I am trying to prove the following statement:
1: ($\forall a,b,c\in\mathbb{Z}|a,b,c\geq 1: a^2 + b^2 \neq 3*c^2 $)
using modular arithmetic. Specifically, two previosly derived properties of $Mod_4$ (notationally used as a function on values in this post):
- $Mod_{4}(a^2 + b^2)\neq 3$
- $Mod_{4}(Mod_{4}(n)^2)\in\{1,0\}, \forall n \in \mathbb{Z}$
I plan to do this by contradiction. The opposite statement of $1$, is
- $(\exists a,b,c\in\mathbb{Z}|a,b,c\geq 1: a^2 + b^2 = 3*c^2) $
Using assumption $a^2 + b^2 = 3*c^2$, $2 $, and modular arithemetic, i now arive at:
$Mod_{4}(3*Mod_{4}(Mod_{4}(c)^2))\neq3$
From $3$ I know that $Mod_{4}(Mod_{4}(c)^2)$ is either $0$ or $1$. One can easily directly check, that if it is $0$ we get: $0\neq3$ which is $true$. If it is $1$ we get $3\neq3$ which is false.
So we have, that if there exists solutions, then some of those solutions lead to contradictions. Logically speaking - as it (seemingly?) involves predicate logic (i only know propositional) i can't really wrap my head around if this proves the statement or not. There exists, leads to there might exists contradictions?
Of course this not enough; getting a disjunction of cases from a hypothesis, and then finding that one of the cases is contradictory does not imply that the hypothesis is wrong, it only implies that you're in the other case.
For instance : is there a solution to $|x|=-x+1$ ? Well either $x<0$, but then this means $-x = -x+11$ so $0=1$: this is absurd; or $x\geq 0$.
But you do see that I can't stop there and say "since one of the cases leads to a contradiction, there is no solution" (this is the same as your situation : there "might"exist negative solutions, and those lead to a contradiction).
You still have to see what happens in the other cases : in my toy example, if $x\geq 0$, the equation implies $x=-x+1$ and so $x=\frac{1}{2}$. We can then check that $\frac{1}{2}$ is a solution : we would have been wrong to stop at the first case.