The problem: how many pairs of integers (j , k) with k > j exist such that 42² + j² = k²?

42² = k² - j² = (k + j)(k - j)

Since k and j are integers and k > j, k + j and k - j must be positive integers which, when multiplied, equal 42². This seems straightforward - find the number of factors of 42², subtract 1 (since we can't have k = 0), and divide by two.

42 = (2)(3)(7)
42² = 2²3²7²

So 42 has (2+1)³ = 27 factors, and 13 factor pairs.

However, not all factor pairs will give us integer values of j and k. Allow the two factors to be F₁ and F₂ so that F₂ > F₁ and F₁F₂ = 42²

k + j = F₂
k - j = F₁
k =
j =

We can therefore allow both factors to be odd or both factors to be even, but we can never have one odd and one even. This means we must eliminate all cases where F₁ OR F₂ = 4k, where k is any factor of 3²7². There are 9 such factors, so only 4 right triangles with integer side lengths and a leg equal to 42 exist. In general, we eliminate 3^n-1 factor pairs. This leads us to our general equation for the bonus question -- (3ⁿ - 1) - 3ⁿ = (3ⁿ-3)

Incidentally, this proves that there are no right triangles with integer side lengths and a leg equal to 2 (that is, for n = 1, (3ⁿ - 3) = 0 ).

For a very interesting approach to this problem, please check out Graeme's solution at
http://mcraefamily.com/MathHelp/GeometryPythagoreanTriplesWithProductOfPrimesLeg.htm