Proof by contradiction : $n^3 −n−6 = 0$ means $m^3 − m − 6 ≠ 0$ when $m≠n$

Question

I have to prove by contradiction that if $n$ is a positive integer such as $n^3-n-6 = 0$, so for any positive integer $m$ such as $m ≠ n$, we have $m^3-m-6 ≠ 0$

I am new to proofs, but I tried and here is what I did.

I first assume the opposite of the conclusion, so $m^3-m-6 = 0$

$(m-2)(m^2+2m+3) = 0$

m=2

If I do the same thing for the n function and find the solution for $n^3-n-6 = 0, n=2$.

This is a contradiction since $m=n$, but we said $m ≠ n$.

I feel like im doing something wrong here and I would like someone to tell me what it is ! (Again, I am a beginner)

Answer 1

It looks like your proof would be assuming $n^{3}-n-6=0, m\neq n$ (where $m,n \in \mathbb{Z}$) showing a contradiction if we had $m^{3}-m-6=0$ as well. You have shown by factoring and the fact that $m,n \in \mathbb{Z}$, that $m, n = 2$. Hence $m=n$, a contradiction.

Answer 2

It's correct, but you're not actually using contradiction and it's indeed hard to write a real proof by contradiction of this particular statement.

What you proved is that if $n^3-n-6=0$, then $n=2$. Hence for no $m\ne2$ one can have $m^3-m-6=0$.

Answer 3

We have

$$n^3-n-6=0\iff$$ $$(n-1)n(n+1)=1.2.3 \iff n=2$$

thus

$$(m-1)m(m+1)\ne 6 \iff m\ne 2$$