# Diophantine equation of a factorial type

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.

• 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. – Gerhard Paseman Jul 25 at 20:58
• @Gerhard squarefull is far from perfect power, so it's not clear to me how much closer Shorey & Tijdeman will get us. – Gerry Myerson Jul 26 at 1:04
• 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. – Gerhard Paseman Jul 26 at 1:29

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.

• Is there literature on consecutive powerful numbers, or intervals containing many powerful numbers? Gerhard "Are Numbers More Powerful Together?" Paseman, 2020.07.25. – Gerhard Paseman Jul 25 at 21:10
• @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) – Joe Silverman Jul 25 at 22:20
• It is conjectured that there are never three consecutive powerful numbers. – Gerry Myerson Jul 26 at 1:09
• Thanks for the help. This is exactly the type of thing I was looking for. – G G Jul 27 at 2:58

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.

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.

• 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. – Gerhard Paseman Jul 26 at 0:36
• @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 – Gerry Myerson Jul 26 at 1:05
• Ah. So you know about 25*27 and 26*26 already. Gerhard "No More Element Of Surprise" Paseman, 2020.07.25. – Gerhard Paseman Jul 26 at 1:25