Elementary Proof Relating to Pythagorean Triples Please could I get some feedback on the correctness and overall quality of the following proof, which is meant to show that all Pythagorean triples contain unique integers.
To Prove: all Pythagorean triples contain unique integers.
First we note that in a right-angled triangle, if $c$ is the hypotenuse and $a,b$ are the remaining sides, $c > a$ and $c > b$ by triangle inequality. Now it remains to prove that
$\forall (a,b,c) \in \mathbb{N}$, where $a^2+b^2=c^2$, $a \ne b$
We use contradiction.
Assume that $a = b$. Then we have $a^2 + a^2 = c^2$
Thus, $c = \sqrt{2a^2} = a\sqrt{2}$
Since an integer multiple of an irrational number remains irrational, $c \notin \mathbb{N}$, which is a contradiction of our original assumption.
Therefore $a \ne b$
Since $a \ne b$ and $c > a$ and $c > b$,
$a,b,c$ are distinct integers and theorem is proven.
$\square$

We are given$\quad A=m^2-n^2\quad B=2mn\quad C=m^2+n^2.$ and we wish to prove that that $A,B,C$ are unique integers. 
If $A=B$ then$\quad(m^2-n^2)^2+(m^2-n^2)^2=C^2\Rightarrow C=(m^2-n^2)\sqrt{2}$ so it is not Pythagorean triple where all sides are integers.
If $B=A$ then$\quad (2mn)^2+(2mn)^2=8m^2n^2=C\Rightarrow C=2mn\sqrt{2}$ so it is also not Pythagorean triple where all sides are integers.
If $A=C$ then$\quad m^2-n^2=m^2+n^2\Rightarrow 0=2n^2$ and any zero value of $n$ will produce a trivial triple where $B=0$ such as $(1,0,1)$.
If $B=C$ then $\quad 2mn=m^2+n^2\Rightarrow m^2-2mn+n^2=(m-n)^2=0$. This can only be true if $m=n$ and that will always produce a trivial triple where $A=0$ such as $(0,1,1)$
None of these conditions produce a Pythagorean triple. Therefore, $A,B,C$ are always distinct integers.
