If $a$ and $b$ be positive integers such that $a^2 - b^2 = 19$, then the value of $a$ is

### Concept & Formula When a difference of two perfect squares equals a prime number, factor the squares and equate the factors to the prime number and 1. $$a^2 - b^2 = (a - b)(a + b)$$ ### Step-by-Step Solution * **Given:** $a^2 - b^2 = 19$, where $a$ and $b$ are positive integers. * Expand the left side using the standard algebraic identity: $$(a - b)(a + b) = 19$$ * Since 19 is a prime number, its only positive integer factors are 1 and 19. * Because $a$ and $b$ are positive integers, the sum $(a + b)$ must be greater than the difference $(a - b)$. * Therefore, we can set up a system of linear equations: $$a - b = 1$$ $$a + b = 19$$ * Add the two equations together to eliminate $b$: $$(a - b) + (a + b) = 1 + 19$$ $$2a = 20$$ $$a = 10$$ ### Exam Strategy & Shortcut Whenever $a^2 - b^2 = P$ (where $P$ is a prime number), the larger number $a$ is always $\frac{P + 1}{2}$ and the smaller number $b$ is always $\frac{P - 1}{2}$. Mentally calculate $\frac{19 + 1}{2} = 10$ to find the answer instantly without writing anything down. ### Common Pitfall Students often try to blindly guess perfect squares whose difference is 19 (e.g., trying $25-6$, $36-17$, $100-81$) which wastes valuable exam time. Always rely on prime factorization properties instead of trial and error. ### Final Answer Therefore, the correct answer is **10**.