$$\triangle ABC$$ is right-angled triangle, $$\angle ACB = 90^\circ$$. The sides are of integral length. If the product of the legs is equal to the sum of all sides, then find the sides of the triangle. How many such triangle are there?

Let the legs be $$a = BC$$ and $$b = AC$$, so the hypotenuse is $$c = \sqrt{a^2 + b^2}$$ and the condition is $$ab = a + b + \sqrt{a^2 + b^2}$$. Squaring $$(ab - a - b)^2 = a^2 + b^2$$ gives $$a^2b^2 - 2a^2b - 2ab^2 + 2ab = 0$$, that is $$ab - 2a - 2b + 2 = 0$$. Adding 2 to both sides gives $$(a - 2)(b - 2) = 2$$. Negative factor pairs make a side zero, so $$\{a - 2, b - 2\} = \{1, 2\}$$, the legs are 3 and 4, and the hypotenuse is 5 (check $$3 \times 4 = 12 = 3 + 4 + 5$$). Hence only one triangle, with sides 3, 4, 5, is possible (2 if the ordered assignments $$(BC, AC) = (3, 4)$$ and $$(4, 3)$$ are counted separately).