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The question is

Prove that product of four consecutive natural numbers can not be a perfect cube .

I really dont know what actually do to proceed with the question. However after seeing this related result for the product of four consecutive numbers Proving any product of four consecutive integers is one less than a perfect square I tried with the same procedure of converting the product into a whole cube with a sum or difference of a constant . But i fail to do so. Any help or hint would be appreciated. Thank you

Edit: I just came to know the catalan's conjecture , so please if anyone could provide a proof for it or use some algebra to prove the question would be a great help.Also i think the question would not hold a very tough or lengthy proof since it is from a book recommended for students preparing for olympiads.

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  • $\begingroup$ This is implied by "no number one less than a perfect square is a perfect cube" because of your link, right? In that case it's a corollary of Catalan's conjecture. en.wikipedia.org/wiki/Catalan's_conjecture $\endgroup$ – 79037662 Sep 27 at 14:43
  • $\begingroup$ at most two will have factor of 3. $\endgroup$ – Roddy MacPhee Sep 27 at 14:44
  • $\begingroup$ Yes but the link was just for any help or idea $\endgroup$ – Akshaj Bansal Sep 27 at 14:44
  • $\begingroup$ @79037662 i see that but do you have any other proof using algebra $\endgroup$ – Akshaj Bansal Sep 27 at 14:46
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    $\begingroup$ Think of $$n(n+1)(n+2)(n+3)$$ prime $p(>3)$ can divide exactly one of the multiplicands So, if $n>1$ at least three of them will have distinct prim factors $\endgroup$ – lab bhattacharjee Sep 27 at 15:10
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First, note if a prime $p>3$ divides one of the numbers, it can't divide any other. So it must occur cubed, in the form $p^{3k}$ for some $k$.

Second, note that at least one of the four numbers must be co-prime to both $2$ and $3$; so this number contains only primes $p>3$, and it must therefore be a perfect cube.

And if one of the numbers is a perfect cube, then the product of the other three must also be a perfect cube. So now we have reduced the problem to showing that the product of three out of four consecutive numbers can't be a perfect cube.

There are three cases: $n(n+1)(n+2), n(n+1)(n+3),$ and $n(n+2)(n+3)$. We show by elementary algebra that each of these lies strictly between two consecutive cubes, except for $n(n+1)(n+3)=8$ when $n=1$; so none of them can be a perfect cube, except for this one case. But this case does not lead to a counter-example of the full statement, because the missing number $n+2=3$ is not a perfect cube.

Case 1: $n(n+1)(n+2)$

$n^3<n(n+1)(n+2)=n^3+3n^2+2n<n^3+3n^2+3n+1=(n+1)^3$

Case 2: $n(n+1)(n+3)$

$n(n+1)(n+3)-(n+1)^3=n^2-1$, so unless $n=1$, we have $(n+1)^3<n(n+1)(n+3)<(n+2)^3$.

Case 3: $n(n+2)(n+3)$

$(n+1)^3=n^3+3n^2+3n+1\underset{3n<2n^2+5}{<}n^3+5n^2+6=n(n+2)(n+3)\underset{n(n+3)<(n+2)^2}{<}(n+2)^3$

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  • $\begingroup$ Nicely done - you also could have $n$ as the cube but $(n+1)(n+2)(n+3)$ can't be a cube either because you've already proved that the product of three successive integers can't be. This is the proof I was fishing for ... $\endgroup$ – Mark Bennet Sep 27 at 16:01

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