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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.

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  • $\begingroup$ Because FLT lol. :) $\endgroup$ – Deepak Nov 26 '15 at 3:05
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$n^3-6n^2-2=0 \implies n^2(n-6)=2$

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

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

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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.

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    $\begingroup$ Why can't the roots be $\pm 1$? $\endgroup$ – Ragnar Aug 16 '14 at 19:54
  • $\begingroup$ ^Thanks for that catch. $\endgroup$ – JimmyK4542 Aug 16 '14 at 19:54
  • $\begingroup$ Thanks for the Fermat, I have been dying to use that one. $\endgroup$ – pseudomarvin Aug 16 '14 at 20:03
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    $\begingroup$ Nitpicking, but we didn't need to check $n=1$ since $n-1$ is not a positive integer then. $\endgroup$ – user26486 Aug 16 '14 at 20:18
  • $\begingroup$ Lol I was thinking of posting 'because FLT lol' as a comment. :) never mind doing it anyway haha. $\endgroup$ – Deepak Nov 26 '15 at 3:03
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This is a big cheating

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

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  • $\begingroup$ I can't believe this is getting up votes (even if it is technically correct). Using FLT here is like using a strategic thermonuclear device to kill a fly. I assumed that all the FLT comments were jokes, and not serious answers. :) $\endgroup$ – Deepak Nov 26 '15 at 3:09
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    $\begingroup$ @Deepak: I think that (especially on a homework question) people occasionally find humor in pulling out the most inappropriately powerful tool to solve it. I suspect the up-votes are being awarded on that account. It's not exactly responsible, but I don't think anyone will be misled in this instance. $\endgroup$ – Brian Tung Nov 26 '15 at 3:12
  • $\begingroup$ @BrianTung Thanks Brian. Which is why jokesy answers should be left as comments at most IMHO. Anyway, point taken. $\endgroup$ – Deepak Nov 26 '15 at 3:13
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The graph of $f(x)=x^3-6x^2-2$ has a zero occurring at a non-integer value.

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    $\begingroup$ I would not consider this as a mathematical proof, since it requires drawing a graph and looking at it. Saying for example that $f(1)<0$ and $f(2)>0$ is a valid proof. $\endgroup$ – Ragnar Aug 16 '14 at 19:55
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    $\begingroup$ I completely agree. I'm still very new at this so my apologies. $\endgroup$ – iamgroot Aug 16 '14 at 20:01

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