A085692 Brocard's problem: squares which can be written as n!+1 for some n.

25, 121, 5041

Offset: 1

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jul 18 2003

Keywords

• nonn
• bref

Comments

Next term, if it exists, is greater than 10^850. - Sascha Kurz, Sep 22 2003

No more terms < 10^20000. - David Wasserman, Feb 08 2005

The problem of whether there are any other terms in this sequence, Brocard's problem, has been unsolved since 1876. The known calculations give a(4) > (10^9)! = factorial(10^9). - Stefan Steinerberger, Mar 19 2006

I wrote a similar program sieving against the 40 smallest primes larger than 4*10^9 and can report that a(4) > factorial(4*10^9+1). In other words, it's now known that the only n <= 4*10^9 for which n!+1 is a square are 4, 5 and 7. C source code available on request. - Tim Peters (tim.one(AT)comcast.net), Jul 02 2006

Robert Matson claims to have verified that 4, 5, and 7 are the only values of n <= 10^12 for which n!+1 is a square. This implies that the next term, if it exists, is greater than (10^12+1)! ~ 1.4*10^11565705518115. - David Radcliffe, Oct 28 2019

Examples

			   5^2 =   25 = 4! + 1;
  11^2 =  121 = 5! + 1;
  71^2 = 5041 = 7! + 1.
		

References

• R. Guy, "Unsolved Problems in Number Theory", 3rd edition, D25
• Clifford A. Pickover, A Passion for Mathematics (2005) at 69, 306.
• James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 19.

Links

• Bruce C. Berndt and William F. Galway, On the Brocard-Ramanujan Diophantine Equation n! + 1 = m^2, The Ramanujan Journal, March 2000, Volume 4, Issue 1, pp 41-42.
• Robert D. Matson, Brocard's Problem 4th Solution Search Utilizing Quadratic Residues, Unsolved Problems.
• Wikipedia, Brocard's problem.

Crossrefs

A085692, A146968, A216071 are all essentially the same sequence. - N. J. A. Sloane, Sep 01 2012

Programs

Mathematica

Select[Range[0,100]!+1,IntegerQ[Sqrt[#]] &] (* Stefano Spezia, Jul 02 2025 *)

PARI

A085692=select( issquare, vector(99,n,n!+1)) \\ M. F. Hasler, Nov 20 2018

Formula

a(n) = A216071(n)^2 = A146968(n)!+1 = A038507(A146968(n)). - M. F. Hasler, Nov 20 2018