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I am kind of stuck on a practice problem relating to proof by contradiction that goes as follows:

"Prove that there do not exist positive integers $m$ and $n$ such that $m^2 - n^2 = 1$"

For the outline of my proof (which is quite long and inefficient): I assumed, to the contrary, that there exists positive integers $m$ and $n$ such that $m^2 - n^2 = 1$"

Then considered four possible cases where:

case 1: $m$ even, and $n$ odd, and showing that $m^2 - n^2 = 1$ is a contradiction.

case 2: $m$ even, $n$ even, and showing that $m^2 - n^2 = 1$ is a contradiction.

case 3: $m$ odd, $n$ odd and showing that $m^2 - n^2 = 1$ is a contradiction.

case 4: $m$ odd, $n$ odd and showing that $m^2 - n^2 = 1$ is a contradiction (things got a little sticky here) so I used a lemma given by:

lemma: The product of two consecutive positive integer, cannot be the square of an integer.

Now where I ran into a problem was proving the lemma (which I am not 100% is true or not but intuition tells me that it seems like it), which basically renders my 4th case invalid.

I am pretty sure that I am doing unnecessary work here and would greatly appreciate feedback since I am trying to practice and the textbook didnt give a solution to this problem.

Thank you.

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All you have to do is factor and assume the statement is true. We see that $$m^2 - n^2 = (m + n)(m - n) = 1.$$

If $m$ and $n$ are positive integers, both $m + n$ and $m - n$ must also be integers. Thus, we must have that $m + n = m - n = 1.$

But clearly, this is not possible - the only solution to the system is $m = 1$ and $n = 0,$ which is not a positive integer!

And we are finished! Hope this helped!

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From $m^2-n^2= (m-n)(m+n)=1$.

The can only happen if $m-n=1$ and $m+n=1$ adding those two we see that $2m=2\Rightarrow m=1$. But then $n=0$.

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Assume there are. Then $$m^2-n^2=(m+n)(m-n)=1$$ Hence, $$m+n=m-n=1$$ or $$m+n=m-n=-1$$ The first case leads to $n=0$ and $m=1$, contradiction. The second case leads to $n=0$ and $m=-1$, contradiction.

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