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?

Let A=-p+q+r+s,B=p-q+r+s,C=p+q-r+s,D=p+q+r-s.
then A<B<C<D, B>0,C>0,D>0
AC=pr+qs. Similarly BD=3*AC--(1)
4*(pr+qs)=(p+r)²-(q-s)²+(q+s)²-(p-r)² (Now use a²-b²=(a+b)*(a-b))
4*AC=BD+AC=>BD=3*AC
Similarly,4*(pq+rs)=AB+CD
Multiply both sides by BC
4*BC*(pq+rs)=B²*AC+C²*BD --(2)(sub BD=3*AC)
pq+rs=(AC/(4*BC))*(B²+3*C²)
pq+rs=(A/(4*B))*(B²+3*C²)
A should be factor of 4*B for pq+rs to be prime
(-p+q+r+s)*c=4*(p-q+r+s) (c is positive integer)
(p-q)=(c-4)*(r+s)/(c+4) --(3)
c>4 because p>q and LHS>0, so RHS should be greater than 0
Now sub AC=1/3*BD in eq(2), we get
pq+rs=(D/(12*C))*(B²+3*C²)  

Now D should be factor of 12*C for pq+rs to be prime
Now (r-s)=(12-k)*(p+q)/(k+12) --(4) (k is positive integer)
LHS>0, So 0<k<12

Now Sub p+q=(k+12)*(r-s)/(12-k) and p-q=(c-4)*(r+s)/(c+4) in BD=3*AC (Resubstitute A=-p+q+r+s,B=p-q+r+s,..)
Now simplify and we will get
2c*24/((c+4)*(12-k))=3*2*2k/((c+4)*(12-k))
4*c=k for k<12,c will be less than 3 which contradicts c>4 the required condition.
So, pq+rs can not be a prime number.

Edited on July 11, 2007, 12:23 pm

Posted by Praneeth Yalavarthi on 2007-07-11 12:16:55