How to show that $7\mid a^2+b^2$ implies $7\mid a$ and $7\mid b$? For my proof I distinguished the two possible cases which derive from $7 \mid a^2+b^2$: 
Case one: $7\mid a^2$ and $7 \mid b^2$
Case two (which (I think) is not possible): $7$ does not divide $a^2$ and $7$ does not divide $b^2$, but their sum.
I've shown that case $1$ implies $7\mid a$ and $7\mid b$,
so I just have show that case 2 isn't possible
- I'll be happy for any input.
MfG,
Karl

a = 7k + m, and b = 7p + n ==> a^2 + b^2 = m^ + n^2 ( mod 7 ).
Case 1: m = 1, n = 1 => m^2  + n^2 = 2. ( mod 7 )
Case 2: m = 1, n = 2 => m^2 + n^2 = 5.
Case 3: m = 1, n = 3 => m^2 + n^2 = 3.
Case 4: m = 1, n = 4 => m^2 + n^2 = 3.
Case 5: m = 1, n = 6 => m^2 + n^2 = 2.
....
All cases yield non-zero mod 7. So it can happen only when both a and b are 0 mod 7.
