Results 1 to 3 of 3

- September 22nd 2009, 10:32 PM #1

- Joined
- Sep 2009
- Posts
- 1

- September 23rd 2009, 02:02 AM #2

- Joined
- Sep 2009
- Posts
- 13

- September 24th 2009, 01:38 AM #3
1. Please state the problem in the text of the thread, not just in the title.

2. Please state the problem correctly.

3. I'm guessing that**(n!(j)+1,n!(j)+1)=1 for 1<=i<j<=n**means gcd(i×n! + 1, j×n! + 1) = 1.

4. If d divides i×n! + 1 and j×n! + 1, then d also divides their difference, which is (j–i)×n!. But all the prime factors of the product (j–i)×n! must lie between 1 and n. It follows that any prime factor of d must lie between 1 and n. So d cannot divide i×n! + 1. Therefore d has no prime factors, so d must be 1.