Proving that for every integer $n, n^3 + n$ is even via contraposition

Question

I know this is trivial with direct proofs, but how would one go about proving:

For every integer n, n^3 + n is even

using contraposition? I'm a little rusty on how to properly use contrapositive proofs. I know that if given $P \Longrightarrow Q$, the contrapositive is $\neg Q\Longrightarrow \neg P$.

So to prove via contraposition, assume $\neg Q$ is true and if $\neg P$ logically follows, then we have shown $P \Longrightarrow Q$.

$Q$ or $\neg Q$ is trivial, however, what is $P$ in this case?

Edit:

This was just out of curiosity in trying to figure out ways to prove this indirectly.

Answer 1

Over-View :

I have given a way to generate the Contra-Positive , then I have suggested a way to Prove that.
That Proof is long & Cumbersome , hence I have suggested a variation with a slight alteration , which gives a Simple Proof By Contra-Positive.

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Statement given is like this ( With $Z$ = Integral Numbers ) :
$P_1$ : $\forall n \in Z : Even(n^3+n)$

It has no Implication , hence Contra-Positive will not arise.

We have to make it with Implication to get a Contra-Positive. Doing that requires using a larger Universal Set ( Eg With $Q$ = Rational Numbers & With $R$ = Real Numbers ) .

These 2 will work :
$P_2$ : $\forall n \in Q : Integer(n) \implies Even(n^3+n)$
$P_3$ : $\forall n \in R : Integer(n) \implies Even(n^3+n)$

Then the Contra-Positives are :
$P_4$ : $\forall n \in Q : \lnot Even(n^3+n) \implies \lnot Integer(n)$
$P_5$ : $\forall n \in R : \lnot Even(n^3+n) \implies \lnot Integer(n)$

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Proof By Contra-Positive :

When $n^3+n$ is not Even , we have $n^3+n=2m+1$ which has the Unique Solution (real) given by Wolfram :

Problem now becomes showing that $N$ is not Integer , no matter what Integer $M$ is.
That can be done , though it require too much unnecessary work.

Thus Proof By Contra-Positive is too Cumbersome here.
Direct Proof is much Easier !

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Now , it is your Curiosity to try this type of Proof , hence we can make it easier to go forward by considering this :
$P_6$ : $\forall n \in R : Integer(n) \implies Even(n^2+n)$

The Contra-Positive is :
$P_7$ : $\forall n \in R : \lnot Even(n^2+n) \implies \lnot Integer(n)$

When $n^2+n$ is not Even , we have $n^2+n=2m+1$ , which gives :

$n=\frac{-1 \pm \sqrt{1+4(2m+1)}}{2}$

$n=\frac{-1 \pm \sqrt{1+8m+4}}{2}$

$n=\frac{-1 \pm \sqrt{8m+5}}{2}$

No matter what $m$ is , $8m+5$ is not a Square ( It is easy to check that $S^2 \equiv 1 \mod 8$ & $S^2 \not \equiv 5 \mod 8$ ) , hence $n$ is irrational , not Integer !

We have the Proof By Contra-Positive here !
Direct Proof will work Easier too.

Answer 2

I interpreted your question as follows :

Prove that, if $n$ is an integer, then $n^3+n$ is always even for all $n\in\mathbb Z$ , using the method of Proof by contradiction .

Suppose that, there exist $n\in\mathbb Z$ , such that $n^3+n$ is an odd integer .

If $n^3+n$ is odd, then $n(n^2+1)$ is odd. This implies that, $n$ and $n^2+1$ can not be even integers . Thus, $n$ is odd . If $n$ is odd, then $n^2+1$ is an even integer .

A contradiction .

Answer 3

If we rewrite your statement for integers modulo $2$, then it says: "for every element $n$ in $\Bbb Z/2\Bbb Z$ we have $n+n^3=0$." But $n^3=n$ in $\Bbb Z/2\Bbb Z$, so we have:

For every $n$ in $\Bbb Z/2\Bbb Z$ we have $n+n=2n=0$.

This is clear, because $2=0$ in $\Bbb Z/2\Bbb Z$. Now we can prove this also by contradiction. Suppose that $2n$ is nonzero. Then $n$ and $2$ are nonzero in $\Bbb Z/2\Bbb Z$. This is a contradiction to $2=0$.

Answer 4

Suppose contradiction i.e. there exists some integer $n$ for which $n^3+n$ is odd.
Thus for some integers $n$ and $k$, you must have
$n^3+n=2k+1$
$\implies n^3+n-2=2k-1$
$\implies (n^3-1)+(n-1)=2k-1$
$\implies (n-1)(n^2+n+2)=2k-1$
$\implies (n-1)[n(n+1)+2]=2k-1$
You can argue why $[n(n+1)+2]$ is always even for $n(n+1)$ being the product of consecutive integers, which makes LHS an even integer; but RHS is odd. Hence the result.