Recently **Faride Firoozbakht** wrote to me the following results of one
of her works:

"A multi-digit prime of the form m^m + m! - 1 or m^m - m! - 1 if it
exists (highly unlikely though) has more than 27000000 digits!!...So, I
think that

**5 is the only prime of the form m^m - m!-1
or m^m + m!-1 (A)**

But I don't know how we can prove this.

In fact by using Euler's Theorem I proved that:

**If m>2 & m^m - m!-1 or m^m + m!-1 is prime
then 1806 divides m (B).**

By using (B) and some computational results we can show that next prime
of the form m^m - m!-1 or m^m + m!-1 has more than 27000000 digits!!...

**Questions:**

**1. Can you demonstrate (A)? If not, can you find
a probable-prime of the forms studied by Faride?**

**2. Do you devise a way to prove (B)?**

The solution for question 1. came unexpectedly easy. But it resulted so
easy that it was found just by three puzzlers: **Mike Oakes, Phil Carmody
**and** Rudolf Knjzek**.

(m-1) | m^m-1, (m-1) | m!, so (m-1) is always a factor of **
m^m - m!-1 or m^m + m!-1. **So
**m^m - m!-1 or m^m + m!-1 is
**never prime for m > 2.

Question 2 remains as unproved by the Puzzlers.

I think it is easy to prove part 2 for the puzzle 294.
Write 1806 = 2*3*7*43.

If m-odd, then m^{m}-1 is even and m^{m}±m!-1
also even and greater than 2 for m>2.

Then m=2k, 2|m.

Suppose 3 does not divide m, then m=6k+2 or 6k+4. Use that
2^{2}
º
4^{2} º
1 (mod 3) (by Fermat) to get in this case m^{m}-1
º
0 (mod 3), so: m^{m}±m!-1
also divisible by 3 and greater than 3 for any m>2. We conclude m=6k, 6|m.

Suppose 7 does not divide m, then m=42k+6, 12, 18, 24, 30,
36. (42k+6i, i = 1 to 6). Use again that 6^{6}
º
12^{6} º
18^{6} º
24^{6} º
30^{6} º
36^{6} º
1 (mod 7) to get in all this cases m^{m}-1
º
0 (mod 7), so m^{m}±m!-1
also divisible by 7 and greater than 7 for any m>6. In order to complete
check for m=3,4,5,6 that m^{m}±m!-1
are both composite. We conclude m=42k, 42|m.

Suppose 43 does not divide m, then m=1806k+42i, where i = 1
to 42. Use that (42i)^{42}
º
1 (mod 43) to get also m^{m}-1
º
0 (mod 43), so m^{m}±m!-1
also divisible by 43 and greater than 43 for any m>42. . Again to complete
check using a computer for m=3,4,…,42 that m^{m}±m!-1
are both composite.

The proof is done.

The technique I used here is: if p-1 already divides m and
p is prime then p also divides m in order to get one of m^{m}±m!-1
as prime number. 2|m, then 3=2+1, 3|m, then 7=2*3+1, 7|m, then 43=2*3*7+1,
43|m.

Since for any other divisor d of 1806 different 2, 6, 42,
we have d+1 as composite, we cannot go further with this kind of proof.