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Algebra Math Olympiad PRMO

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 \(a + b – c = 1\) 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

quadratic equation

Factorization

Check the Answer


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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