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How to find all positive integers $m,n$ such that $n(n+1)(n+2)=6m^3$ ?

I can see that $m=n=1$ is a solution, but is it the only solution ?

user26857
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    A numerical check finds no small non-trivial solutions. – joriki Jul 01 '15 at 13:31
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    This paper (https://projecteuclid.org/download/pdf_1/euclid.pja/1195510942) speaks of a theorem that proves that there are only finitely many solutions to your equation. It goes further to put a bound on these solutions. The bound is still way too large to check all the cases. – Pjotr5 Jul 01 '15 at 13:46
  • according to wikipedia the answer is yes, but they do not provide a reference. They say: "The only tetrahedral number that is also a square pyramidal number is 1 (Beukers, 1988), and the only tetrahedral number that is also a perfect cube is 1." Wolfram gives the reference to Beukers but does not mention the latter result. – Mirko Jul 01 '15 at 14:31
  • That statement on Wikipedia was introduced in 2007 by this edit, which was rather inappropriately marked as "minor". No source is given in the edit summary. The user who added the statement was active on Wikipedia for 12 days and hasn't contributed since. I wouldn't put too much stock in that statement. (Also I'd expect that if this were known it would be recorded in the OEIS sequence of the tetrahedral numbers, which notes the three solutions in the square case.) – joriki Jul 01 '15 at 14:54
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    it looks the result may go to back to Euler ... or Lucas, or Hayashi, though I have not yet seen the proof itself. See p.25, p.34 of Dickson, History of the theory of numbers v.2 Diophantine Analysis, ch.1 Polygonal, Pyramidal and Figurate numbers. In that book it seems that "pyramidal number" means "triangular pyramidal number" which is the same as "tetrahedral number" https://books.google.com/books?id=eNjKEBLt_tQC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false – Mirko Jul 01 '15 at 15:19
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    In particular p.25 line12-13 "No pyramidal [understood triangular pyramidal, i.e. tetrahedral, I believe] number is a cube, or fifth power..." (no ref given). Re p.34 lines 15-17 "Hayasi proved that the quadruple of a number $a(a+b)(a+2b)/6$ and hence of a pyramidal number is not a cube, by making use of the known impossibility of $x^3+y^3=3z^3$." Ref given Nouv.Ann.Math.(4) 10, 1910, 83, but this latter result is about the quadruple of a pyramidal number (not the pyramidal number itself). http://archive.numdam.org/ARCHIVE/NAM/NAM_1910_4_10_/NAM_1910_4_10__83_1/NAM_1910_4_10__83_1.pdf – Mirko Jul 01 '15 at 16:33

1 Answers1

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The paper Rational approximation to algebraic numbers of small height: the Diophantine equation $\left|ax^n-by^n\right|=1$ by Michael A. Bennett (J. reine angew. Math. 535 (2001),1–49) shows that the equation in the title has at most one solution in positive integers $x,y$ for given positive integers $a,b,n$ with $n\ge3$.

It follows that $x=y=1$ is the only solution both of $\left|2x^3-y^3\right|=1$ and of $\left|2x^3-3y^3\right|=1$.

From now on, $n$ is the $n$ in the question, not in the paper.

$n$, $n+1$ and $n+2$ have no factors other than $2$ in common. It follows that except for factors of $2$ and one factor of $3$ to cancel the factor of $3$ in the factor $6$, $n$, $n+1$ and $n+2$ must all be perfect cubes. This allows us to exclude non-trivial solutions by case analysis:

If $n+1$ is even, it does not contain the single uncubed factor of $3$, since otherwise $n$ and $n+2$ would both be perfect cubes. It has a single uncubed factor of $2$ to cancel the factor of $2$ in the factor $6$, since $n$ and $n+2$ are odd. Thus $n+1=2x^3$, and one of $n$ and $n+2$ must be a perfect cube $y^3$, and with $\left|2x^3-y^3\right|=1$ it follows that $x=y=1$.

If $n$ and $n+2$ are even, one of them, say, $n$, is divisible by $2$ but not by $4$. It cannot also contain the single uncubed factor of $3$, since otherwise the other two would have to be perfect cubes. Thus $n=2x^3$, and $n+1$ is either $y^3$ or $3y^3$. Again it follows that $x=y=1$.

joriki
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  • P.S.: I see now that the paper I cited actually gives references (Delone 1922, Nagell 1925) that had proved the paper's result for $n=3$ long ago. – joriki Jul 01 '15 at 15:41
  • Very nice! I was half-way there myself, but I didn't know that result of Bennett's. – TonyK Jul 01 '15 at 15:44
  • @TonyK: I hadn't known it either; I came across it looking for what's known on diophantine approximations of $\sqrt[3]2$. Another interesting paper I found was Rational approximations to $\sqrt[3]2$ and other algebraic numbers revisited by Paul M. Voutier (Journal de Théorie des Nombres de Bordeaux 19 (2007), 263–288), which gives the bound

    $$\left|2^{1/3}-\frac pq\right|\gt\frac{0.25}{\left|q\right|^{2.4325}}$$

    for integers $p$, $q$ with $q\neq0$, which I suspect could also be used to resolve the present question.

     – joriki Jul 01 '15 at 15:57
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    The book Figurate Numbers By Elena Deza, M. Deza https://books.google.com/books/about/Figurate_Numbers.html?id=cDxYdstLPz4C gives the result on p.98, V. "It is known that no pile of bullets with a triangular or a square base contains a number of bullets equal to a cube, biquadrate, or a fifth power [Dick05]. In other words there are no non-trivial cubic tetrahedral numbers, biquadratic tetrahedral numbers, and five-dimensional hypercube tetrahedral numbers, as well as cubic square pyramidal numbers, biquadratic square pyramidal numbers and five-dimensional hypercube square pyramidal numbers." – Mirko Jul 01 '15 at 17:37