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A Perplexing (Prime) Puzzle (Posted on 20070605) 

Consider the quadruplets (p,q,r,s) of positive integers with p>q>r>s, and satisfying pr+qs= (q+s+pr)(q+sp+r).
Is it ever the case that pq+rs is a prime number?

Submitted by K Sengupta

Rating: 4.0000 (3 votes)


Solution:

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If possible, let (pq+rs) be prime.
Now, pq+rs =
(p+s)r + (qr)p
= t*gcd(p+s, qr), where t is a positive integer as each of p, q, r and s are positive......(#).
Since, (pq+rs) is prime, it follows that: Either, t = 1; Or, gcd(p+s, qr) = 1
Case (a): t=1
We observe that:
gcd(p+s, qr) =
pq+rs
> (pq+rs) (pq+ r+s)
= (p+s)(r1) +(qr)(p+1)
> = gcd(p+s, qr)
This is a contradiction.
Case(b): gcd(p+s, qr) = 1
We note that pr+qs = (p+ s)q (qr)p
Then, by conditions of the problem:
pr+rs=(q+s+pr)(q+sp+r)
Or, (p+s)q  (qr)p = (q+s+pr)(q+sp+r)
Or, (p+s)(prs) = (qr)(q + r +s)
Accordingly, there exists a positive integer n such that:
prs = n(qr) and q+rs = n(p+s)
This gives, p+q = n(p + q  r + s, so that:
n(rs) = (n1)(p+q)
We now recall that p> q > r> s.
if n =1, then r=s, which is a contradiction.
If n> 2, then:
2> = n/(n1) = (p+q)/(rs) > (p+q)/r > (r+r)/r =2.
This is a contradiction.
Accordingly, our original assumption that (pq+rs) corresponds to a prime number is false.
Consequently, it is NEVER the case that (pq+rs) is a prime number.
.

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