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# Problem based on Integer | PRMO-2018 | Problem 6 Try this beautiful problem from Algebra based on Integer

## Algebra based on Integer PRMO Problem 6

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
$a^2 + b^2 + c^2$ ?

• $24$
• $18$
• $34$

### Key Concepts

Algebra

Factorization

Answer:$18$

PRMO-2018, Problem 6

Pre College Mathematics

## Try with Hints

Use $(a + b – 1 )= c$ this relation

Can you now finish the problem ..........

$(a + b – 1 )^2= c^2$ (squaring both sides.......)

Can you finish the problem........

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

Now

$(a + b – c )= 1$

$\Rightarrow (a + b – 1 )= c$

$\Rightarrow (a + b – 1 )^2= c^2$ (squarring both sides.......)

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

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

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

$\Rightarrow ab=a+b$

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

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 $(a + b – c )= 1$ ,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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Try this beautiful problem from Algebra based on Integer

## Algebra based on Integer PRMO Problem 6

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
$a^2 + b^2 + c^2$ ?

• $24$
• $18$
• $34$

### Key Concepts

Algebra

Factorization

Answer:$18$

PRMO-2018, Problem 6

Pre College Mathematics

## Try with Hints

Use $(a + b – 1 )= c$ this relation

Can you now finish the problem ..........

$(a + b – 1 )^2= c^2$ (squaring both sides.......)

Can you finish the problem........

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

Now

$(a + b – c )= 1$

$\Rightarrow (a + b – 1 )= c$

$\Rightarrow (a + b – 1 )^2= c^2$ (squarring both sides.......)

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

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

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

$\Rightarrow ab=a+b$

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

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 $(a + b – c )= 1$ ,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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