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Grade: 12

                        

..........................,..................(just to reach minimum character limit

4 years ago

Answers : (2)

mycroft holmes
272 Points
							
We are given that pqr|(p^2q^2+q^2r^2+r^2p^2).
 
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 k_1 \le k_2.
 
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 k_1+k_2 \le 2k_1 and hence we must have k1 = k2. So if p = p’sand q = q’ sk we can divide throughout by s2and 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, with
vt(p) = n1 , vt(q) = nand vt(r) = n3. and WLOG n_1 \le n_2 \le n_3
 
So, vt(pqr) = n_1 +n_2 + n_3 and vt(pqr) = (p2q2+q2r2+r2p2) = min [2(n1+n2), 2(n2+n3), 2(n3+n1)] = 2(n1+n2).
 
So, we need that n_1 +n_2 +n_3 \le 2(n_1+n_2) \Rightarrow n_3 \le n_1+n_2.
 
We already have n_1 \le n_2+n_3 and n_2 \le n_1+n_3. And hence it easily
follows that r|pq and q|pr and p|rq
 
i.e. that \frac{pq}{r}, \frac{qr}{p}, \frac{pr}{q} \in \mathbb{Z}
4 years ago
mycroft holmes
272 Points
							
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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