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4 years ago

							We are given that . Consider a prime that divides p. Then it has to divide q2r2 i.e. it also divides at least one of q and r.  Suppose the prime s divides p and q, but not r. Further let the highest power of s that divides p (denoted by vs(p) ) be k1 and vs(q)  = k2 . WLOG . Then we have vs(pqr) = k1+k2 and vs(p2q2+q2r2+r2p2) = min (2k1+2k2, 2k2, 2k1) = 2k1.And from the fact that pqr divides (p2q2+q2r2+r2p2) we have  and hence we must have k1 = k2. So if p = p’sk and q = q’ sk we can divide throughout by s2k and hence obtain that we need p’q’r’|(p’2q’2+q’2r2+r2p’2). So let us therefore assume that all primes that divide precisely two of p,q,r have been divided out this way. So, we now have a prime t that divides all three, withvt(p) = n1 , vt(q) = n2 and vt(r) = n3. and WLOG  So, vt(pqr) =  and vt(pqr) = (p2q2+q2r2+r2p2) = min [2(n1+n2), 2(n2+n3), 2(n3+n1)] = 2(n1+n2). So, we need that . We already have  and . And hence it easilyfollows that r|pq and q|pr and p|rq i.e. that

4 years ago
							I forgot to add that since any prime that divides precisely two of p,q,r appears to eqyal power in the two numbers they will cancel out even in pq/r, qr/p, and pr/q. Hence only need to concern ourselves with primes that divide all three of p,q,r

4 years ago
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