Answer to Question #174023 in Discrete Mathematics for Tracy Kristoff

Question

Prove that 2/n⁴-3 if and only off 4/n²+3

Answer 1

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 $4|(n^2+3).$