If the sum of two squares is divisible by $7$, both numbers are divisible by $7$ 
How do I prove that if $7\mid a^2+b^2$, then $7\mid a$ and $7\mid b$? 

I am not allowed to use modular arithmetic. 
Assuming $7$ divides $a^2+b^2$, how do I prove that the sum of the squares of the residuals is 0?

Since you said that you are not allowed to use modular arithmetic, argue by the contrapositive. If 7 does not divide $a$ or $b$. Then,
$a = 7k \pm 1, 7k \pm 2, 7k \pm 3 \implies$
$a^2 = 7(7k^2 \pm 2k) + 1, 7(7k^2 \pm 4k) +4, 7(7k^2 +8k + 1)+2$, $k \in \Bbb Z$
Now, look at any linear combination of $a^2$ (with integer weights, or course) and it is easy to convince oneself that 7 will never divide the sum.
