Question

if a+b-c=1 and a^2+b^2-c^2=-1 then find a^2+b^2+c^2

Answer 1

$\large \: RMO \: \: 2018 \: \: [ \: Question - 6 \: ] :D \\ \\ \\ a + b - c = 1 \\ a + b = c + 1 \\ \: \\ Squaring \: \: both \: \: sides \: \: , \\ {a}^{2} + {b}^{2} + 2ab = {c}^{2} + 2c + 1 \\ {a}^{2} + {b}^{2} - {c}^{2} = 2c + 1 - 2ab \\ \\ {a}^{2} + {b}^{2} - {c}^{2} = - 1 \\ \\ 2ab = 2c + 2 \\ ab = c + 1 \\ \\ \\ \boxed{ \bf \: a + b = ab} \\ \\ \\ \boxed{ \bf \frac{1}{a} + \frac{1}{b} = 1 \: \: or \:(a - 1)(b - 1) = 1 }\\ \\ \\ Solving \: for \: a \: and \: b \: , \: possible \: \\ values \: are : \\ \\ \boxed {a = b = 0 \: \: \: \: \: \: \: \: c = - 1} \\ \\ \boxed{ \: a = b = 2 \: \: \: \: \: \: \: \: c \: = \: 3} \\ \\ \\ For \: first \: set \: of \: values \: , \\ {a}^{2} + {b}^{2} + {c}^{2} = 1 \\ \\ For \: second \: set \: of \: values \: , \\ {a}^{2} + {b}^{2} + {c}^{2} = \: 17 \\ \\ \\ \large \boxed{Sum \: of \: all \: possible \: values \: of \: {a}^{2} + {b}^{2} + {c}^{2} \: = 18}$

Answer 2

Answer:

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