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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, 2019 at 14:43
  • $\begingroup$ at most two will have factor of 3. $\endgroup$
    – user645636
    Sep 27, 2019 at 14:44
  • $\begingroup$ Yes but the link was just for any help or idea $\endgroup$ Sep 27, 2019 at 14:44
  • $\begingroup$ @79037662 i see that but do you have any other proof using algebra $\endgroup$ Sep 27, 2019 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 $\endgroup$ Sep 27, 2019 at 15:10

4 Answers 4

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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 ... $\endgroup$ Sep 27, 2019 at 16:01
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Assume for a contradiction that $n(n+1)(n+2)(n+3)$ is a cube, where $n$ is a positive integer; of course $n\gt1$ since $1\cdot2\cdot3\cdot4=24$ is not a cube.

Of the two middle factors $n+1$ and $n+2$, whichever one is odd is relatively prime to the other three factors, so it is a cube, and the product of the three remaining factors is a cube; i.e., either $n(n+1)(n+3)$ or $n(n+2)(n+3)$ must be a cube.

But you can easily verify that, since $n\gt1$, $$(n+1)^3\lt n(n+1)(n+3)\lt n(n+2)(n+3)\lt(n+2)^3.$$ Since there are no cubes between $(n+1)^3$ and $(n+2)^3$, we have arrived at a contradiction.

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    $\begingroup$ This is a distinct improvement on my answer. $\endgroup$
    – TonyK
    Feb 24, 2023 at 15:38
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In fact, by generalizing the methods used in the previous two answers, it can be proved that product of $4$ consecutive positive integers can not be a perfect power (exponent $\geq 2$).

For the $4$ positive integers namely $x,x+1,x+2,x+3$, $x+1$ is coprime with the other three when $2|x$ while $x+2$ is coprime with the other three when $2\nmid x$. Assume for some $m,n\geq 2$, $x(x+1)(x+2)(x+3)=m^n$. Notice that $m^n=(x^2+5x+5)^2-1$, we have $n\neq 2$, i.e. $n\geq 3$.

(1) If $2|x$, $\operatorname{gcd}(x+1,x(x+2)(x+3))=1$ and thus there exists $a,b\in\mathbb{Z}^{+}$ such that $x+1=a^n,x^3+5x^2+6x=x(x+2)(x+3)=b^n$. However, as $x\geq 1$, $$(a^3)^n=(x+1)^3<x^3+5x^2+6x<(x+1)^3+2(x+1)^2\\<a^{3n}+2a^{2n}+1<a^{3n}+na^{3(n-1)}+1\leq (a^3+1)^n$$, drawing a contradiction.

(2) If $2\nmid x$, $\operatorname{gcd}(x+2,x(x+1)(x+3))=1$ and thus there exists $a,b\in\mathbb{Z}^{+}$ such that $x+2=a^n,x^3+4x^2+3x=x(x+1)(x+3)=b^n$. However, as $x\geq 1$, $$(a^3-1)^n\leq a^{3n}-a^{3(n-3)}(3a^{6}-2a^3+1)<a^{3n}-a^{3(n-3)}(2a^{6}+a^3)\\\leq a^{3n}-2a^{2n}-a^{n}+2=x^3+4x^2+3x<(x+2)^3=(a^3)^n$$, drawing a contradiction.

Remark. The inequality $(a^3-1)^n\leq a^{3n}-a^{3(n-3)}(3a^{6}-2a^3+1)$ follows from $(a^{3})^n-(a^3-1)^n\geq (a^{3})^{n-1}+(a^{3})^{n-2}(a^3-1)+(a^{3})^{n-3}(a^3-1)^2$.

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Ok so let's see , I am taking a bit different approach here : n(n+1)(n+2)(n+3) For the above equation if we try to write the whole equation in the form of a single factor , (n+1) we can rewrite it as , (n+1)^4 + 2(n+1)^3 - (n+1)^2 - 2(n+1) Now replacing (n+1) by t(where t is a natural number) we get , t^4 + 2 t^3 - t^2 - 2t

One taking factors common , (t^2 - 1)(t^2 + 2t) =(t^2 - 1)[(t + 1)^2 - 1 ]

For this factor to be a cube root let's assume it to equal to p^3 (where p is a natural number).

Now by observation we can see that t^2 -1 < (t+1)^2 -1 , so the on comparing we can easily state that , (t^2 - 1) = p and (t+1)^2 - 1 = p^2 ( p^2 > p for any natural number) Putting the original values again we get ,

(a) (n+1)^2 - 1 = p , which is not satisfied by any natural number values of n and p respectively .

(b) (n+2)^2 -1 = p^2 , which is again not satisfied by any natural number values of n and p respectively , which contradicts aur assumption.

Hence , it is proved that it cannot be a perfect cube.

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    $\begingroup$ It's not possible to assume that the factorization of $p^3$ is $p\cdot p^2$ unless $p$ is prime. $\endgroup$ Feb 23, 2023 at 6:44
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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Feb 23, 2023 at 6:45
  • $\begingroup$ "we can easily state that" $\;-\;$ No, that does not follow. Just because a product of two different numbers is a perfect cube $ab=p^3$ does not mean that the two numbers must be $p$ and $p^2$. Try for example $p=6$, $a=9$, $b=24$. $\endgroup$
    – dxiv
    Jul 7, 2023 at 5:33

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