Prove by contradiction that $(n+1)^3 \not= n^3 +(n-1)^3$ for $3$ consecutive positive integers

Question

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

Assume that: $$ (n+1)^3 = (n-1)^3+n^3 $$ $$n^3+3n^2+3n+1=n^3-3n^2+3n-1+n^3$$ $$3n^2+1=-3n^2-1+n^3$$ $$n^3-6n^2-2=0$$

I don't know how to move from here. Instead of this solution considered that two of the integers are either odd or even but this idea didn't help much.

Answer 1

Now, you just have to show that the cubic $x^3-6x^2-2 = 0$ has no positive integer roots.

By the rational root theorem, the only possible rational roots are $x = \pm 1, \pm 2$. Hence, the only possible positive integer roots are $x = 1,2$. Is $x = 1$ or $x = 2$ a root? If not, then there are no positive integer solutions $x = n$ to $x^3-6x^2-2 = 0$.

Alternatively, you can use Fermat's Last Theorem, but that is overkill.

Answer 2

The graph of $f(x)=x^3-6x^2-2$ has a zero occurring at a non-integer value.

Answer 3

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

• $n<6 \implies n^2(n-6)<0$
• $n=6 \implies n^2(n-6)=0$
• $n>6 \implies n^2(n-6)\geq49$

Therefore, $\forall{n}\in\mathbb{N}:n^2(n-6)\neq2$

Answer 4

This is a big cheating

But $(n+1)^3 = n^3+(n-1)^3$ is in contradiction with Fermat's Last Theorem