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

a+b-c=1

c=a+b-1 (1)

a^2+b^2-c^2=-1

a^2+b^2+1=c^2 (2)

put value of c from (1) in (2)

a^2+b^2+1=(a+b-1)^2

a^2+b^2+1=a^2+b^2+1+2ab-2a-2b

2ab-2a-2b=0

ab=a+b

or b(a-1)=a

now it is given that a,b and c are integers, so only two integers can be expressed as a multiple of the integer one less than them, which are 2 and 0.

so, a=2 or a=0

putting these values of a in (3)

if a=2 =>b=2 =>c=3 from equation(1)

so value of a^2+b^2+c^2=17...

if a=0 =>b=0 => c=-1

so a^2+b^2+c^2=1 ,

hence sum of all possible values satisfying given conditions is 1+17=18...