(1) p+qrs=2
(2) q+rsp=2
(3) r+pqs=2
(4) s+pqr=2
solve (1) for p and substitute in (2)
q+rs(2qrs)=2
q+2rsqr^2s^2=2
q(rs+1)(rs1)=2(rs1) now if rs=1 we have zero on both sides
so lets consider all solutions with rs=1
r=1/s
but 1/s<=s
s^2>=1
s>=1 or s<=1
now p=2qrs but rs=1 so
p=2q
p<=q
2q<=q
2q>=2
q>=1
so if s<=1 then q>=1 contradicts q<=s thus s>=1 q>=1 q<=s gives s=q=1
which then in turn gives p=r=1 thus if rs=1 we have 1 solution (1,1,1,1)
now lets assume rs is not 1, then
q(rs+1)(rs1)=2(rs1)
q=2/(rs+1)
substite this in to p=2qrs=2/(rs+1)=q
substitute p=q=2/(rs+1) in both (3) and (4)
(5) r+4s/(rs+1)^2=2
(6) s+4r/(rs+1)^2=2
solving (5) for r gives 3 solutions
r=(2s1)/s or r=(1+sqrt(8s+1))/(2s)
putting each of these into (6) and solving for s gives the following real
solutions for (r,s)
(1,1), (3,1), (1,3)
r=s=1 contradicts rs not being 1
r=3 s=1 contradicts r<=s
that leaves r=1 s=3
this gives q=2/(3+1)=2/2=1
q=1 r=1 s=3 gives
p=2qrs=2(1)(1)3=23=1
thus another solution is (1,1,1,3)
thus in conclusion the only real solutions also happen to be integer solutions
and they are (1,1,1,1) and (1,1,1,3)

Posted by Daniel
on 20080901 13:06:26 