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I'm interested in nontrivial solutions of Diophantine equations of the type $$a^2b^3 = \frac{c!}{(c-k)!} $$

For various values of k fixed, and of course $a,b,c \in \mathbb{Z^+}$

Does anyone have any insight into this type of equation or a good reference for further reading? My search is being swamped by irrelevant results.

Edit: I changed n to c to emphasize that I am looking for a,b,c that solve this equation. Thus for k= 1, the equation becomes $a^2b^3 = c$, which clearly has infinity many solutions.

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  • $\begingroup$ Look up Shorey Tijdeman and related on perfect powers of (products of) consecutive integers. It will get you a step closer, and most likely your results of interest will reference their work. Laishram has related material as well. Gerhard "Is Researching Sylvester And Schur" Paseman, 2020.07.25. $\endgroup$
    – Gerhard Paseman
    Jul 25, 2020 at 20:58
  • $\begingroup$ @Gerhard squarefull is far from perfect power, so it's not clear to me how much closer Shorey & Tijdeman will get us. $\endgroup$
    – Gerry Myerson
    Jul 26, 2020 at 1:04
  • $\begingroup$ No, but if anyone has published on this problem recently, I can't think of any other paper they would reference. Gerhard "Maybe You Know Of One?" Paseman, 2020.07.25. $\endgroup$
    – Gerhard Paseman
    Jul 26, 2020 at 1:29

3 Answers 3

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You may already know this, but numbers of the form $a^2b^3$ are called powerful numbers. A closely related question that might provide information on your question is to ask for binomial coefficients that are powerful. A Google search of "powerful number" and "binomial coefficient" brought up the following paper of Granville:

On the scarcity of powerful binomial coefficients

Andrew Granville

https://dms.umontreal.ca/~andrew/PDF/powerful.pdf

He proves that there are only finitely many powerful binomial coefficients, contingent on the abc conjecture.

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  • $\begingroup$ Is there literature on consecutive powerful numbers, or intervals containing many powerful numbers? Gerhard "Are Numbers More Powerful Together?" Paseman, 2020.07.25. $\endgroup$
    – Gerhard Paseman
    Jul 25, 2020 at 21:10
  • $\begingroup$ @GerhardPaseman Here's a reference with some information on that question: POWERFUL NUMBERS IN SHORT INTERVALS, JEAN-MARIE DE KONINCK, FLORIAN LUCA AND IGOR E. SHPARLINSKI, BULL. AUSTRAL. MATH. SOC. 71 (2005) $\endgroup$
    – Joe Silverman
    Jul 25, 2020 at 22:20
  • $\begingroup$ It is conjectured that there are never three consecutive powerful numbers. $\endgroup$
    – Gerry Myerson
    Jul 26, 2020 at 1:09
  • $\begingroup$ Thanks for the help. This is exactly the type of thing I was looking for. $\endgroup$
    – G G
    Jul 27, 2020 at 2:58
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You might be interested in extensions to the Sylvester Schur theorem, which by your constraints shows that c is bigger than k^2 as the set of consecutive integers in the product must have a single multiple of q^2 for some prime q bigger than k. A paper of Saradha and Shorey from 2003, Almost Squares and Factorizations in Consecutive Integers, shows the sparsity of solutions to your equation where k-1 of the numbers on the right hand side multiply to a square. This may be useful for you in a citation search .

Gerhard "Not Quite Almost Powerful Numbers" Paseman, 2020.07.25.

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The smallest interesting case of $k=2$ reduces to a family of Pell equations paramaterized by $b$: $$(2c-1)^2 - b^3(2a)^2 = 1.$$ This gives infinitely many solutions.

For example, for $b=2$, we have a series of solutions indexed by $n$: $$c_n + a_n\sqrt{8} = \frac{(17+6\sqrt{8})^n+1}2.$$ Numerical values of $c_n$ are listed in OEIS A055792.

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  • $\begingroup$ Also, given powerful numbers m and m+1, we have 4m(m+1) and (2m+1)^2 also powerful. Gerhard "Finally Remembered That Contest Problem" Paseman, 2020.07.25. $\endgroup$
    – Gerhard Paseman
    Jul 26, 2020 at 0:36
  • $\begingroup$ @Gerhard, yes, where the trick is finding consecutive powerful numbers. Numbers $n$ such that $n$ and $n+1$ are both powerful are tabulated at oeis.org/A060355 $\endgroup$
    – Gerry Myerson
    Jul 26, 2020 at 1:05
  • $\begingroup$ Ah. So you know about 25*27 and 26*26 already. Gerhard "No More Element Of Surprise" Paseman, 2020.07.25. $\endgroup$
    – Gerhard Paseman
    Jul 26, 2020 at 1:25

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