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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:

enter image description here

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?

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  • $\begingroup$ $+1$ for the diagram! (and overall well-presented question) $\endgroup$
    – user384138
    Jan 22, 2017 at 19:11
  • $\begingroup$ Everything is very cumbersome. And if the equation will be different? Or is it proof specifically for a single equation? The solvability of quadratic forms is reduced to finding the solvability of some equivalent of Pell's equation. Take the formula and find out when it has at least one full rate. math.stackexchange.com/questions/1513733/… $\endgroup$
    – individ
    Jan 23, 2017 at 4:21
  • $\begingroup$ @individ, thank you for your advice. I fully agree with you regarding the usefulness of solving the general case of the problem. However, the main accent of my question is focused on the correctness of the reasoning in the 8th case of the provided proof (with a diagram). I would like to know, if there are any flaws in the reasoning inside the 8th case. $\endgroup$
    – stemm
    Jan 23, 2017 at 20:57
  • $\begingroup$ Anyone can ask a question. You specify the form of the solution. You equation is not solved. Why did you decide that the form of the solution needs to look like this? Maybe the solution needs to have a different view? If there is a solution you can always obtain another by multiplying or dividing by a common divisor. What is the evidence? The equations to be solved, rather than guessing what the solution might be. $\endgroup$
    – individ
    Jan 24, 2017 at 4:17
  • $\begingroup$ @individ as I have already pointed, the main emphasis of the question is on the correctness of reasoning in the 8th case of the presented proof. Was that a correct logical step - to rely on the graph of dependencies between the cases of the proof? If yes, I would appreciate the references to examples of other proofs, which make use of a similar technique. If approach is wrong, I would like to know the reason why. $\endgroup$
    – stemm
    Jan 24, 2017 at 7:45

2 Answers 2

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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).

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  • $\begingroup$ Hagen von Eitzen, thanks a lot for the explanation! $\endgroup$
    – stemm
    Jan 29, 2017 at 16:55
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Suppose a solution with at least one of $a,b\neq 0$ exists. If $(a,b,c)$ is a solution then $(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)},\frac{c}{\gcd(a,b)})$ is also a solution (and all terms are integers), and $a$ and $b$ are coprime.

Now notice that $a^2+b^2$ cannot be a multiple of $3$ unless both are multiples of $3$ (because $a^2\equiv 0$ or $1\bmod 3$). We conclude that if $a^2+b^2=3c^2$ we have $a=b=0$.

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    $\begingroup$ This proof is basically infinite descent, only that instead of using infinite descent we prove that a coprime solution must exist if a non-trivial solution exists. $\endgroup$
    – Asinomás
    Jan 22, 2017 at 19:16
  • $\begingroup$ Jorge Fernández Hidalgo, thank you for the suggested proof. However, I would really appreciate any comments regarding the correctness of the proof, which described in the question. $\endgroup$
    – stemm
    Jan 29, 2017 at 16:43
  • $\begingroup$ where does $(a,b)\mid c$ come from? :o $\endgroup$
    – AlvinL
    Aug 2, 2019 at 19:08
  • $\begingroup$ You have to prove it but it is pretty straight forward $\endgroup$
    – Asinomás
    Aug 3, 2019 at 0:20
  • $\begingroup$ @JorgeFernándezHidalgo Well, that's why I asked, doesn't look that obvious at all. We could analyse two cases: $(3,c^2) =1$ and $(3,c^2)=3$, but it's not clear cut. How do you approach this? $\endgroup$
    – AlvinL
    Aug 3, 2019 at 8:10

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