Brocard’s Problem

Brocard’s Problem by Thomas McClure I

Introduction

Brocard’s Problem concerns the equation

n! + 1 = m^2

where n and m are both integers, and n! Is the factorial of n, that is, the product of all integers less than or equal to n (so, for example, 3! = 3*2*1 = 6, and 4! = 4*3*2*1 = 24, etc).

Only three solutions to this equation are known, when n = 4 or 5 or 7.

The problem is to find other values of n for which this equation is true, or to prove that there are no other solutions..

http://www.unsolvedproblems.org/index_files/Brocard.htm For further information, please see:

[1] http://mathworld.wolfram.com/BrocardsProblem.html

[2] http://blogs.ams.org/mathgradblog/2013/08/08/problembrocard/#sthash.Sl87FtZe.dpbs

[3] http://en.wikipedia.org/wiki/Brocard%27s_problem

II

Restatement

n! = P(x) = (x - a)(x - b) = x^2 - (a + b)x + ab, given the abc conjecture {c, c} = (a + b) c < rad(abc) usually and c > rad(abc)^(1 + ε) n! = x^2 - (a + b)x + ab

III

Examples

321 = 6 = 0 - 0 + ab a=3

b=2

This is a counter example 4321 = 24 = 1 - {c,c}*1 + ab 54321 = 120 = 0 - 0 + ab = 8*15 7654321 = 42*120 = 4 - {c,c}*2 + ab = (2 - a)(2 - b) (2 - a) = 42 a = -40 (2 - b) = 120 b = -118

IV

Conclusion

The result of n! equals a polynomial of degree two, given the abc conjecture is true. Hence the product of consecutive numbers has a finitely number of solutions with integer coefficients, as a Diophintine equation.

Appendix

Brocard's problem From Wikipedia, the free encyclopedia Not to be confused with Brocard's conjecture. Brocard's problem is a problem in mathematics that asks to find integer values of n for which

n!+1 = m^2, where n! is the factorial. It was posed by Henri Brocard in a pair of articles in 1876 and 1885, and independently in 1913 by Srinivasa Ramanujan.

List of unsolved problems in mathematics Do integers, n,m, exist such that n!+1=m^2 other than n=4,5,7?

Brown numbers Pairs of the numbers (n, m) that solve Brocard's problem are called

Brown numbers. There are only three known pairs of Brown numbers:

(4,5), (5,11), and (7,71). Paul Erdős conjectured that no other solutions exist. Overholt (1993) showed that there are only finitely many solutions provided that the abc conjecture is true. Berndt & Galway (2000) performed calculations for n up to 109 and found no further solutions.

Variants of the problem Dabrowski (1996) generalized Overholt's result by showing that it would follow from the abc conjecture that

n!+A = k^2 has only finitely many solutions, for any given integer A. This result was further generalized by Luca (2002), who showed (again assuming the abc conjecture) that the equation

n! = P(x) has only finitely many integer solutions for a given polynomial P(x) of degree at least 2 with integer coefficients.

References Berndt, Bruce C.; Galway, William F. (2000), "The Brocard– Ramanujan diophantine equation n! + 1 = m2" (PDF), The Ramanujan Journal 4: 41–42, doi:10.1023/A:1009873805276. Brocard's conjecture

Not to be confused with Brocard's problem. In number theory, Brocard's conjecture is a conjecture that there are at least four prime numbers between (pn)2 and (pn+1)2, for n > 1, where pn is the nth prime number.[1] It is widely believed that this conjecture is true. However, it remains unproven as of March 2015. https://en.wikipedia.org/wiki/Brocard%27s_conjecture n

p_n p_n^2

Prime numbers \Delta

1

2

4

5, 7 2

2

3

9

11, 13, 17, 19, 23

3

5

25

29, 31, 37, 41, 43, 47 6

4

7

49

53, 59, 61, 67, 71…

5

11

121 127, 131, 137, 139, 149…

5

15 9

\Delta stands for \pi(p_{n+1}^2) - \pi(p_n^2). The number of primes between prime squares is 2, 5, 6, 15, 9, 22, 11, 27, ... OEIS A050216. Legendre's conjecture that there is a prime between consecutive integer squares directly implies that there are at least two primes between prime squares for pn ≥ 3 since pn+1 - pn ≥ 2.

Notes Jump up ^ Weisstein, Eric W., "Brocard's Conjecture", MathWorld.