ and $y$:** $$a = \frac{x+y}{2}, \quad b = \frac{y-x}{2}.$$ Since $a$ and $b$ are positive integers, both $\frac{x+y}{2}$ and $\frac{y-x}{2}$ must be positive integers. 5. **Conditions on $x$ and $y$:** - $x$ and $y$ are positive divisors of 2026 with $x < y$. - $x$ and $y$ must be both even or both odd for $a$ and $b$ to be integers (because their sum and difference must be even). 6. **Factorize 2026:** $$2026 = 2 \times 1013,$$ where 1013 is prime. 7. **List positive divisor pairs $(x,y)$ of 2026 with $x < y$:** - $(1, 2026)$ - $(2, 1013)$ 8. **Check parity of pairs:** - For $(1, 2026)$: 1 is odd, 2026 is even (different parity) → discard. - For $(2, 1013)$: 2 is even, 1013 is odd (different parity) → discard. 9. **No pairs satisfy parity condition, so no solutions?** Wait, check if $a$ and $b$ can be integers if $x$ and $y$ have different parity. 10. **Check parity condition carefully:** Since $$a = \frac{x+y}{2}, b = \frac{y-x}{2},$$ for $a,b$ to be integers, $x+y$ and $y-x$ must be even. - $x+y$ even means $x$ and $y$ have the same parity. - $y-x$ even means $x$ and $y$ have the same parity. Since all divisor pairs have different parity, no integer solutions for $a,b$. 11. **Conclusion:** No ordered positive integer pairs $(a,b)$ satisfy $a^2 - b^2 = 2026$. **Final answer:** $$\boxed{0}.$$