There's the trivial observation that N! - 1 /= K^2 when N >= 4, since 4 divides N! and squares are only = 0 or 1 (mod 4). This general idea can be carried a little further, but I don't think it solves the general problem: Looking at N! = K^2-1 = (K-1)(K+1), We know that N! is divisible by 2^(N-hammingweight(N)), K must be odd, one of K+-1 contributes one factor of 2, so that K-+1 must be divisible by 2^(N-smidgen). Similarly, all the factors of 3 must divide one of K+-1, though not necessarily the same choice of +- as the factors of 2. And so on for all the other primes < N. Getting concrete with N=7, 7! = 5040 = 70 * 72 = 2.5.7 * 2^3.3^2 Reasoning that "there's some subset of the prime powers that must multiply out to be too large or too small" doesn't seem to go anywhere. It might be possible to attack the K^3 problem this way. Half the primes have x->x^3 (mod p) as a 1-1 map, so the only cube root of 1 is 1, and the only cube root of -1 is -1. These primes are 2(mod 3), and 3. If we try to analyze 20!+1 = K^3, we get that K=1 mod 2, 5, 11, and 17. Going a bit deeper, we require K=1 mod 2^18, 3^7, 5^4. The product of the primes & prime powers exceeds cbrt(20!), leaving K=1 as our only (absurd) choice. For this to work for general large N, we'd need to get good bounds on the fraction of primes = 2 (mod 3). Rich ---------------- Quoting Dan Asimov <dasimov@earthlink.net>:
Brocard's problem (Brocard 1876, 1885; Ramanujan 1913) is to decide whether
4! + 1 = 5^2
5! + 1 = 11^1
7! + 1 = 71^2
are the only factorial-plus-ones equal to a square:
(*) n! + 1 = K^2
. It remains unsolved.
Question: --------- What is known about generalizing to any exact power:
n! + 1 = K^r (r >= 2)
and what if we let +1 be replaced by -1 here or in (*) ???
Note: It was shown by Luca (2002) that the abc conjecture (https://en.wikipedia.org/wiki/Abc_conjecture) implies that at most finitely many solutions exist to the still more general equation
n! = P(n)
where P is any integer polynomial of degree at least 2.
?Dan
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