Show that $gcd(n!-1, (n+k)!-1)=1$

Question

How can I proceed? I've tried Bezout's identity but didn't work.

Answer 1

$$5! - 1 = 7 \cdot 17$$ $$11! - 1 = 13 \cdot 17 \cdot 23 \cdot 7853 $$

with common factor $17$

$$ \gcd( 39916799, 119 ) = 17 $$
$$ 39916799 \cdot 3 - 119 \cdot 1006306 = -17 $$

If you turn it around, Wilson's Theorem is relevant; $(p-2)! - 1$ is always divisible by $p.$ This $p-2$ is the largest possible, it is possible to have several smaller $n$ such that $n! \equiv 1 \pmod p$

Here are the first occurrences of each "count" of $n$ such that $n! \equiv 1 \pmod p.$ For prime $3011$ there are $9$ such $n.$

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3 : count 0
5 : 3! - 1; count 1
29 : 10! - 1; 27! - 1; count 2
17 : 5! - 1; 11! - 1; 15! - 1; count 3
23 : 4! - 1; 8! - 1; 11! - 1; 21! - 1; count 4
199 : 81! - 1; 89! - 1; 109! - 1; 117! - 1;
      197! - 1; count 5
619 : 111! - 1; 189! - 1; 294! - 1; 429! - 1; 
       507! - 1; 617! - 1; count 6
3313 : 107! - 1; 1662! - 1; 1886! - 1; 2084! - 1; 
      2970! - 1; 3205! - 1; 3311! - 1; count 7
4093 : 557! - 1; 575! - 1; 871! - 1;
       2312! - 1;  3221! - 1; 3517! - 1; 3535! - 1;
      4091! - 1; count 8
3011 : 611! - 1; 723! - 1; 749! - 1; 
       805! - 1; 2205! - 1; 2261! - 1; 
       2287! - 1; 2399! - 1; 3009! - 1; count 9
52163 :    3924! - 1 ;  7291! - 1 ;  7427! - 1 ;
          18519! - 1 ;  24931! - 1 ;  26081! - 1 ;
          27231! - 1 ;   33643! - 1 ;  44735! - 1 ;
          44871! - 1 ;  52161! - 1 ;  count  11    

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