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# Problem based on Integer | PRMO-2018 | Problem 6

Try this beautiful problem from Algebra based on Quadratic equation from PRMO 8, 2018. You may use sequential hints to solve the problem.

Try this beautiful problem from Algebra based on Integer

## Algebra based on Integer PRMO Problem 6

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

• $24$
• $18$
• $34$

### Key Concepts

Algebra

Factorization

Answer:$18$

PRMO-2018, Problem 6

Pre College Mathematics

## Try with Hints

Use $(a + b â€“ 1 )= c$ this relation

Can you now finish the problem ……….

$(a + b â€“ 1 )^2= c^2$ (squaring both sides…….)

Can you finish the problem……..

Given that a,b,c are integer satisfy $a + b â€“ c = 1$ and $a^2 + b^2 â€“ c^2 = â€“1$.

Now

$(a + b â€“ c )= 1$

$\Rightarrow (a + b â€“ 1 )= c$

$\Rightarrow (a + b â€“ 1 )^2= c^2$ (squarring both sides…….)

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

$\Rightarrow (a^2 + b^2 â€“c^2+1+2ab)= 2(a+b)$

$\Rightarrow (-1)+1+2ab= 2(a+b)$ ( as $a^2 + b^2 â€“ c^2 = â€“1$.

$\Rightarrow ab=a+b$

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

Therefore the possibile cases are $(a-1)=\pm 1$ and $(b-1)=\pm 1$

Therefore $a=1$,$b=1$ or $a=0$,$b=0$

From the equation $(a + b â€“ c )= 1$ ,C =3 (as a=b=2) and C=-1(as a=b=0)

Therefore $(a^2 +b^2 +c^2)=4+4+9=17$ and $(a^2 +b^2 +c^2)=0+0+1=1$

sum of all possible values of $a^2 + b^2 + c^2=17+1=18$

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