A small logic is to be found!

Level 1

Integers a , b , c a,b,c satisfy a + b c = 1 a+b-c = 1 and a 2 + b 2 c 2 a^{2} + b^{2} - c^{2} = 1 = -1 . What is the sum of all possible values of a 2 + b 2 + c 2 a^{2} + b^{2} + c^{2} ?

12 5 10 18

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1 solution

Tom Engelsman
Sep 14, 2018

Let c = a + b 1 c = a + b - 1 and c 2 = ( a + b ) 2 2 ( a + b ) + 1 c^2 = (a+b)^2 - 2(a+b) + 1 . Substitution of the latter squared expression into a 2 + b 2 c 2 = 1 a^2 + b^2 - c^2 = -1 yields:

a 2 + b 2 [ ( a + b ) 2 2 ( a + b ) + 1 ] = 1 ; a^2 + b^2 - [ (a+b)^2 - 2(a+b) + 1 ] = -1;

or a 2 + b 2 ( a 2 + 2 a b + b 2 ) + 2 a + 2 b 1 = 1 ; a^2 + b^2 - (a^2 + 2ab + b^2) + 2a + 2b - 1 = -1;

or 2 a + 2 b 2 a b = 0 ; 2a + 2b - 2ab = 0;

or a + b a b = 0 ; a + b - ab = 0;

or b = a a 1 = 1 + 1 a 1 . b = \frac{a}{a-1} = 1 + \frac{1}{a-1}.

In order for b b to be an integer, we must have ( a 1 ) 1 a = 0 , 2 (a-1) | 1 \Rightarrow a = 0, 2 . This ultimately produces the integral triplets ( a , b , c ) = ( 0 , 0 , 1 ) ; ( 2 , 2 , 3 ) (a,b,c) = (0,0,-1); (2,2,3) , and the sums of a 2 + b 2 + c 2 = [ 0 2 + 0 2 + ( 1 ) 2 ] + [ 2 2 + 2 2 + 3 2 ] = 1 + 17 = 18 . a^2 + b^2 + c^2 = [0^2 + 0^2 + (-1)^2] + [2^2 + 2^2 + 3^2] = 1 + 17 = \boxed{18}.

Nice, @Tom Engelsman !

Prem Chebrolu - 2 years, 8 months ago

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My pleasure, Prem!

tom engelsman - 2 years, 8 months ago

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