## Task B-1.4.

How many right triangles are there with the lengths of the legs being integers $a, b$, and the length of the hypotenuse $b+1$, where $b<100$?
## Solution.

For a right-angled triangle, we have $(b+1)^{2}=a^{2}+b^{2}$, which means $b^{2}+2 b+1=a^{2}+b^{2}$, from which it follows that $2 b=a^{2}-1, \text{i.e., } b=\frac{a^{2}-1}{2}$.

Since $b<100$, we have $\frac{a^{2}-1}{2}<100, a^{2}<201$.

The squares of integers less than 201 are 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196.

Since $b \in \mathbb{N}$, from $b=\frac{a^{2}-1}{2}$ it follows that $a^{2}$ is odd.

Then $a \in\{3,5,7,9,11,13\}$. It follows that $b \in\{4,12,24,40,60,84\}$.

Therefore, there are 6 triangles with sides $(a, b, b+1)$ and the given property.