What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:

Theorem:  $\sqrt{2}$ is irrational.
Proof:
  $3^2-2\cdot 2^2 = 1$.

(That's it)
That is a corollary of
this result:

Theorem:
  If $n$ is a positive integer
  and there are positive integers
  $x$ and $y$ such that
  $x^2-ny^2 = 1$,
  then
  $\sqrt{n}$ is irrational.

The proof is in two parts,
each of which
has a one line proof.
Part 1:

Lemma: If 
  $x^2-ny^2 = 1$,
  then there are arbitrarily large integers
  $u$ and $v$ such that
  $u^2-nv^2 = 1$.

Proof of part 1:

Apply the identity
  $(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
  as many times as needed.

Part 2:

Lemma:  If
  $x^2-ny^2 = 1$
  and
  $\sqrt{n} = \frac{a}{b}$
  then
  $x < b$.

Proof of part 2:

$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
  or
  $b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
  so
  $x < b$.

These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
 is that
if $n > 0$
then
$n \ge 1$.

$$\boxed{\text{If the boxed statement is true, then the square root of two is irrational.}}$$
Lemma. The boxed statement is true.
Proof. Assume for a contradiction that the boxed statement is false. Then it has the form "if $S$ then $T$" where $S$ is false, but a conditional with a false antecedent is true.
Theorem. The square root of two is irrational.
Proof.


*

*The boxed statement is true. (By the Lemma.)

*If the boxed statement is true, then the square root of two is irrational. (This is the boxed statement itself.)

*The square root of two is irrational. (Modus ponens.)
