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A Perplexing (Prime) Puzzle (Posted on 2007-06-05) Difficulty: 4 of 5
Consider the quadruplets (p,q,r,s) of positive integers with p>q>r>s, and satisfying pr+qs= (q+s+p-r)(q+s-p+r).

Is it ever the case that pq+rs is a prime number?

  Submitted by K Sengupta    
Rating: 4.0000 (3 votes)
Solution: (Hide)
If possible, let (pq+rs) be prime.

Now, pq+rs =
(p+s)r + (q-r)p
= t*gcd(p+s, q-r), 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, q-r) = 1

Case (a): t=1

We observe that:
gcd(p+s, q-r) =
pq+rs
> (pq+rs) -(p-q+ r+s)
= (p+s)(r-1) +(q-r)(p+1)
> = gcd(p+s, q-r)
This is a contradiction.

Case(b): gcd(p+s, q-r) = 1

We note that pr+qs = (p+ s)q -(q-r)p

Then, by conditions of the problem:
pr+rs=(q+s+p-r)(q+s-p+r)
Or, (p+s)q - (q-r)p = (q+s+p-r)(q+s-p+r)
Or, (p+s)(p-r-s) = (q-r)(q + r +s)

Accordingly, there exists a positive integer n such that:

p-r-s = n(q-r) and q+r-s = n(p+s)
This gives, p+q = n(p + q - r + s, so that:
n(r-s) = (n-1)(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/(n-1) = (p+q)/(r-s) > (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.

.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Might be correct!Praneeth Yalavarthi2007-07-11 12:16:55
Hints/TipsHintK Sengupta2007-06-29 12:52:26
Helpful Observation?Tommy2007-06-28 14:23:15
More number crunchingBrian Smith2007-06-07 22:08:55
Some number crunchingBrian Smith2007-06-07 00:34:43
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