NSEJS 2016! Part 2

a²+b²– 8c = 3 .......(1) We know that perfect square of any positive integer is in the form of 4n or 4n + 1 Case - I : a² = 4n 1 and b²=4n _2 then put in (1) 4n 1 + 4n 2– 8c = 3 when we divide LHS by 4 we get rem = 0 but on RHS we get rem = 3 LHS \neq RHS Case - II : If a²=4n 1+ 1 and b² = 4n 2 then, again put in equation (1) 4n 1 + 1 + 4n 2– 8c = 3 Divide the above equation by 4. On LHS, we will get rem. 1 but on RHS, we will get 3. LHS \neq RHS. Case- III : If a²= 4n 1 + 1 and b² = 4n 2 + 1 then put in equation (1) 4n 1+ 4n_2 + 1 – 8c = 3 divide by 4. On LHS, we will get rem = 2 On RHS, we will get rem = 3 LHS \neq) RHS Hence there are no possible value of a, b, c .

One of the a or b is odd and the other is even.

$a^2 + b^2 \equiv$ 1 (mod 4)

$a^2 + b^2 - 3 \equiv$ 2 (mod 4)

LHS cannot be divisible by 8.