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I have to use proof by contradiction to solve the problem and I believe that proving $m^2 - n^2 = 1$ is the way to go with this, but where do I start?

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For the sake of a contradiction, assume $m^2-n^2=1$.

$$\begin{array}{rcl} m^2-n^2 &=& 1 \\ (m-n)(m+n) &=& 1 \end{array}$$

So, $m-n=1$ and $m+n=1$.

However, since $m$ and $n$ are positive integers, $m \ge 1$ and $n \ge 1$, so $m+n \ge 2$.

Hence, a contradiction is reached, since $1 \not \ge 2$.

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You don't need contradiction, by the way: it can be done directly as "If $m \leq n$ then $m^2-n^2 \leq 0$ so isn't $1$. Otherwise, $(m+n)(m-n)$ is something-bigger-than-$1$ times something-bigger-than-or-equal-to-$1$, so is bigger than $1$."

However, if you really must use contradiction: start by supposing $m^2-n^2=1$. Then $(m+n)(m-n) = 1$. Since both brackets are integers, and the first is positive, they are both equal to $1$. What can you deduce?

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If you have to do this by a proof by contradiction, your start is:

Assume that for some positive integers $m$ and $n$: $m^2 - n^2 = 1$

... and now you need to derive a contradiction from this!

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$m^2-n^2=1$ implies there exist two perfect squares with a difference of $1$. It is obvious that none exist, since squaring even the first two positive integers and taking the difference gives $4-1=3>1$, and the difference of any two squares is bounded below by the difference of the smallest two.

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  • $\begingroup$ Proof be example is not acceptable. This helps develop the intuition, but in no way proves anything other than your particular example. $\endgroup$ – Laars Helenius Apr 14 '17 at 18:00
  • $\begingroup$ This isn't proof by example. The difference of two squares is bound by the difference of the smallest two. I added that, although I feel like it was obvious what I was saying... $\endgroup$ – The Count Apr 14 '17 at 18:01
  • $\begingroup$ Why? Again I am not saying you are wrong, but this is incomplete until you prove your claim. $\endgroup$ – Laars Helenius Apr 14 '17 at 18:04
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    $\begingroup$ I suppose I see what you are saying, but I guess that I've learned over time on this site that for such a small question with little effort demonstrated, completeness is not required. Hints are acceptable. People do that all the time, and I don't see why that's okay but this hint isn't. $\endgroup$ – The Count Apr 14 '17 at 18:08
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Obviously, $m$ and $n$ can not be simultaneously even or odd.

$m$ must be odd and $n$ even or vice versa.

Case 1: $m=2a$ and $n=2b+1$. Then:

$(2a)^2-(2b+1)^2=1 \Rightarrow 2a^2-2b^2-2b=1$, which is impossible.

Case 2: $m=2a+1$ and $n=2b$. Note: $m>n>0 \Rightarrow a\ge b>0$. Then:

$(2a+1)^2-(2b)^2 =1 \Rightarrow a^2+a-b^2=0\Rightarrow (a-b)(a+b)=-a$, which is impossible.

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