Question

Integers a,b,c satisfy a+b-c=1 and a2+b2-c2= -1. What is the sum of all possible values of a2+b2+c2 ?

Answer 1

it is riddle math . this type of questions are very conceptual. well, let's try to solve it .

given, a + b - c = 1........(i)
and a² + b² - c² = -1

a² + b² - c² = -1
=> (a + b)² - 2ab - c² = -1
=> (a + b)² - c² - 2ab = -1
=> (a + b + c)(a + b - c) - 2ab = -1
=> a + b + c - 2ab = -1 [ from eq. (i) ]
=> 1 + c + c - 2ab = -1 [ from eq. (i) ]
=> 2c - 2ab = -2
=> c = ab - 1
=> c = ab - (a + b - c) [ from eq. (i)]
=> c = ab - a - b + c
=> ab - a - b = 0
=> a( b - 1) - b = 0
=> a(b - 1) - b + 1 = 1
=> a(b - 1) - (b - 1) = 1
=> (a - 1)(b - 1) = 1
as we know, a , b and c are integers .
a = b = 2 then c = a + b - 1 = 2 + 2 - 1 = 3
then, a² + b² + c² = 2² + 2² + 3² = 17

if a = b = 0 , c = a + b - 1 = -1
then, a² + b² + c² = 1

then sum of possible value = 1 + 17 = 18
answer should be 18

Answer 2

It is enigma math.

This sort of inquiries is exceptionally applied. Indeed, how about we attempt to comprehend it.

Given, a + b - c = 1 and a² + b² - c² = - 1 if a = b = 0 ,

c = a + b - 1 = - 1 at that point, a² + b² + c² = 1 at that point aggregate of conceivable esteem

= 1 + 17 = 18 answer ought to be 18.