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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.

Bill Dubuque
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Akshaj Bansal
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    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. https://en.wikipedia.org/wiki/Catalan's_conjecture – 79037662 Sep 27 '19 at 14:43
  • at most two will have factor of 3. –  Sep 27 '19 at 14:44
  • Yes but the link was just for any help or idea – Akshaj Bansal Sep 27 '19 at 14:44
  • @79037662 i see that but do you have any other proof using algebra – Akshaj Bansal Sep 27 '19 at 14:46
  • @79037662 In fact the unique solution in positive integers to $y^3= x^2-1$ being $x=3,y=2$ goes back to Levi ben Gershon in 1343 – Henry Sep 27 '19 at 14:53
  • @RoddyMacPhee: but one of those multiples of 3 can be a multiple of 9, so that doesn't help. – TonyK Sep 27 '19 at 14:53
  • @Henry Thanks for that, do you happen to know Levi ben Gershon's proof? If so (and if the proof isn't too long/hard), you can pretty easily write the answer to this question. – 79037662 Sep 27 '19 at 14:56
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    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

     – lab bhattacharjee Sep 27 '19 at 15:10
  • @79037662 I may have got this backwards: he in fact showed that the only solutions to $2^n-3^m=\pm1$ are $(1,2)$, $(2,3)$, $(3,4)$ and $(8,9)$ – Henry Sep 27 '19 at 15:20
  • You get a near miss for $35\cdot 36\cdot 37\cdot 38 = 121^3-1$. – paw88789 Sep 27 '19 at 15:26
  • If $n$ is not divisible by $3$ you can easily show that $n(n+2)=(n+1)^2-1$ and $(n+1)(n+3)=(n+2)^2-1$ have no prime factors in common and would both have to be cubes. But they are too close together to be successive cubes. I can't though see a similar way to do the divisible by $3$ case - I'm trying to avoid appealing to other theorems. – Mark Bennet Sep 27 '19 at 15:49

4 Answers4

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

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

TonyK
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  • 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 ... – Mark Bennet Sep 27 '19 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.

user14111
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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 $(a^{3})^n-(a^3-1)^n\geq (a^{3})^{n-1}+(a^{3})^{n-2}(a^3-1)+(a^{3})^{n-3}(a^3-1)^2$.

Wembley Inter
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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.

Rajorshi Chowdhury
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    It's not possible to assume that the factorization of $p^3$ is $p\cdot p^2$ unless $p$ is prime. – Andrew Dudzik Feb 23 '23 at 06:44
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    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. – Community Feb 23 '23 at 06:45
  • "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 $p=6$, $a=9$, $b=24$. – dxiv Jul 07 '23 at 05:33