is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

You will get a detailed answer to your question or assignment in the shortest time possible.

## Here's the Solution to this Question

Direct Proof:

Assume that $2|n^4-3.$ This means that there is some integer $y$ such that $n^4-3=2y.$ Then $n^4=2y+3=2(y+1)+1.$

Since $y+1$ is an integer, we see that $n^4$ is odd.

Let $x\in \Z.$ Then $x^2$ is even if and only if $x$ is even.

Taking the contrapositive of both directions of the biconditional gives: $x^2$ is odd if and only if $x$ is odd.

Let $x=n^2.$ Then $n^4$ is odd if and only if $n^2$ is odd. Then $n^2$is odd if and only if $n$ is odd.

Therefore there is an integer $k$ such that $n=2k+1.$

We write

$n^2+3=(2k+1)^2+3=4k^2+4k+4$

$=4(k^2+k+1)$

Since $k^2+k+1$ is an integer, we see that $4|(n^2+3).$

Contrapositive:

Assume that $4\not| (n^2+3).$ This means that for $k\in \Z,$ either

(i)$n^2+3=4k+1,$ (ii) $n^2+3=4k+2,$ or (iii) $n^2+3=4k+3.$

If $n$ is even then there is an integer $m$ such that

$n^2+3=(2m)^2+3=4m^2+3$

which implies that we must be in case (iii).

If $n$ is odd then there is an integer $m$ such that

$n^2+3=(2m+1)^2+3=4m^2+4m+4$

$=4(m^2+m+1)$

In this case we see that $4|(n^2+3).$This is contrary to our assumption that $4\not| (n^2+3)$ so under our assumption this case can never occur.

Thus we only have to consider case (iii) $n^2+3=4m+3.$

Rearranging we can write $n^2=4k.$ Hence

$n^4-3=(4m)^2-3=16m^2-3$

$=2(8m^2-1)+1.$

This means that the remainder is $1$ when $n^4-3$ is divided by $2.$Therefore,

$2\not|(n^4-3).$

Therefore for $n\in \Z$ $2|(n^4-3)$ if and only if $4|(n^2+3).$