Update:
(because of the length of the question, I put an update at the top)
I appreciate recommendations regarding the alternative proofs. However, the main emphasis of my question is about the correctness of the reasoning in the 8th case of the provided proof (with a diagram).
Original question:
I would like to know, whether the following proof, is a valid way to prove that $a^2 + b^2 \neq 3c^2$ for all $a, b, c \in Z$ (except the trivial case, when $a=b=c=0$). More formally, we have to prove the correctness of the following statement:
$$P: (\forall a,b,c \in Z, a^2 + b^2 \neq 3c^2 \lor (a=b=c=0))$$
Proof. (by contradiction)
For the sake of contradiction let's assume, that there exist such $a, b, c \in Z$, that $a^2 + b^2 = 3c^2$ (and the combination of $a,b,c$ is not a trivial case). More, formally, let's assume that $\neg P$ is true:
$$\neg P: (\exists a,b,c \in Z, a^2 + b^2 = 3c^2 \land \neg (a=b=c=0))$$
There are $2^3$ possible combinations of different parities of $a,b,c$ (8 disjoint cases, which cover entire $Z^3$). So, in order to prove the original statement, we have to consider each case, and show that the true-ness of the $\neg P$ always leads to some sort of contradiction.
Let's consider 8 possible cases (7 of which are simple, whereas the 8th case looks a bit intricate, and I am not sure regarding its correctness):
Case 1) $a$ is odd, $b$ is odd, $c$ is odd
Thus:
$a = (2x + 1)$, $b = (2y + 1)$, $c = (2z + 1)$ for some $x, y, z \in Z$
So:
$$
a^2 + b^2 = 3c^2 \\
\implies (2x + 1) ^2 + (2y + 1)^2 = 3 \cdot (2z + 1)^2 \\
\implies 2 \cdot (2x^2 + 2x + 2y^2 + 2y + 1) = 2 \cdot (6z^2 + 6z + 1) + 1 \\
\implies even\ number = odd\ number \\
$$
However, the derived result contradicts to the fact that odd numbers and even numbers can't be equal.
Hence: $(even\ number = odd\ number) \land (even\ number \neq odd\ number)$, or equivalently: $(even\ number = odd\ number) \land \neg (even\ number = odd\ number)$. Contradiction.
Case 2) $a$ is odd, $b$ is odd, $c$ is even
Thus:
$a = (2x + 1)$, $b = (2y + 1)$, $c = 2z$ for some $x, y, z \in Z$
So:
$$
a^2 + b^2 = 3c^2 \\
\implies (2x + 1) ^2 + (2y + 1)^2 = 3 \cdot (2z)^2 \\
\implies 2 \cdot (2x^2 + 2x + 2y^2 + 2y + 1) = 12z^2 \\
\implies 2 \cdot (x^2 + x + y^2 + y) + 1 = 6z^2 \\
\implies odd\ number = even\ number
$$
Contradiction.
Case 3) $a$ is odd, $b$ is even, $c$ is odd
Thus:
$a = (2x + 1)$, $b = 2y$, $c = (2z + 1)$ for some $x, y, z \in Z$
So:
$$
a^2 + b^2 = 3c^2 \\
\implies (2x + 1) ^2 + (2y)^2 = 3 \cdot (2z + 1)^2 \\
\implies 4x^2 + 4x + 1 + 4y^2 = 12z^2 + 12z + 3 \\
\implies 4\cdot(x^2 + x + y^2) = 2 \cdot (6z^2 + 6z + 1) \\
\implies 2\cdot(x^2 + x + y^2) = 6z^2 + 6z + 1 \\
\implies even\ number = odd\ number
$$
Contradiction.
Case 4) $a$ is odd, $b$ is even, $c$ is even
The square of an odd number is odd (so, $a^2$ is odd).
The square of an even number is even (so, $b^2$ and $3c^2$ are even).
Fact: the sum of an even number and an odd number is odd.
However, equality: $a^2 + b^2 = 3c^2$ leads to the conclusion, that: $odd\ number + even\ number = even\ number$
Contradiction.
Case 5) $a$ is even, $b$ is odd, $c$ is odd
Symmetric to the Case 3 (because $a$ and $b$ are mutually exchangeable), which shows the contradiction.
Case 6) $a$ is even, $b$ is odd, $c$ is even
Symmetric to the Case 4, which shows the contradiction.
Case 7) $a$ is even, $b$ is even, $c$ is odd
Thus:
$a = 2x$, $b = 2y$, $c = (2z + 1)$ for some $x, y, z \in Z$
So:
$$
a^2 + b^2 = 3c^2 \\
\implies 4x^2 + 4y^2 = 12z^2 + 12z + 3 \\
\implies even\ number = odd\ number
$$
Contradiction.
Case 8) $a$ is even, $b$ is even, $c$ is even
Thus:
$a = 2x$, $b = 2y$, $c = 2z$ for some $x, y, z \in Z$
So:
$$
a^2 + b^2 = 3c^2 \\
\implies 4x^2 + 4y^2 = 3 \cdot 4z^2 \\
\implies x^2 + y^2 = 3z^2
$$
Now, we are faced with the similar instance of the problem, however, the size of the problem is strictly smaller ($x = {a \over 2}$, $y = {b \over 2}$, $z = {c \over 2}$).
At first glance, it seems that we have to consider again the eight possible parities of $x, y, z$. However, if we analyze all dependencies between the cases of the problem, we will notice that the only possible outcomes are either contradiction or the trivial case:
We have shown the contradiction in all cases, hence we have subsequently proved the original statement.
$\blacksquare$
So, I would like to know, if there is any problem with reasoning in the 8th case?
Best Answer
There is truth in your method for case 8. It is called inifinte descent and equivalent to induction (i.e., alternatively you might start away with assuming that $(a,b,c)$ is the smallest non-trivial solution, and then $(a/2,b/2,c/2)$ cannot be a solution).