Answer( Problem 26):

$a^{2}+b^{2}+c^{2} \equiv 7(\text { mod8 })$ or, $a^{2}+b^{2}+c^{2}+1 \equiv 0(\text { mod8 })$ This further implies that 8 divides the sum of the remainders of $a^{2}, b^{2}, c^{2}$ and 1 on dividing by $8 .$ Now, square of any natural number gives remainders 0,1 or 4 on dividing by $8 .$ By trial and error, we see that the sum of the remainders is never divisible by $8,$ for any combination of remainders. Hence, proved.