given that I=a2+b2+c2; where a and b are consecutive integer and c= ab. So,b=a+1 andc=a(a+1)=a2+a I=a2+(a+1)2+a2(a+1)2 =a2+a2+2a+1+a4+2a3+a2 =a4+2a3+3a2+2a+1 =(a2+a+1)2 =[a(a+1)+1]2 =(ab+1)2 =(c+1)2 The product of any two consecutive number is always a even number, soc is a even number and I is a square of an odd integer.