Problems need HardWork #1

$\ 7w^2 \equiv 0,3 mod(4)$ so either $\ x^2,y^2,z^2 \equiv 0 mod(4)$ or $\ x^2,y^2,z^2 \equiv 1 mod(4)$ ( # ). Case1: $\ x = 2x_1 \ ~~ \ etc:$ $\ 4x_1^2 + 4y_1^2 + 4z_1^2 = 7w^2$ Clearly w is even so we use a similar substitution ( $\ w = 2w_1$ ) to obtain: $\ x_1^2 + y_1^2 + z_1^2 = 7w_1^2$ Hence, (using proof by descent) there are no positive integer solutions. Case 2: All variables are odd (from # and parity) so we have $\ (2x_1 +1)^2 + (2y_1 +1)^2 + (2z_1 +1)^2 = 7(2w_1 +1 )^2$ This can be rearranged then divided by 4 to give: $\ x_1 (x_1 +1) + y_1 (y_1 +1) + z_1 (z_1 +1) -1 = 7w_1 (w_1 +1)$ $\implies\ odd = even$ This contradiction together with case 1 proves that there are no positive integer solutions.