a*(a-1)=p*c*(c-1)/q
set up an array : m[i] = i*(i-1)
find c that m[c]%q==0
solve the equation a^2-a-p*m[c]/q=0
if a exists (a,c-a) is the answer, or find next c;

 p/q = a(a-1)/(a+b)(a+b-1)
令a = x, a+b = y;
则：            p/q = (x^2-x)/(y^2-y)
               x^2-x-np = 0
               y^2-x-np = 0
由二次方程根与系数的关系，以及 x<y 得到的