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divisibility
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since a,b & c,d are roots of x2+px+1=0 & x2+qx+1=0 respectively, {c2+qc+1=0 d2+qd+1=0 a+b=(-p) c+d=(-q) ab=1 cd=1} using the above relations, (a-c)(b-c)(a+d)(b+d)=[ab-c(a+b)+ c2][ab+(a+b)d+d2]=[1+pc+c2][1-pd+d2] = (q-p)(q+p) =(a+b-c-d)(-1)(a+b+c+d) hence divisible by [ a+b-c-d] &[a+b+c+d]
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