Integers satisfy and . What is the sum of all possible values of ?
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Let c = a + b − 1 and c 2 = ( a + b ) 2 − 2 ( a + b ) + 1 . Substitution of the latter squared expression into a 2 + b 2 − c 2 = − 1 yields:
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 ;
or 2 a + 2 b − 2 a b = 0 ;
or a + b − a b = 0 ;
or b = a − 1 a = 1 + a − 1 1 .
In order for b to be an integer, we must have ( a − 1 ) ∣ 1 ⇒ a = 0 , 2 . This ultimately produces the integral triplets ( 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 + 1 7 = 1 8 .