Proof that $a^2+b^2=3c^2$ has no non-trivial solutions I am following Dummit and Foote's "Abstract Algebra" and stumbled upon the following Question:

For (6) it occurrs to me that brute force should suffice:
$$
\begin{aligned}
0^2 (mod\;4) = 0\\
1^2 (mod\;4) = 1\\
2^2 (mod\;4) = 0\\
3^2 (mod\;4) = 1\\
\end{aligned}
$$
For (7) it seems to me, a direct result of (6); as no summand in the equation $a^2+b^2$
can be greater than one than their sum cannot be greater than two (modulo 4).
For (8) His explanation somewhat makes sense to me however when I consider the terms on each side of the equation: $a^2+b^2=3c^2$, from (6) the left-hand side cannot be greater than two while the right-hand side is an expression taking values in the set $\{0,3\}$ this leaves only the trivial solution as a solution for the equation. Is this a valid solution?
I am trying to understand his suggested solution - Firstly, I dont follow why a,b,c would need to be divisible by two, I guess its b/c this is the only congruence class that would lead to a valid eq. - But that is just another way of saying that $a^2,b^2,c^2$ must be the trivial solution. Isn't it?
 A: Your solutions are mostly fine, but you do need to be careful since the concept of "greater than" isn't necessarily well-defined once you're talking about the equivalence classes modulo $n$.
For 6, you're right that brute force is sufficient - the equivalence classes mod $4$ are $\{[0], [1], [2], [3]\}$ and you can show that the square of any number congruent to 0 or 2 will be congruent to 0, and similarly for 1 and 3.
Then for 7, it would be more accurate to say that since all squares are congruent to either 0 or 1 mod 4, the possibilities for the sum of two squares mod 4 are that they're congruent to either 0, 1 or 2 (which you could again show through exhaustion of the possibilities).
The problem with your solution to 8 is that the question has asked you to consider $a, b, c$ as non-zero integers, and to use congruence modulo 4 as a tool to get you there. You've managed to prove that $a^2 \equiv b^2 \equiv c^2 \equiv 0 \mod 4$ is necessary for a solution to exist, but that only gets you so far as to say that $a, b, c \equiv 0, 2 \mod 4$ which is far from proving that $a = b = c = 0$.
A: It is clear that we cannot have $a=4n+1$, $a=4n+3$, $b=4n+1$, or $b=4n+3$, because if we did then the LHS would equal $1$ or $2$ modulo $4$, while the RHS must equal $0$ or $3$ modulo $4$. It follows that $a$ and $b$ must both be even, and so $c$ must also be even.
Let $t$ be the largest power of 2 which is a factor of all three numbers $a$, $b$ and $c$. This number exists unless $a$, $b$ and $c$ are all zero. If $t$ exists we can write $a=ta'$, $b=tb'$ and $c=tc'$, where $a', b', c'$ are integers, and at least one of them is odd.
The equation $a^2+b^2=3c^2$ becomes $t^2a'^2+t^2b'^2=3t^2c'^2$. We can divide both sides by $t^2$, giving $a'^2+b'^2=3c'^2$.
As we have shown above, all three of $a', b', c'$ are even. But one of them is odd. This is a contradiction, so $t$ cannot exist. In this case $a=b=c=0$, as required.
