p-1/p=q²

q(p+1/p)=4

square the second eq, q²*(p+1/p)²=16

(p-1/p)*((p-1/p)²+4)=16

let p-1/p=x

x³+4x-16=0 => (x-2)*(x²+2x+8)=0

x²+2x+8=0 has no real roots as its discriminant is less than 0

x=2 is the only solution.

Solving p-1/p=2 yields p=(1(+/-)√2)

substitute these in eq(2) to get q=+√2 for p=(1+√2)

and q=-√2 for p=(1-√2)

*Edited on ***July 10, 2007, 11:48 am**