Integers p,q,r satisfy p+q-r=-1 and p^2+q^2-r^2=-1 What is the sum of all positive values of p+q+r.
Answers
Given : Integers p,q,r satisfy p+q-r=-1 and p^2+q^2-r^2=-1
To find : sum of all positive values of p+q+r
Solution:
p+q-r=-1
=> p + q = r - 1
Squaring both sides
=> p² + q² + 2pq = r² + 1 - 2r
p²+q²-r²=-1
=> p²+q² = r² -1
Substituting this value
=> r² -1 + 2pq = r² + 1 - 2r
=> 2pq = 2 - 2r
=> pq = 1 - r
=> pq = - (r - 1)
=> pq = - (p + q)
=> p + q + pq = 0
=> p = -q/(q + 1) & q = -p/(p + 1)
p , q & r are integer so q + 1 or p + 1 = ± 1
only possible Values
p = q = 0 or p = q = -2
=>pq = 0 or pq = 4
pq = 1 - r => r = 1 - pq
p = q = 0 => r = 1 - 0 = 1
=> p + q + r = 1
p = q = -2 => r = 1 - 4 = - 3
p + q + r = -2 - 3 = - 5 ( negative Value)
positive values of p+q+r. = 1 only
sum of all positive values of p+q+r. = 1
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