2011/109) How do you solve 1/x^2 + 1/y^2 = 1/z^2 in integers

1/x, 1/y . 1/z form numbers of (Pythagorean triplet assuming them to be integers and if they are not: which is generally true my multiplying by sutable nnumber ) so they are in the ratio

1/x : 1/y : 1/z = 2pq : p^2-q^2 : p^2 + q^2 (p and q are rational)

(knowing that (2pq)^2 + (p^2-q^2)^2 = (p^2+q^2) and you can find it in number theory book

or x: y : z : = 1/(2pq) : 1/ (p^2 - q^2) : 1/(p^2 + q^2)

or x: y: z = (p^2+q^2) (p^2-q^2) : 2pq(p^2+q^2) : 2pq(p^2-q^2)

putting p = m/r , q = n/r ( m,.n,.r are integers)

x:y:z = (m^2+n^2)(m^2-n^2) /r^4 : 2mn(m^2+n^2)/r^4 : 2mn(m^2-n^2)/r^ 4
or (m^2+n^2)(m^2-n^2) : 2mn(m^2+n^2) : 2mn(m^2-n^2)/


in a trivial case 1/x : 1/y: 1/z = 1/3: 1/4: 1/5 = 20:15: 12