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A Reciprocal And Square Problem (Posted on 2007-07-10) Difficulty: 2 of 5
Find all real pairs (p, q) satisfying the following system of equations:
                      p - 1/p - q2 = 0

                      q/p + pq = 4

See The Solution Submitted by K Sengupta    
Rating: 3.5000 (2 votes)

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Solution | Comment 1 of 7
p-1/p=q^2 ---(1)
q(p+1/p)=4 ---(2)
(p+1/p)^2 - (p-1/p)^2 = 4
=> (p+1/p)^2 - q^2 = q(p+1/p) (from eq(1)&eq(2))
Let (P+1/p)=x, then x^2-x*q-q^2=0
yields x=q(1+root(5))/2 or q(1-root(5))/2
sub these in eq(2),
q^2 = 8/(root(5)+1) or 8/(1-root(5))
but 1-root5 is negative, so this cant be equal to q^2
so q = (+/-)√(2*(√5-1))
substiute q^2 = 2*(√5-1) in eq(1) to solve for p
p=(+/-)(√5-1)(+/-)*√(7-2*√5)
That means p can take 4 values and q can take 2 values

  Posted by Praneeth Yalavarthi on 2007-07-10 10:15:27
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