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 Some Reals Sum Two (Posted on 2008-09-01)
Determine all possible real quadruplet(s) (P, Q, R, S) with P ≤ Q ≤ R ≤ S that satisfy this system of equations:

P + Q*R*S = 2, and:

Q + R*S*P = 2, and:

R + P*Q*S = 2, and:

S + P*Q*R = 2.

 See The Solution Submitted by K Sengupta No Rating

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 solution (spoiler) | Comment 1 of 6

(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(2-qrs)=2
q+2rs-qr^2s^2=2
q(rs+1)(rs-1)=2(rs-1)  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=2-qrs but rs=1 so
p=2-q
p<=q
2-q<=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)(rs-1)=2(rs-1)
q=2/(rs+1)

substite this in to p=2-qrs=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=(2s-1)/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
that leaves r=-1 s=3
this gives q=2/(-3+1)=2/-2=-1
q=-1 r=-1 s=3 gives
p=2-qrs=2-(-1)(-1)3=2-3=-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 2008-09-01 13:06:26

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