Thread: if (K^2 -1) is even, then either k+1 or k-1 is even

This has to do with positive integers.

I'm stating that since k^2-1 is even, then 2 must divide it, so by the definition:

k^2 -1 = 2c , for some integer c.

Then i figures since k^2 -1 is (k+1)(k-1) that I could use this. The only problem is that I end up with
k-1 = 2*(c/k+1) , which doesn't work because of the fraction.


I am also confused because of the OR in the statement I am trying to prove.

Originally Posted by Th3sandm4n

This has to do with positive integers.

I'm stating that since k^2-1 is even, then 2 must divide it, so by the definition:

k^2 -1 = 2c , for some integer c.

Then i figures since k^2 -1 is (k+1)(k-1) that I could use this. The only problem is that I end up with
k-1 = 2*(c/k+1) , which doesn't work because of the fraction.


I am also confused because of the OR in the statement I am trying to prove.

A few things to note first:

2 even numbers multiply to give an even number.
1 odd number and one even number multiply to give an even number.

In other words, for a product to be even, at least one of the factors has to be even!

Consider



WE know the LHS if even because 2 is even. And any number times an even number is even!

So we know the RHS must also be even.

And if a product is to be even, at least one of the factors has to be even. So either (k+1) or (k-1) or both is even.

Ah, so simple! This is the tricky part about number theory, you have to think about these things! I'm way too used to just algebra and working things out.

Thanks!

Originally Posted by Th3sandm4n

Ah, so simple! This is the tricky part about number theory, you have to think about these things! I'm way too used to just algebra and working things out.

Thanks!

Yes. The ultime result is that they are both even.



, which is odd.

which is even.

So k+1 AND k-1 are even. K is odd.

Yeah that's what I figured just because it seems so common sense. But that is what makes number theory so difficult, you have to prove the common sense stuff.