Question

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

Answer 1

$a+b-c=1\dots \left(i\right)$

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

$\Rightarrow (a+b{)}^2-2ab-c^2=-1$

$\Rightarrow (a+b{)}^2-c^2-2ab=-1$

$\Rightarrow (a+b+c)(a+b-c)-2ab=-1$

Now by using equation $\left(i\right),$

$a+b+c-2ab=-1$

$\Rightarrow 1+c+c-2ab=-1$

$\Rightarrow 2c-2ab=-2$

$\Rightarrow c=ab-1$

Again, substitute $a+b-c$ for $1$ by using $\left(i\right),$

$c=ab-(a+b-c)$

$\Rightarrow c=ab-a-b+c$

$\Rightarrow ab-a-b=0$

Now, on factorization of above obtained expression,

$a(b-1)-b=0$

$\Rightarrow a(b-1)-b+1=1$

$\Rightarrow a(b-1)-(b-1)=1$

$\Rightarrow (a-1)(b-1)=1$

So, $a=b=2$ and $a=b=0$ are two possible solutions,

Case 1: $a=b=2$

$c=a+b-1$

$=2+2-1$

$=3$

So,

$a^2+b^2+c^2=17$

Case 2: $a=b=0$

$c=a+b-1$

$=0+0-1$

$=-1$

So,

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

Hence, sum of all possible values of $a^2+b^2+c^2$ is equal to $1+17=18$.