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

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 v_s(p) ) be k₁ and v_s(q)  = k₂ . WLOG .

 

Then we have v_s(pqr) = k₁+k₂ and v_s(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’s^k and q = q’ s^k we can divide throughout by s^2k 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, with

v_t(p) = n₁ , v_t(q) = n₂ and v_t(r) = n₃. and WLOG 

 

So, v_t(pqr) =  and v_t(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 easily

follows that r|pq and q|pr and p|rq

 

i.e. that 

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