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Proof that all integers are equal
*June 1, 2017*

*Posted by Ezra Resnick in Math.*

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1 comment so far

In particular, we prove that ** n = -1 for any integer n**.

First, assume that *n* is odd. That means that *n* – 1 is even, so *n* – 1 = 2*k* for some integer *k*.

The following identity is trivially true for any *n*:

*n*^{2} – 1 = (*n* – 1)(*n* + 1)

We can substitute 2*k* for (*n* – 1):

*n*^{2} – 1 = 2*k*(*n* + 1) = 2*k**n* + 2*k*

Next, we subtract (2*kn* + *n* – 1) from both sides of the equation, yielding:

*n*^{2} – 2*kn* – *n* = 2*k* – *n* + 1

Factoring out (*n* – 2*k* – 1) produces:

*n*(*n* – 2*k* – 1) = –(*n* – 2*k* – 1)

Now we simply divide both sides of the equation by (*n* – 2*k* – 1), and voila:

*n* = -1

Alternatively, assume that *n* is even. That means that *n* – 1 is odd, so *n* – 1 = 2*k* + 1 for some integer *k*.

Once again, we begin with the identity:

*n*^{2} – 1 = (*n* – 1)(*n* + 1)

This time we substitute (2*k* + 1) for (*n* – 1):

*n*^{2} – 1 = (2*k* + 1)(*n* + 1) = 2*k**n* + 2*k* +* n* + 1

We subtract (2*kn* + 2*n* – 1) from both sides of the equation:

*n*^{2} – 2*kn* – 2*n* = 2*k* – *n* + 2

We factor out (*n* – 2*k* – 2):

*n*(*n* – 2*k* – 2) = –(*n* – 2*k* – 2)

And now we simply divide both sides of the equation by (*n* – 2*k* – 2) to complete the proof:

*n* = -1

*Q.E.D.*