# Beginning Discrete Proof

• Mar 10th 2009, 08:12 AM
jthrock
Beginning Discrete Proof
Hello everyone,
Got stuck on this one and wanted to see if anyone could help me here.
Thank you.

For all integers n, if n is a perfect square, then n+2 is NOT a perfect square.

I keep running in circles, can anyone help me prove this one?
Thanks!
• Mar 10th 2009, 09:50 AM
Plato
Quote:

Originally Posted by jthrock
For all integers n, if n is a perfect square, then n+2 is NOT a perfect square.

If $N = K^2 \;\& \;N + 2 = J^2 \; \Rightarrow \;J > K$. These are positive integers.
That means $2 = J^2 - K^2 = \left( {J - K} \right)\left( {J + K} \right)$.
There is contradiction hiding there. Where is it?
• Mar 10th 2009, 09:58 AM
jthrock
I know, this is where we got stuck in class. The prof was unsure about it either so we decided this would be a "quiz" in the future.
• Mar 10th 2009, 10:13 AM
Plato
Both $(J-K)\;\&\;(J+K)$ are positive integers.
Both factors of 2. What are the factors of 2?
• Mar 10th 2009, 10:26 AM
jthrock
I'm sorry but I'm a bit confused, the factors of 2 are 2 *1. But I do not understand where you are going with this. By the way, thank you for your help with this.