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

Grade:12

## 2 Answers

mycroft holmes
272 Points
4 years ago
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 $\frac{pq}{r}, \frac{qr}{p}, \frac{pr}{q} \in \mathbb{Z}$
mycroft holmes
272 Points
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

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