Problem based on Integer | PRMO-2018 | Problem 6

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 \(a^2 + b^2 + c^2=17+1=18\)

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 \(a^2 + b^2 + c^2=17+1=18\)