If $n − 1$, $n$, and $n + 1$ are consecutive positive integers, then the cube of the largest cannot be equal to the sum of the cubes of the other two.

Question

Prove, by contradiction, that, if $n − 1$, $n$, and $n + 1$ are consecutive positive integers, then the cube of the largest cannot be equal to the sum of the cubes of the other two.

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Proposition: If $n − 1$, $n$, and $n + 1$ are consecutive positive integers, then the cube of the largest cannot be equal to the sum of the cubes of the other two.

Hypothesis: $n − 1$, $n$, and $n + 1$ are consecutive positive integers.

Conclusion: The cube of the largest cannot be equal to the sum of the cubes of the other two.

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My workings

A (Hypothesis): $n − 1$, $n$, and $n + 1$ are consecutive positive integers.

A1 ($\neg B$): The cube of the largest is equal to the sum of the cubes of the other two.

A2: There exists $n \in \mathbb{Z} > 1$ such that $(n + 1)^3 = (n - 1)^3 + n^3$.

A3: $(n+1)^3 = (n - 1)^3 + n^3 $

$\implies (n + 1)(n^2 + 2n + 1) = (n-1)(n^2 - 2n + 1) + n^3$

$\implies n^3 + 2n^2 + n + n^2 + 2n + 1 = n^3 - 2n^2 + n - n^2 + 2n - 1 + n^3$

$\implies n^3 + 3n^2 + 3n + 1 = 2n^3 - 3n^2 + 3n - 1$

$\implies n^3 - 6n^2 - 2 = 0$

A4: $n^3 - 6n^2 - 2 = 0$ where $n \in \mathbb{Z} > 1$.

$\implies 1 - \dfrac{6}{n} - \dfrac{2}{n^3} = 0$

$\implies 1 - \dfrac{6}{n} = \dfrac{2}{n^3}$ where $n \in \mathbb{Z} > 1$.

But there is no $n \in \mathbb{Z} > 1$ such that $1 - \dfrac{6}{n} = \dfrac{2}{n^3}$. Contradiction.

$Q.E.D.$

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EDIT: My workings #2

A (Hypothesis): $n − 1$, $n$, and $n + 1$ are consecutive positive integers.

A1 ($\neg B$): The cube of the largest is equal to the sum of the cubes of the other two.

A2: There exists $n \in \mathbb{Z} > 1$ such that $(n + 1)^3 = (n - 1)^3 + n^3$.

A3: $(n+1)^3 = (n - 1)^3 + n^3 $

$\implies (n + 1)(n^2 + 2n + 1) = (n-1)(n^2 - 2n + 1) + n^3$

$\implies n^3 + 2n^2 + n + n^2 + 2n + 1 = n^3 - 2n^2 + n - n^2 + 2n - 1 + n^3$

$\implies n^3 + 3n^2 + 3n + 1 = 2n^3 - 3n^2 + 3n - 1$

$\implies n^3 - 6n^2 - 2 = 0$

A4: $n^3 - 6n^2 - 2 = 0$ where $n \in \mathbb{Z} > 1$.

$\implies n^2(n - 6) - 2 = 0$

$\implies n^2(n - 6) = 2$

A5: $n^2(n - 6) = 2$

$n^2(n - 6) \le 0$ when $6 \ge n > 1$.

$n^2(n - 6) \ge 0$ when $n > 6$

$\therefore$ When $6 \ge n > 1$, $n^2(n - 6) \not = 2$.

A6: $n^2(n - 6) = 2 > 0$

$n^2(n - 6) > 0$

$\implies n - 6 > 0$

$\implies n > 6$

$\therefore \inf(n - 6) = 1$ where $n = 7 > 6$

But $\inf(n^2) = 7^2$ where $n = 7 > 6$

$= 49$

The function $f(n) = n^2(n - 6)$ is continuous, positive, and increasing on the interval $[7, \infty)$.

$\therefore \inf[n^2(n - 6)] = 49(1)$

$= 49 \not = 2$ Contradiction.

$Q.E.D.$

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I would greatly appreciate it if people could please take the time to review my proof and provide feedback on its correctness.

Answer 1

You can use the last Fermat theorem.

Set $a=n-1,$ $b=n$ and $c=n+1$. Then you want to prove that: $$a^3+b^3=c^3.$$

We know that this equation has only the trivial solution. So $a=b=c=0$, but this imply that $n$ is equal to $-1$, $0$ and $1$ at the same time which is clearly impossible.

Answer 2

Well, you need to show that there is no integer $n>1$ such that $$ 1-\frac{6}{n}=\frac{2}{n^3} $$ For instance, if $n>6$, then both sides are positive and less than $1$, so they could be equal.

However, for $n\ge7$, we have $$ 1-\frac{6}{n}\ge1-\frac{6}{7}=\frac{1}{7} $$ whereas $$ \frac{2}{n^3}\le\frac{2}{7^3}<\frac{1}{7} $$ For $1\le n\le 6$, the left hand side is $\le0$.

Alternatively, the only rational roots of $n^3-6n-2$ can be $\pm1$ and $\pm2$; none of them is a root.