Let the numbers be x and (x+2) respectively ⇒x2+(x+2)2=290 ⇒x2+x2+4+(2×x×2)=290 ⇒2x2+4+4x=290 ⇒2x2+4x=290−4 ⇒2x2+4x=286 ⇒x2+2x−143=0 ⇒x2+2x−143=0 ⇒x2+13x−11x−143=0 ⇒x(x+13)−11(x+13)=0 ⇒(x−11)(x−13)=0 ⇒x=11,−13 The number can't be negative so value of x is 11 Second number =11+2=13 ∴ The consecutive odd integers are 11 and 13