Question

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.

Answer 1

Given :

p + q - r = -1

p² + q² - r² = -1

To find :

Sum of all positive values of (p + q + r)

Solution:

p + q - r = -1

p + q = r - 1

(p + q)² = (r - 1)²

p² + q² + 2pq = r² - 2r + 1

p² + q² - r² = -2pq - 2r + 1 ------(1)

Another equation has been given as,

p² + q² - r² = -1 -------(2)

From equation (1) and (2),

-1 = -2pq - 2r + 1

2pq + 2r = 2

pq + r = 1 --------(3)

Since (p + q) = r - 1

r = p + q + 1

By substituting the value of r in equation (3)

pq + p + q + 1 = 1

p + q + pq = 0

p(1 + q) = -q

p = $-\frac{q}{q+1}$

And q = $-\frac{p}{(p+1)}$

Since p, q and r are the integers,

Therefore, denominators (q + 1) = ± 1

q = 0, -2

And (p + 1) = ±1

p = 0, -2

Since, r = p + q + 1

For p = q = 0

r = 1

For p = q = -2

r = -2 - 2 - 1

r = -5

Now, p + q + r = 0 + 0 + 1 = 1

And p + q + r = -2 - 2 - 5 = -9 [Negative value]

Therefore, p + q + r = 1 will be the sum of all positive values of (p + q + r).