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?
 A: HINT : In mod $7$,
$$a^2\equiv 0,1,2,4.$$
A: More generally: $\ p\equiv 3\pmod{\! 4},\ \, p\mid a^2+b^2\,\,\Rightarrow\,\, p\mid a,b$.    
By contradiction: if not, then mod $p$: $\ a^2\equiv -b^2\iff (a/b)^2\equiv -1$
$1)\ \stackrel{(p-1)/2}\Rightarrow(a/b)^{p-1}\equiv \color{#00F}{{(-1)^{(p-1)/2}}}\equiv \color{#00F}{-1}\, $ contradicts little Fermat.   
$2)\ \stackrel{2}\Rightarrow\, (a/b)^4\equiv 1\,$ and $\,\text{ord}_p (a/b)=4\mid p-1$,$\,$ contr $\,p\equiv 3\,\,$ mod $4$.    
It is a consequence of $\,p\nmid a,b,\,\,p\mid a^2+b^2\,\Rightarrow\, p\equiv 1\,\,$ mod $4\,$ (see here).
A: There is a theorem in number theory which asserts

If $p|a^2+b^2$, and $p\equiv 3\pmod4$, then $p|a,p|b$.

Proof. Suppose not,then by $p|a^2+b^2$, we have $a^2\equiv -b^2\pmod p$, and by using Fermat's little theorem we have
$$
1\equiv a^{p-1}\equiv (a^2)^{(\frac{p-1}2)}\equiv (-b^2)^\frac{p-1}2\equiv(-1)^{\frac{p-1}2}=-1\pmod p
$$
Which contradicts...
Now observe that $7\equiv3\pmod4$
A: 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.
