Show that the integer $Q_n = n! + 1$ where n is a positive integer, has a prime
divisor greater than n.
if n! + 1 is prime we are done, if n! + 1 is not prime then it is composite this argument doesnt seem lead anywhere which points to induction.
base case n=1 1+1=2 2>1 true for base case assume true for kth case WTS true for k+1.
$(k+1)k! +1 = kk! +k! +1 =a $
$kk! +k! +1 =a $ now by IH $k! +1$ has a divsior greater than or equal to k+1 now i want to assume that k+1 is relatively prime to kk! to show that the divisor must be at least k+2 which is > k+1 but i dont think i can?
Edit: each number $Q_n$ is the product of the first n integers +1 somewhere it says something to the effect that $Q_n$ is realtivly prime to every one of the integers in that list.
Best Answer
Hint: Can it have any prime divisors $\leq n$?