Prove the following statement using two methods of proof: If $n$ and $m$ are non-zero integers, then $n^2-m^2≠1$.

Question

I have already seen the proof by contradiction for this problem, but I can't figure out any other way to prove it. This was on a test, so far we have only learned direct method of proof, contrapositive method, contradiction method, and briefly cases method (which is just a combination of the others). The book we use also has this statement, but the only proof it provides (which we have to fill in ourselves) is by contradiction. On the board my professor also included the statement If ab=1 and a,b are integers, then a=b=±1. That statement is used in the contradiction method at least, but I don't know what to do for any other method.

Any ideas are appreciated.

Answer 1

You can calculate the difference between two successive squares explicitly, say for $n$ and $n-1$.

What can you say about this difference (for the integers you're interested in)?

What can you then say about the difference if the integers are not successive?

Answer 2

For $n = m$ the statement is obviously true. Now consider $n \neq m$, without loss of generality we can assume that $n > m$. Since every square is a sum of subsequent odd numbers, we have $$ n^2 = \sum_{i=0}^{N} (2i+1) \quad\text{and}\quad m^2 = \sum_{i=0}^{M} (2i+1) $$ for some $N,M\in\mathbb{N}$, $N > M$. Then $$ n^2 - m^2 = \!\sum_{i=M+1}^{N} \!(2i+1), $$ where each of components of the sum is bigger that one.

Answer 3

Note that $n^2-m^2=(n+m)(n-m)$ is the product of two distinct numbers (using $m\neq 0$) and cannot be $1\times 1$ or $-1\times -1$.

You can do this by contradiction. Or you can do it by cases:

Suppose the first factor is $n+m=1$, then the product is $n-m\neq 1$

Suppose the second factor is $n+m=-1$ then the product is $m-n$, but this can only be equal to $1$ if $m=0,n=-1$ and we are given $m\neq 0$

Suppose the first factor is $n+m\neq \pm 1$, then the product is $\neq 1$ since it is a multiple of $r=n+m$ and either $r=0$ (product is zero) or $|r|\gt 1$

Now this may conceal a contradiction (cases often do - in saying "this case doesn't work" you can be saying "this case leads to a contradiction" - and the work behind the case argument may have involved a contradiction). General statements about integers usually depend on induction somewhere in the background as well. So I don't know whether this is the argument you already have, but I offer it as a possible way of using the information you gave.