If $n$ is a positive integer such that $n^3 − n − 6 = 0$, then, for every positive integer $m$ with $m \not = n$, $m^3 − m − 6 \not = 0$.

Question

Prove, by contradiction, that, if $n$ is a positive integer such that $n^3 − n − 6 = 0$, then, for every positive integer $m$ with $m \not = n$, $m^3 − m − 6 \not = 0$.

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Proposition: If $n$ is a positive integer such that $n^3 − n − 6 = 0$, then, for every positive integer $m$ with $m \not = n$, $m^3 − m − 6 \not = 0$.

Hypothesis: $n$ is a positive integer such that $n^3 − n − 6 = 0$.

Conclusion: For every positive integer $m$ with $m \not = n$, $m^3 − m − 6 \not = 0$.

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My workings

A (Hypothesis): $n$ is a positive integer such that $n^3 − n − 6 = 0$.

A1 ($\neg B$): There exists a positive integer $m$ with $m \not = n$ such that $m^3 − m − 6 = 0$.

A2: $n^3 - n = 6$

$\implies n(n^2 - 1) = 6$ where $n \in \mathbb{Z}^+$

A3: $m^3 - m = 6$

$\implies m(m^2 - 1) = 6$ where $m \in \mathbb{Z}^+$ and $m \not = n$.

A4: $n(n^2 - 1) = m(m^2 - 1)$ where $m \not = n$.

$\implies \dfrac{n(n^2 - 1)}{m(m^2 - 1)} = 1$ where $m \not = n$.

Since $m,n \in \mathbb{Z}^+$, $(n^2 - 1) \ge 0$ and $(m^2 - 1) \ge 0$. Therefore, $n(n^2 - 1) = m(m^2 - 1)$ iff $n = m$.

Contradiction.

$Q.E.D.$

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I would greatly appreciate it if people could please take the time to provide feedback on the correctness of my proof. Given that $m \not = n$, did I construct my proof correctly? Or is it incorrect to structure it this way if $m \not = n$?

Answer 1

Let $f(x)=x^3-x-6.$ Let $n$ be any real number such that $f(n)=0$. Then $f(x)=(x-n)(x^2+Ax+B)$ for all $x,$ for some constants $A, B.$

$$\text {So }\quad x^3-x-6=x^3+x^2(A-n)+x(B-nA)+(-nB)$$ for all $x.$ This requires $0=A-n$ and $-1=B-nA ,$ so $A=n$ and $B=nA-1=n^2-1.$

Now if $f(m)=0$ with $m\ne n$ then $0=m^2+Am+B=m^2+nm+(n^2-1),$ implying $m=(-n\pm \sqrt {4-3n^2}\;)/2.\;$ In order for $m$ to be a real number we must have $4-3n^2\geq 0,$ which implies $|n|\leq 2/\sqrt 3.$

But if $|n|\leq 2/\sqrt 3$ then $|n^3-n|\leq |n|^3+|n|\leq 8/(3\sqrt 3\;)+2/\sqrt 3<6.$

So $f(n)=0$ has at most one real-number solution.

Answer 2

Although that contradiction is correct, it is not entirely clear why you assumed that is obvious. You still have to prove that that cannot hold.

A much easier way would be to prove that $m^3-m-6<0$ if $m0$ if $m>n$, by simple algebra.

Answer 3

We can prove a stronger result for essentially no extra effort.

Let positive integer $n $ be such that $n^3-n-k=0$

Suppose another positive integer $m\ne n$ exists such that $m^3-m-k=0$. Then we see that $n^3-n = k$ and $m^3-m = k$ so $n^3-n = m^3-m $.

Now we can factor to see that this implies $(n-1)n(n+1) = (m-1)m(m+1)$. We can see that if $n>m$ then clearly also $(n-1)>(m-1)$ and also $(n+1)>(m+1)$, and since none of these are negative, it holds that $(n-1)n(n+1) > (m-1)m(m+1)$, a contradiction. The same process holds for $n