# Proofs-Divisibility of Integers

• May 14th 2009, 11:22 PM
utopiaNow
Proofs-Divisibility of Integers
Let $n \in \mathbb{Z}$. Prove that $2\ |\ (n^4 - 3)$ if and only if $4\ |\ (n^2 + 3)$.

Attempt at Proof:
First I just dealt with the forward implication: if $2 \ |\ (n^4 - 3)$ then $4 \ |\ (n^2 + 3)$. To prove this I took the contrapositive of this statement which is: If $4 \not\vert\ (n^2 + 3)$, then $2 \not\vert\ (n^4 - 3)$.
$4 \not\vert\ (n^2 + 3)$ can be broken down into 3 cases, the case I'm having trouble with is: let $(n^2 + 3) = 4q + 2$, then $n^2 = 4q - 1$ and then $n^4 - 3 = (4q - 1)^2 - 3 = 16q^2 - 8q - 2 = 2(8q^2 -4q - 1)$ which is an even number and so $2 \ |\ (n^4 - 3)$, but that's not what I wanted for the contrapositive.

What am I doing wrong?
• May 14th 2009, 11:45 PM
CaptainBlack
Quote:

Originally Posted by utopiaNow
Let $n \in \mathbb{Z}$. Prove that $2\ |\ (n^4 - 3)$ if and only if $4\ |\ (n^2 + 3)$.

Attempt at Proof:
First I just dealt with the forward implication: if $2 \ |\ (n^4 - 3)$ then $4 \ |\ (n^2 + 3)$. To prove this I took the contrapositive of this statement which is: If $4 \not\vert\ (n^2 + 3)$, then $2 \not\vert\ (n^4 - 3)$.
$4 \not\vert\ (n^2 + 3)$ can be broken down into 3 cases, the case I'm having trouble with is: let $(n^2 + 3) = 4q + 2$, then $n^2 = 4q - 1$ and then $n^4 - 3 = (4q - 1)^2 - 3 = 16q^2 - 8q - 2 = 2(8q^2 -4q - 1)$ which is an even number and so $2 \ |\ (n^4 - 3)$, but that's not what I wanted for the contrapositive.

What am I doing wrong?
$2\ |\ (n^4 - 3)$ implies that $n^4$ is odd which implies $n$ is odd.
If $n$ is odd there exists a $k \in \mathbb{Z}$ such that $n=2k+1$
$n^2+1=(2k+1)^2+3=4k^2+4k+4$
The other direction of the proof proceeds by observing that $4\ |\ (n^2 + 3)$ implies $n$ odd ...