How do I complete this proof? Prove the following statement by contradiction:
For  each two positive integers $x$ and $y$, $x^2-y^2 \neq 1$
Proof:
We use proof by contradiction.
1) Suppose $x^2-y^2 = 1$
2) Assuming $x,y\in\mathbb{Z^+}$, let $x = \frac { a }{ b }$ and let $y=\frac {c}{d}$ such that: $a,b,c,d \in\mathbb{Z}$ and $b \neq 0$, and $d \neq 0$  
3) $$(\frac { a }{ b } )^{ 2 }-(\frac { c }{ d } )^{ 2 }=1\Rightarrow \frac { a^{ 2 } }{ b^{ 2 } } -\frac { c^{ 2 } }{ d^{ 2 } } =1\Rightarrow \frac { a^{ 2 }d^{ 2 }-b^{ 2 }c^{ 2 } }{ b^{ 2 }d^{ 2 } } =1$$
4) ...
I'm not sure if I approached this proof the right way, but I am not sure how to convince my reader that $\frac { a^{ 2 }d^{ 2 }-b^{ 2 }c^{ 2 } }{ b^{ 2 }d^{ 2 } } =1$ is a contradiction. I'm using MIT OCW to study discrete math and I read this in one of the chapters and am trying to adhere to some of the basic proof writing principles laid out in the text made available by MIT. 
Here is what it said: "Most especially, don’t use phrases like “clearly” or “obviously” in an attempt to bully the reader into accepting something you’re having trouble proving. Also, go on the alert whenever you see one of these phrases in someone else’s proof."
How do I avoid this in this proof?
 A: HINT:
$$1=x^2-y^2=(x+y)(x-y)$$
$\iff\dfrac1{x+y}=x-y$ which is an integer $\implies(x+y)$ divides $1$
As $x,y>0; x+y$ must be $1\iff x-y=1$
A: There are some $x,y \in \Bbb{Z}^{+}$ such that $x^{2} - y^{2}=1$ only if $(x-y)(x+y) = 1$, only if $x-y = 1$, and only if $x+y = 1+2y > 1$, $\to \gets$.
A: Sometimes introducing an additional arbitrary quantity, or even several
such quantities, will help complete a proof, but the more likely
outcome is like a lot of smoke and maybe some mirrors: it will prevent seeing things, or maybe lead to seeing things that are not there.
That kind of outcome is to be avoided as carefully as words such as
"obviously".
You have $x = \frac ab$ and let $y=\frac cd$.
Therefore $a = bx$ and $c = dy$.
You also found that
$$ \frac { a^2d^2 - b^2c^2 }{ b^2d^2 } =1. $$
Since $b^2d^2 \neq 0$, this is equivalent to
$$ a^2d^2 - b^2c^2 = b^2d^2, $$
and since $a = bx$ and $c = dy$ this is equivalent to
$$ b^2d^2x^2 - b^2d^2y^2 = b^2d^2. $$
So all you really accomplished by the substitutions 
$x = \frac ab$ and $y=\frac cd$ was to show you could
multiply both sides of $x^2 - y^2 = 1$ by a constant.
Instead, try to deduce useful properties of the numbers you have.
Assume that $x^2 - y^2 = 1$.
You could next factor $x^2 - y^2$ as other answers do,
or you could observe that if the assumption implies that
$x^2 = y^2 + 1$, therefore $x^2 > y^2$, and so $x > y$.
(If you haven't already seen a proof that $x^2 > y^2$ implies $x > y$,
that fact is also easy to prove by contradiction.)
Since $x$ and $y$ are integers, their difference $x - y$ is also an integer,
and since $x > y$, the difference $x - y$ is positive.
Let $n = x - y$. Then $n$ is a positive integer and $x = y + n$.
Therefore
$$ x^2 - y^2 = (y + n)^2 - y^2 = (y^2 + 2ny + n^2) - y^2 = 2ny + n^2, $$
where $n$ and $y$ both are positive integers, that is,
$n \geq 1$ and $y \geq 1$.
Can you draw a contradiction from the statements $x^2 - y^2 = 1$,
$x^2 - y^2 = 2ny + n^2$, $n \geq 1$, and $y \geq 1$?

By the way, $n$ in the argument above is what I would consider an example of a useful introduction of a named quantity; we could have just kept on writing "$x - y$" everywhere, but it was easier to write $n$.
