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Show that $m+3$ and $m^2 + 3m +3$ cannot both be perfect cubes.

I've done so much algebra on this, but no luck. Tried multiplying, factoring, etc.

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    $\begingroup$ Jared, why on earth are you trying to post this in meta? It is off-topic there. Meta is not for discussion of math - it is for discussions on how to run the site math.stackexchange.com $\endgroup$ – Jyrki Lahtonen Mar 15 '15 at 15:33
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    $\begingroup$ Total accident. I have no idea how I wound up in Meta. I don't even know what Meta is. Sorry! $\endgroup$ – King Henry V Mar 15 '15 at 15:35
  • $\begingroup$ After substituting $n=m+1$, this appears the same as math.stackexchange.com/questions/71412 $\endgroup$ – punctured dusk Mar 15 '15 at 16:33
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If they were, $$(m+3)(m^2+3m+3)=(m+2)^3+1$$ would be a perfet cube. There aren't many perfect cubes that differ by $1$:
$n^3-(m+2)^3=1$ means $m+2-n\mid1$ and hence $n=m+2\pm1$, reducing the equation to two quadratics: $3(m+2)^2+3(m+2)=0$ and $-3(m+2)^2+3(m+2)-2=0$. This shows that the only value where the statement fails is $m=-2$.

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  • $\begingroup$ Nice! ${}{}{}{}$ $\endgroup$ – Jyrki Lahtonen Mar 15 '15 at 15:46
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    $\begingroup$ So annoying. I had it! I multiplied them both to see what they would produce, but didn't recognize the factorization. I should have used an algebra program next time to check before asking here I guess. It's depressing. $\endgroup$ – King Henry V Mar 15 '15 at 15:49
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    $\begingroup$ The only perfect cubes that differ by $1$ are $-1,0,1$ and it's trivial if you notice that $$(n+1)^3-n^3=3n^3+3n+1=1\iff 3n(n+1)(n-1)=0$$ Just showing a different approach. $\endgroup$ – user236182 May 18 '16 at 18:25
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The first observation is that $m=-2$ is a counterexample. Presumably $m$ should be non-negative.

Assume that $m+3=n^3$ is a perfect cube. Then $n>1$ and $$ m^2+3m+3=n^6-3n^3+3. $$ This is strictly less than $n^6=(n^2)^3$. The previous cube below this is $$ (n^2-1)^3=n^6-3n^4+3n^2-1, $$ which is $<n^6-3n^3+3$ as $n\ge2$. Therefore $m^2+3m+3$ is strictly between two consecutive perfect cubes, and cannot be a perfect cube itself.

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  • $\begingroup$ Clever. Is there an immediate way to see $n^6-3n^4+3n^2-1<n^6-3n^3+3$, for $n\geq2$? I can prove it by taking a derivative, but is there a more obvious way to see it? $\endgroup$ – Gregory Grant Mar 15 '15 at 15:57
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    $\begingroup$ @GregoryGrant: Admittedly I did something similar. Checking it with smallest possible values of $n$, and then relying on the quartic growing faster than the cubic. $\endgroup$ – Jyrki Lahtonen Mar 15 '15 at 22:48
  • $\begingroup$ @GregoryGrant The claimed inequality follows from $n^4>n^3+n^2$ which is true for $n\ge 2$ because $n^4\ge2n^3>n^3+n^2$. $\endgroup$ – Mario Carneiro Mar 16 '15 at 6:54
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The statement is false: for $m=-2$, we have $m+3=m^2+3m+3=1$.

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  • $\begingroup$ (This follows from reflecting on @barto's answer.) $\endgroup$ – Théophile Mar 15 '15 at 15:43

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