2008 AMC 10B Problem 15

The Pythagorean theorem gives $a^2 + b^2 = (b + 1)^2 = b^2 + 2b + 1$, so
$$a^2 = 2b + 1.$$
Thus $a^2$ is odd, so $a$ is odd, and $b = \dfrac{a^2 - 1}{2}$ is automatically an integer for odd $a$. Each odd $a$ therefore yields exactly one triangle, provided $b$ is a positive integer below $100$.

Positivity: $a = 1$ gives $b = 0$, which is not a triangle, so $a \ge 3$.

Bound: $b < 100$ means $a^2 - 1 < 200$, so $a^2 < 201$, giving $a \le 14$. The odd values are $a = 3, 5, 7, 9, 11, 13$, with $b = 4, 12, 24, 40, 60, 84$.

That is $6$ triangles. The answer is $\boxed{\textbf{(A)}\ 6}$.

Hypotenuse one more than a leg is the classic family $(a, \frac{a^2-1}{2}, \frac{a^2+1}{2})$ of Pythagorean triples ($3$-$4$-$5$, $5$-$12$-$13$, $7$-$24$-$25$, ...). Expanding $(b+1)^2$ cancels the $b^2$ terms and leaves a linear relation, so parity and a bound finish the count. When a problem fixes a difference between two sides, expand and cancel first.

• Including a = 1 (which gives b = 0, not a triangle) to get 7, or including a = 15 even though its b = 112 exceeds 100.
• Bounding with $a < 100$ instead of $b < 100$ and counting far too many odd values.