ONLY SOLVE THIS EDUCATION IF U ARE A GENIUS
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atharv9890:
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since 'a' and 'b' are symmetrical we assume a>b
let k = (a2+b2) / (ab+1) ab+1 divides a2+b2
===> ab+1 also divides (a2+b2)*b
(a2+b2)*b= (ab +1)*a+b3-a
let c = (b3-a)/ (ab+1)
(a2+b2)*b=(ab+1)*a+(ab+1)*c
(a2+b2)/(ab+1)*b=a+c
bk=(a+c)
substitute a =bk - c to c
c=(b3-a)/(ab+1)
=(b3-bk+c)/(bk-c)b+1)
= (b3-bk+c)/(b2*k-bc-1)
c(b2*k-bc+1)=( b3-bk+c)
b2*c*k-b*c2=b3-bk
b*c*k-c2=b2-k
b*c*k+k-b2+c2
k=(b2+c2)/(bc+1)
let k = (a2+b2) / (ab+1) ab+1 divides a2+b2
===> ab+1 also divides (a2+b2)*b
(a2+b2)*b= (ab +1)*a+b3-a
let c = (b3-a)/ (ab+1)
(a2+b2)*b=(ab+1)*a+(ab+1)*c
(a2+b2)/(ab+1)*b=a+c
bk=(a+c)
substitute a =bk - c to c
c=(b3-a)/(ab+1)
=(b3-bk+c)/(bk-c)b+1)
= (b3-bk+c)/(b2*k-bc-1)
c(b2*k-bc+1)=( b3-bk+c)
b2*c*k-b*c2=b3-bk
b*c*k-c2=b2-k
b*c*k+k-b2+c2
k=(b2+c2)/(bc+1)
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